# 11: Thermochemistry

These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

## Q11.1

Calculate the amount of heat (in Joules) that is:

1. achieved when raising the temperature of 5 g. of water from 22°C to 50°C?
2. required to raise the temperature of 10 g. of water from 10°C to 100°C?

## S11.1

1. $q=m c_{p} \Delta T$

$q= (5 \cancel{g}) \times \frac{4.184J}{1~ \cancel{gºC}} \times (28 \cancel{ºC})$

$q= 586J$

1. $q=m c_{p} \Delta T$

$q= (10 \cancel{g}) \times \frac{4.184J}{1~ \cancel{gºC}} \times (90 \cancel{ºC})$

$q= 3.77 \times 10^3 J$

## Q11.1

Calculate the quantity of heat, in kilojoules, that is:

1. required to raise the temperature of 12.5 L of water from 24.9 ºC to 35.8 ºC
2. associated with a 53.3 ºC decrease in temperature in a 9.66 kg aluminum bar (specific heat of aluminum = 0.903 J/gºC)

## S11.1

1. $q~=~m c_{p} \Delta T$

$m~=~(12.5~\cancel{L}) \times (\dfrac{1,000~\cancel{mL}}{\cancel{L}}) \times (\dfrac{1~\cancel{cm^3}}{1~\cancel{mL}}) \times (\dfrac{1~g}{1~\cancel{cm^3}})~=~12500~g$

$q~=~(12,500~ \cancel{g}) \times \frac{4.184~J}{1~ \cancel{gºC}} \times (10.9~\cancel{ºC})$

$q~=~570070J~=~570.1~kJ$

1. $q=m c_{p} \Delta T$

$q~=~(9.66~\cancel{kg.}) \times (\dfrac{1000~\cancel{g.}}{1~\cancel{kg.}}) \times (\dfrac{0.903J}{1~\cancel{gºC}}) \times (-53.3~\cancel{ºC})$

$q= -464934.834J = -464.9kJ$

## Q11.1

Calculate the quantity of heat, in kilojoules, that is:

1. associated with 4.9 L of water decreasing in temperature from 42.0°C to 33.0°C.
2. required to raise the temperature of a 3.82 kg lead bar by 51.2°C (specific heat of lead = 0.128 J / g °C).

## S11.1

a) $q = m c_{p} \Delta T$

$m = (4.9~\cancel{L}) (\dfrac{1,000~\cancel{mL}}{1~\cancel{L}}) (\dfrac{1~\cancel{cm^3}}{1~\cancel{mL}}) (\dfrac{1~g}{1~\cancel{cm^3}}) = 4900 ~\text{g}$

$q = (4900~\cancel{g}) \frac{4.184~J}{1 \cancel{g°C}} (33.0 – 42.0) \cancel{°C}$

$q = -184514~\text{J} = -185~\text{kJ}$

b) $q = m c_{p} \Delta T$

$q = (3.82~\cancel{kg}) (\dfrac{1000~\cancel{g}}{1~\cancel{kg}}) (\dfrac{0.128~\text{J}}{1~\cancel{g°C}}) (51.2~\cancel{°C})$

$q = 25034.752~\text{J} = 25.03~\text{kJ}$

## Q11.1

Calculate the final temperature that results from a:

1. 2.48 g sample of nickel at an initial temperature of 16.7°C that absorbs 59 J of heat (Note: The specific heat capacity of Ni is 0.440 J/g°C)
2. 72.8 g of water at an initial temperature of 72.5°C that releases 270 J of heat (Note: The specific heat capacity water = 4.184 J/g°C)

## S11.1

a)

$q=m c_{p} \Delta T$

$q=m c_{p} (T_f - T_i)$

$59J = (2.48g.) \times \frac{0.44~J}{1~g{^{\circ}C}} \times (T_f - 16.7{^{\circ}C})$

$T_{f} = \dfrac{59~\cancel{J}}{(2.48~\cancel{g}) \times \frac{0.44~\cancel{J}}{1~\cancel{g}{^{\circ}C}}} + 16.7^{\circ}C$

$T_f = 70.8{^{\circ}C}$

b)

$q=m c_{p} \Delta T$

$q=m c_{p} (T_f - T_i)$

$-270J = (72.8g.)\, (4.184 \text J/g{^{\circ}C})\, (T_f - 72.5{^{\circ}C})$

$T_{f} = \dfrac{-270~\cancel{J}}{(72.8~\cancel{g}) \times \frac{4.184~\cancel{J}}{1~\cancel{g}{^{\circ}C}}} + 72.5^{\circ}C$

$T_f = 71.61{^{\circ}C}$

## Q11.1

Calculate the quantity of heat, in joules, required in order to;

1. lower the temperature of 50.0 g water from 25 °C to 15 °C?
2. raise the temperature of a 4 kg iron bar from 25 °C to 100 °C, given that the specific heat of iron= 0.449 J/g℃

## S11.1

1. $q=m c_{p} \Delta T$

$q= (50g.) (4.184J/gºC) (-10ºC)$

$q= -2.092 \times 10^3 J$

1. $q=m c_{p} \Delta T$

$q= (4~\cancel{kg.}) (\dfrac{1000g}{1~\cancel{kg}}) (0.449J/gºC) (75ºC)$

$q= 1.347 \times 10^5 J$

## Q11.1

Find the amount of heat required to:

1. raise the temperature of 10 L of water from 23 ˚C to 30 ˚C;
2. associated with a 33 ˚C decrease in temperature in a 6.80 kg aluminum bar (specific heat of aluminum = 0.903 J/(g ˚C)).

## S11.1

1. $q = m c_{p} \Delta T$

$q = (10,000g) (\dfrac{4.184~\cancel{J}}{1~gºC}) (7ºC) (\dfrac{1kJ}{1000~\cancel{J}})$

$q = 292.9kJ$

1. $q = m c_{p} \Delta T$

$q = (6.80~\cancel{kg}) (\dfrac{1000g}{1~\cancel{kg}}) (0.903~\cancel{J}/gºC) (-33ºC) (\dfrac{1kJ}{1000~\cancel{J}})$

$q = -202.6kJ$

Video Solution for part b

## Q11.1

Calculate the quantity of heat, in kilojoules, that is:

1. needed to raise the temperature of 6.00L of water from 20.0 to 28.5°C;
2. related to a decrease in temperature from 45.0 °C to 11.0 °C in a 4.47 kg copper bar (specific heat of copper = 0.385 J/g°C).

## S11.1

1. $q = m c_{p} \Delta T$

$q = (6~\cancel{L}) (\dfrac{1000~\cancel{mL}}{1~\cancel{L}}) (\dfrac{1~g}{1~\cancel{mL}}) (4.184 J/gºC) (8.5ºC)$

$q = 213384J = 213.4 kJ$

1. $q = m c_{p} \Delta T$

$q = (4.47~\cancel{kg}) (\dfrac{1000~g}{1~\cancel{kg}}) (0.385J/gºC) (-34ºC)$

$q = -58512.3J = -58.5 kJ$

## Q11.2

What is the final temperature resulting from a:

1. 15.0 g sample of water at 8 ºC absorbing 345 J of heat;
2. 2.67 kg piece of Iron at 80 ºC giving off 1.15 kcal of heat (sp heat of Fe = 0.449 J/gC)

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f}–T{i})$

$345J = (15g) (4.184J/gºC) (T_{f}–8ºC)$

$T_{f} = \dfrac {345J}{(15g) (4.184J/gºC)} + 8ºC$

$T_{f} = 13.5ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f}-T{i})$
$(1.15~\cancel{kcal})(\dfrac{4184~J}{1~\cancel{kcal}}) = (2.67~\cancel{kg}) (\dfrac{1000~g}{1~\cancel{kg}}) (0.903J/gºC) (T_{f}–80ºC)$

$T_{f} = \dfrac {-4811.6J}{(2670g) (0.903J/gºC)} + 80ºC$

$T_{f} = -75.99ºC$

## Q11.2

Calculate the resulting temperature when:

1. a 203.2 g sample of aluminum at 43.5 °C gives off 432 cal of heat (specific heat of aluminum = 0.903 cal /g °C);
2. a 320 kg sample of gold at 93.4 °C absorbs 324 kJ of heat (specific heat of gold = 0.126 J / g °C).

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$-432cal = (203.2g) (0.903cal/gºC) (T_{f} – 43.5ºC)$

$T_{f} = \dfrac {-432cal}{(203.2g) (0.903cal/gºC)} + 43.5ºC$

$T_{f} = -41.15ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$(324~\cancel{kJ})(\dfrac{1000J}{1~\cancel{kJ}}) = (320~\cancel{kg}) (\dfrac{1000g}{1~\cancel{kg}}) (0.126J/gºC) (T_{f} – 93.4ºC)$

$T_{f} = \dfrac {324,000J}{(320,000g) (0.126J/gºC)} + 93.4ºC$

$T_{f} = 101.44ºC$

## Q11.2

What is the final temperature when:

1. 12 g of water at 23 ˚C absorbs 900J of heat;
2. 1.60 kg of platinum at 79.1 ˚C gives off 1.10 kcal of heat ($$c_{sp}$$ of Pt is 0.032 cal/(g°C))

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac {900J}{(12g) (4.184J/gºC)} + (23ºC)$

$T_{f} = 40.92ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac{(1.1~\cancel{kcal}) (\dfrac{1000cal}{1~\cancel{kcal}})}{(1.6~\cancel{kg}) (\dfrac{1000g}{1~\cancel{kg}}) (0.032cal/gºC)} + (79.1ºC)$

$T_{f} = 57.62ºC$

## Q11.2

Calculate initial temperature when

1. 26 g of water absorbed 1000 J at 56°C
2. 26 kg of water absorbed 50 kJ at 273.15 K?

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$-T_{i} = \dfrac{q}{(m)(c_{p})} - T_{f}$

$-T_{i} = \dfrac {1000J}{(26g) (4.184J/gºC)} - 56ºC$

$-T_{i} = -46.81ºC$

$T_{i} = 46.81ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$-T_{i} = \dfrac{q}{(m)(c_{p})} - T_{f}$

$-T_{i} = \dfrac {(50~\cancel{kJ})(\dfrac{1000J}{1~\cancel{kJ}})}{(26~\cancel{kg})(\dfrac{1000g}{1~\cancel{kg}})(4.184J/gºC)} - 0ºC$

$-T_{i} = 0.4597ºC$

$T_{i} = -0.4597ºC$

## Q11.2

Calculate the final temperature that results when

1. a 13.8 g sample of water at 21.7°C absorbs 679 J of heat;
2. a 2.43 kg sample of aluminum at 69.5 °C gives off 2.04 kcal of heat ($$c_{sp}$$ of Al = 0.902 \dfrac{J}{g°C}).

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac {679J}{(13.8g) (4.184J/gºC)} + 21.7ºC$

$T_{f} = 33.46ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac{(2.04~\cancel{kcal})(\dfrac{4184J}{1~\cancel{kcal}})}{(2.43~\cancel{kg})(\dfrac{1000g}{1~\cancel{kg}})(0.902J/gºC)} + 69.5ºC$

$T_{f} = 73.39ºC$

## Q2

Calculate the final temperature that is reached;

1. by a 3 kg copper wire at 30℃ absorbs 4.5 \times 10^4 J
2. When 6 kg of Al(s) at 25℃ emits 6.3 \times 10^2 kJ

## S11.2

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac {45000J}{(3~\cancel{kg}) (\dfrac{1000g}{1~\cancel{kg}}) (0.385J/gºC)} + 30ºC$

$T_{f} = 68.96ºC$

1. $q=m c_{p} \Delta T$

$q = m c_{p} (T_{f} – T{i})$

$T_{f} = \dfrac{q}{(m)(c_{p})} + T_{i}$

$T_{f} = \dfrac {(-630~\cancel{kJ})(\dfrac{1000J}{1~\cancel{kJ}})}{(6~\cancel{kg})(\dfrac{1000g}{1~\cancel{kg}})(0.903J/gºC)} + 25ºC$

$T_{f} = -91.28ºC$

## Q11.2

Calculate the quantity of heat, in kJ required to:

1. Raise the temperature of a 32.2 g bar of pure gold from 22.0°C to 100°C (Note: Csp of gold = 0.129 J/g \times C)
2. Decrease the temperature of 45.5 L of water 35.0°C

## S11.2

a)

$q=m\, c_p \, \Delta T$

$q = (32.2\; g) (0.129 \dfrac{J}{g°C}) (100 °C - 22.0 °C)$

$q = 324\; J = 0.324\; kJ$

b)

$45.5~\cancel{L} \times (\dfrac{1000~\cancel{cm^3}}{1~\cancel{L}}) \times \dfrac{1~g}{1~\cancel{cm^3}} =45,500 g$

$q = (45,500 \;g) (4.184 \dfrac{J}{g°C})(-35.0 °C)$

$q = -6.66 \times 10^6 J=-6.66 \times 10^3 kJ$

## Q11.3

1. 150 g K and 15.9°C as the final temperature
2. 27 g Na and 12°C as the final temperature
3. 2 kg Mg and 273.15 K as the final temperature

## S11.3

1. $q = m \times \dfrac{q}{m \Delta T}$

$q_{K} = 150g_{K} \times \dfrac{-1.4 \times 10^{3}J}{(150g_{K})(15.9^{\circ}C - 100.0^{\circ}C)}$

$q_{K} = 16.64J/{g^{\circ}C}$

1. $q = m \times \dfrac{q}{m \Delta T}$

$q_{Na} = 27g_{Na} \times \dfrac{-1.4 \times 10^{3}J}{(27g_{Na})(12^{\circ}C - 100.0^{\circ}C)}$

$q_{Na} = 15.91J/{g^{\circ}C}$

1. $q = m \times \dfrac{q}{m \Delta T}$

$m = 2kg = 2000g \hspace{30pt} T_{f} = 273.15K = 0^{\circ}C$

$q_{Mg} = 2000g_{Mg} \times \dfrac{-1.4 \times 10^{3}J}{(2000g_{Mg})(0^{\circ}C - 100.0^{\circ}C)}$

$q_{Mg} = 14J/{g^{\circ}C}$

## Q11.4

When 13.6 g of sucrose is burned in a bomb calorimeter, the temperature increases by 5.6°C. The heat capacity of the calorimeter is calculated at 3.40 kJ/°C. Calculate the heat of combustion of the sugar per mole.

## S11.4

$q_{cal} = (3.40kJ/ºC) \times (5.6ºC) = 19.04kJ$

$q_{cal} = -q_{sys} = -19.04kJ$

Convert gas to moles:

$13.6~\cancel{g C_{12}H_{22}O_{11}} \times (\dfrac{1~mol C_{12}H_{22}O_{11}}{342.3~\cancel{g C_{12}H_{22}O_{11}}}) = 0.0397 mol$

Heat of Combustion = $(19.04kJ/0.0397 mols C_{12}H_{22}O_{11}) = -479.6kJ/mol$

## Q11.3

A 300 g sample of lead is heated in boiling water (100°C). The lead is then dropped inside a calorimeter with 100 g of water with an initial temperature of 22.0°C The final temp was observed to be 28.8°C. What is the specific heat of lead?

## S11.3

$m_{water} \times c_{p_{water}} \times \Delta T_{water} ~=~ q_{water}$

$100g_{water} \times 4.184J/g°C \times (28.8°C - 22°C) = 2842.4J$

$q_{lead}~=~-q_{water}~=~-2842.4~J$

$m_{lead} \times c_{p_{lead}} \times \Delta T_{lead} ~=~ q_{lead}$

$300g_{lead} \times c_{p_{lead}} \times (28.8°C - 100°C) = -2842.4J$

$\dfrac{-2842.4J}{300g_{lead} \times -71.2°C} = 133J/g°C$

## Q11.3

A 96.5 g chunk of an unknown metal heated to 98.0°C is dropped into a coffee cup calorimeter holding 45.0 g of pure water. Before the addition of the metal, the temperature of the water is measured at 25.0 °C. After the addition of the metal, the temperature of the water raises to 33.3°C. What is the specific heat of the given metal?

## S11.3

Solve for the q of water:

$q = (45.0 \;g) \left (4.184 \dfrac{J}{g˚C}\right) (33.3 ˚C - 25.0 ˚C) = 1,563\; J$

$q_{water} = -q_{metal}= -1,563\; J$

$c_{sp} = \dfrac{q_{metal}}{m \times \Delta T} = \dfrac{-1,563\; J}{95.5\; g \times (33.3˚C - 98.0 ˚C)} = 0.250 \;\dfrac{J}{g ˚C}$

## Q11.3

A 140 g sample of Zn, Pt, and Al are heated individually in water at its boiling point. 40 g of water is then added to a closed system where the water is 23°C. What is the specific heat, in J/(g°C), of:

1. Pt at 30˚C,
2. Al at 52.7˚C, and
3. Al at Zn at 40˚C.

## S11.3

(a)

$q=m c_{p} \Delta T$

$q = (40g)( 4.184J/gºC)( 7ºC)$

$q = 1171.52J$

$q_{water} = -q_{metal} = -1171.52J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-1171.52 J}{140g. \times -70ºC}$

$c_{p} = 0.1195 J/gºC$

(b)

$q=m c_{p} \Delta T$

$q = (40g)( 4.184J/gºC)( 29.7ºC)$

$q = 4970.592 J$

$q_{water} = -q_{metal} = -4970.592 J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-4970.592 J}{140g. \times -47.3ºC}$

$c_{p} = 0.7506 J/gºC$

(c)

$q=m c_{p} \Delta T$

$q = (40g)( 4.184J/gºC)( 17ºC)$

$q = 2845.12 J$

$q_{water} = -q_{metal} = -2845.12 J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-2845.12 J}{140g. \times -60ºC}$

$c_{p} = 0.3387 J/gºC$

## Q11.3

Calculate the specific heat of:

1. a 119.5 g sample of zinc that give off 1.02 kJ of heat when it cools from 65.7 °C to 43.5 °C;
2. a 0.893 kg sample of mercury increases in temperature by 21.2 °C when it absorbs 0.69 kcal of heat.

## S11.3

a)

$c_p=\dfrac{q}{m∆T}$

$c_p(zinc) = \dfrac{(-1.02\; \cancel{kJ}) \left(\dfrac{1000\; J}{1\; \cancel{kJ}}\right)}{(119.5\; g)(43.5 °C \times –65.7 °C)} = 0.384\; \dfrac{J}{g °C}$

b)

$c_p = \dfrac{q}{m∆T}$

$c_p(mercury) = \dfrac{(0.69\; \cancel{kcal}) \left(\dfrac{1000\; cal}{1\; \cancel{kcal}} \right)}{(0.893 \;\cancel{kg}) \left( \dfrac{1000\; g}{1\; \cancel{kg}} \right)(21.2 °C)} = 0.0364 \;\dfrac{cal}{g °C}$

## Q11.3

A 110.0g sample of titanium is heated to the temperature of 98°C. A 41.0g sample of water is added to a calorimeter, and its temperature is found to be 23.0°C. The hot titanium is dumped into the cold water, and the temperature of the final titanium-water mixture is 29.4°C. Using the same set up, find the specific heat of each metal expressed in J g-1 °C-1, of the metals at the following final temperatures

a) Ti, 29.4°C;

b) Ni, 38.6°C;

c) Cu, 49.2°C.

## S11.3

(a)

$q=m c_{p} \Delta T$

$q = (41g)( 4.184J/gºC)( 6.4ºC)$

$q = 1097.8816 J$

$q_{water} = -q_{metal} = -1097.8816 J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-1097.8816 J}{110g. \times -68.6ºC}$

$c_{p} = 0.1455 J/gºC$

(b)

$q=m c_{p} \Delta T$

$q = (41g)( 4.184J/gºC)( 15.6ºC)$

$q = 2676.0864 J$

$q_{water} = -q_{metal} = -2676.0864 J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-2676.0864 J}{110g. \times -59.4ºC}$

$c_{p} = 0.4096 J/gºC$

(c)

$q=m c_{p} \Delta T$

$q = (41g)( 4.184J/gºC)( 26.2ºC)$

$q = 4494.4528 J$

$q_{water} = -q_{metal} = -4494.4528 J$

$c_{p} = \dfrac{q}{m \Delta T}$

$c_{p} = \dfrac{-4494.4528 J}{110g. \times -48.8ºC}$

$c_{p} = 0.8373 J/gºC$

## Q11.5

Calculate the work, in joules, associated with an ideal gas expansion from 3.04 L to 5.60 L against an external pressure of 1.4 atm.

## S11.5

$w = -P_{ext} \Delta V$

$w = -(1.4\; atm) (5.60\; L - 3.04\; L) = 3.58\; L \cdot atm$

However, this not the correct unit requested for energy so we must convert

$w = (3.58\; \cancel{ L \cdot atm} )\left( \dfrac {101\; J}{1 \cancel{L \cdot atm}} \right)= -362\; J$

## Q11.6

In compressing a gas, 489 J of work is done on the system, while 239 J escape the system as heat. Calculate $$\Delta U$$ for the system for this compression.

## S11.6

$\Delta U=q+w$

$∆U = (-239 \;J)+(489\; J)=250\; J$

## Q11.7

Which of the following is NOT true?

1. When heat is absorbed by a system, $$q<0$$
2. If work is done on a system, $$w<0$$
3. When heat is released by a system, $$q<0$$
4. $$∆U$$ of an isolated system is $$0$$

## S11.7

When q is negative, heat has flowed from the system. When q is positive, heat has been absorbed by the system.

Therefore, the correct answer is A.

## Q11.22

Cold water and a piece of hot metal come into contact in an isolated container. When the final temperature of the metal and water are identical, the total energy change in this process is:

1. zero
2. negative
3. positive
4. not enough information

## S11.22

The law of conservation of energy informs you that if the cold water gained thermal energy through the transfer of heat (as indicated by its rise in temperature), then the hot metal must have lost an equal amount of thermal energy. This can be displayed by:

Heat change of metal + Heat change of water = 0

(mcpΔT)metal + ( mcpΔT)water = 0

## Q11.29

How many moles of $$H_2O_{(l)}$$ is left with the same quantity of heat released from 7 g $$H_2O_{(s)}$$ at 279.15 K? (Heat of Fusion = 6.01 kJ/mol)

0.029 mole

## Q11.29

Ethanol, C2H5OH boils at 78.29 ºC. How much energy, in Joules, is required to raise the temperature of 1.00 kg of ethanol from 20.0 ºC to the boiling point of 78.29 ºC. (Heat capacity of liquid ethanol is 2.44 J/g K)

## S11.29

$m = 1kg = 1000g; \Delta T = 58.29ºC = 58.29K$

$q=m c_{p} \Delta T$

$q = (1000g)( 2.44J/gK)(58.29K)$

$q = 142,227.6J = 142,000J$

## Q11.29

What mass of ice can be melted with the same quantity of heat as required to raise the temperature of 4.5 mol $$H_{2}O_{(l)}$$ by 35.0 °C? $∆H°_{fusion} =6.01 kJ/mol H_{2}O_{(s)}$

## S11.29

First identify how much heat is available

$q= mc_{sp}\Delta T$

$q_{water} = 4.5\; mol × \left(\dfrac{18.015\; g H_2 O}{1\; mol\; H_2O} \right) \left(4.18 \dfrac{J}{g°C} \right) ( 35.0°C) = 11.9 \times 10^3 \;J$

and equate to the heat of fusion equation

$q_{heat}= q_{fus} = m \Delta H_{fus}$

$q_{fus} = m \left(\dfrac{1\; mol\; H_2O}{18.015\; g H_2 O} \right) (6.01 \times 10^3\; J/mol)$

$11.9 \times 10^3 \;J =m (333.6\; J/g)$

$m= \dfrac{11.9 \times 10^3\; J}{333.6\; J/g} = 35.6\; g\; H_2O$

## Q11.33

If 300.0 J of heat is transferred to a block of dry ice at -80.5°C, what volume of CO2 gas ($$\rho$$ = 1.88 g/L) will be generated? The specific enthalpy of sublimation for dry ice (CO2) is 581 kJ/kg at -80.5°C.

## S11.33

$581\; kJ/kg \times \dfrac{1000\;J}{1\;kJ} = 5.81 \times 10^5 \;J/kg$

$300\; J \times \dfrac{1\; kg}{5.81 \times 10^5\; J} \times \dfrac{1000\;g}{1\;kg} \times \dfrac{1\;L}{1.88\;g} = 0.274\; L$

## Q11.35

A bomb calorimeter with a sample compound has a specific heat capacity of 2.37 kJ/˚C. The temperature of the system increases by 12.80˚C during the experiment. Calculate the heat of combustion of the sample in calories.

## S11.35

$\Delta H°_{comb} = c \Delta T$

$\Delta H°_{comb} = (2.37 kJ/°C)(12.8°C)$

$\Delta H°_{comb} = 30.336kJ = 7250.478 cal$

## Q11.35

A sample releases 4,678 cal. of heat when combusted in a bomb calorimeter which increased its temperature by 10.1°C. Calculate the heat capacity of the calorimeter in kilojoules per degree Celsius.

## S11.35

$C_v = \dfrac{q}{\Delta T}$

$C_v = \dfrac{4,678\; \cancel{cal} \left( \dfrac{ 4.184\; \cancel{J}}{1\; \cancel{cal}} \right) \left( \dfrac{1 \;kJ}{1000\; \cancel{J}} \right) }{10.1 °C} = 1.94 \; kJ/°C$

## Q11.35

A bomb calorimeter determined that a combustion reaction of a random substance gave off 10.2kcal and raised the temperature by 373.15K. What is the specific heat capacity of this 10.0g sample? (Hint: 1 cal = 4.18J)

## S11.35

$\Delta T = 373.15K = 100ºC; 10.2kcal = 10,200cal = 42636J$

$q=m c_{p} \Delta T$

$c_{p} = \dfrac{q}{ m \Delta T }$

$c_{p} = \dfrac{42636J}{ (10g)(100ºC}$

$c_{p} = 42.6 J/gºC$

## Q11.39

A student has 250mL of 1M NaOH and 250mL of 1M HCl at atmospheric pressure. After mixing the two compounds the temperature rose from 23ºC to 31ºC. Find the change in energy of the system and is this an exothermic process or endothermic?

## S11.39

$\dfrac{moles}{L} = M$

$\dfrac{moles}{0.25L NaOH} = 1M NaOH$

$= 0.25 mol NaOH$

$0.25 mol NaOH \times \dfrac{40g NaOH}{1mol NaOH} = 10g. NaOH$

$\dfrac{moles}{L} = M$

$\dfrac{moles}{0.25L_{HCl}} = 1M_{HCl}$

$= 0.25 mol_{HCl}$

$0.25 mol HCl \times \dfrac{36.5g HCl}{1mol HCl} = 9.13g. HCl$

$q=m c_{p} \Delta T$

$q = (10g + 9.13g)(4.184J/gºC)( 8ºC)$

$q = 640.3J$

$\dfrac{640.3J}{0.25mol} = 2561.2 J/mol$

Because of the positive q value, we can state that this reaction was endothermic.

## Q11.41

A 1.432 g sample of glucose, C6H12O6, is completely burned in a bomb calorimeter, and an increase of 6.7°C is noted. If the heat of combustion of glucose is −2805 kJ/mol C6H12O6, what is the heat capacity of the bomb calorimeter?

## S11.41

$1.432g. C_{6}H_{12}O_{6}= 0.00795mol C_{6}H_{12}O_{6}$

$q = \Delta Hº_{comb} \times m$

$q = -2805kJ/mol C_{6}H_{12}O_{6} \times 0.00795mol C_{6}H_{12}O_{6}$

$q = 22.2998kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{22.2998kJ}{6.7ºC}$

$C_{cal} = -3.3 kJ/ºC$

## Q11.41

A 1.5 g sample of propane $$C_3H_8$$ is burned in a calorimeter. The temperature increases by 10 ℃ If the heat of combustion of propane is 49.9 kJ/g, what is the heat capacity of the calorimeter?

## S11.41

$q = \Delta Hº_{comb} \times m$

$q = 49.9kJ/g C_{3}H_{8} \times 1.5g C_{3}H_{8}$

$q = 74.85kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{74.85kJ}{10ºC}$

$C_{cal} = 7.485kJ/ºC$

## Q11.41

A 1.5 g sample of C6h6 (s) completely burns in a bomb calorimeter assembly and a temperature increase of 13°C is noted. If the heat of combustion of benzene is –41.7KJ/gl what is the heat capacity of the bomb calorimeter?

## S11.41

$q = \Delta Hº_{comb} \times m$

$q = -41.7kJ/g C_{6}H_{6} \times 1.5g C_{6}H_{6}$

$q = -62.55kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{-62.55kJ}{10ºC}$

$C_{cal} = -7.485kJ/ºC$

## Q11.41

A 2.70 g sample of phenol, $$C_6H_5OH$$, is completely burned in a bomb calorimeter assembly and a temperature increase of 10 °C is noted. If the heat of combustion of phenol is – 5156 kJ/mol $$C_6H_5OH$$, what is the heat capacity of the bomb calorimeter?

## S11.41

$q = \Delta Hº_{comb} \times m$

$q = -5156kJ/mol C_{6}H_{5}OH \times 0.0287mol C_{6}H_{5}OH$

$q = -147.9772kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{-147.9772kJ}{10ºC}$

$C_{cal} = -14.8kJ/ºC$

## Q11.41

The heat of combustion of C10H8(s) is -5156 kJ/mol $$C_{10}H_8$$. What is the heat capacity of a bomb calorimeter when a 2.13 g sample of $$C_{10}H_{8(s)}$$ is burned fully in the bomb calorimeter and the temperature increases by 8.50 ˚C.

## S11.41

$q = \Delta Hº_{comb} \times m$

$q = -5156kJ/mol C_{10}H_{8} \times 0.0166mol C_{10}H_{8}$

$q = -85.5896kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{-85.5896kJ}{8.5ºC}$

$C_{cal} = -10.07kJ/ºC$

## Q11.41

A 2.53 g sample of phenanthrene (s) (C_14 H_10) is completely combusted in a bomb calorimeter, and you notice the temperature raises 4.8 ℃. If the heat of combustion of phenanthrene (s) is -7052.6 kJ/mol, what is the heat capacity of the bomb calorimeter?

## S11.41

$q = \Delta Hº_{comb} \times m$

$q = -7052kJ/mol C_{14}H_{10} \times 0.0142mol C_{14}H_{10}$

$q = -100.14692kJ$

$C_{cal} = \dfrac{q}{\Delta T}$

$C_{cal} = \dfrac{-100.14692kJ}{4.8ºC}$

$C_{cal} = -20.86kJ/ºC$

## Q11.47

Calculate the quantity of work associated with a 3.5 L expansion of a gas (ΔV) against a pressure 372 mmHg in the units (a) atm L; (b) joules (J); (c) calories (cal)

## S11.47

(a)

$w=-P_{ext}ΔV$

$w = -(372mmHg) \times ( \dfrac{1atm}{760mmHg}) \times (3.5L)$

$w = -1.7Latm$

(b)

$(-1.7Latm) \times ( \dfrac{101.325J}{1Latm}) = 1.7 \times 10^2 J$

(c)

$(-1.7 \times 10^2 J) \times ( \dfrac{1cal}{4.184J}) = -41cal$

## Q11.47

Calculate the work done by a gas that expands 4.5 L against a pressure of 770 mmHg in Joules.

## S11.47

$w = -P \Delta V$

$w = -(850mmHg) \times ( \dfrac{1atm}{760mmHg}) \times (4.5L)$

$w = -5.03Latm$

$(-5.03Latm) \times ( \dfrac{101.325kPa}{1atm}) \times ( \dfrac{1J}{1LkPa})$

$= -510J$

## Q11.47

Calculate the quantity of work associated with a 4000 mL expansion of a gas against pressure of 5 atm in the units: (a) atm L (b) mmHg (c) kilocalories (kcal).

## S11.47

(a)

$w= -P \Delta V$

$w = (-5atm) \times (4000mL) \times ( \dfrac{1L}{1000mL})$

$w = -20 L atm$

(b)

$(-20Latm) \times ( \dfrac{101.325J}{1Latm})$

$= -2026.5 J$

(c)

$(-2026.5J) \times ( \dfrac{.239005736cal}{1J}) \times ( \dfrac{1kcal}{1000cal})$

$= -.4843 kcal$

## Q11.47

What is the work when the system has a 10.0L gas expansion while going against a 600 torr pressure. What is the work in terms of

1. kJ
2. kcal
3. L mmHg

## S11.47

$w = -P \Delta V$

$w = (-600torr) \times (10L)$

$w = -6,000Ltorr$

(a)

$(-6,000Ltorr) \times ( \dfrac{1J}{7.5Ltorr}) = -800J$

$(-800J) \times ( \dfrac{1kJ}{1000J}) = -0.8kJ$

(b)

$(-800J) \times ( \dfrac{0.239005763cal}{1J}) \times ( \dfrac{1kcal}{1000cal}) = 0.191kcal)$

(c)

$(-6,000Ltorr) \times ( \dfrac{1mmHg}{1torr}) = -6,000LmmHg$

## Q11.47

Calculate the quantity of work when a gas expands 4.7 L against a pressure of 844 mmHg in:

1. units of atm L
2. units of joules (J)
3. units of calories (cal)

## S11.47

1. $-P∆V=4.7L × (844 mmHg)(1atm/760mmHg)=-5.22 L atm$
2. Remember, we don’t have a conversion from atm to Joules, but we have a conversion from atm to kPa and we know that 1 L kPa = 1 Joule: $-5.22 L atm × ((101.325 kPa)/1atm)× ((1 J)/(1 L kPa))= -5.3 × 10^2 J$
3. $-5.3 × 10^2 J × ((1 cal)/(4.184 J))=-126.7 cal$

## Q11.47

How much work is done when a gas expands 3 L against a pressure of 753 mmHg in:

1. $$L \cdot atm$$
2. joules
3. calories

## S11.47

$w= -P_{ext}\Delta V=(753\;\cancel{ mmHg}) \left( \dfrac{1\; atm}{760\; \cancel{mmHg}} \right) (3\; L) = -2.97 \; L \cdot atm$

$w= (-2.07\; \cancel{ L \cdot atm} ) \left( \dfrac{101.3 \;J }{1 \; \cancel{ L \cdot atm}} \right)= -301\; J$

$w= 301 \; \cancel{J} \times \left(\dfrac{1 \;cal}{4.184\;\cancel{J}} \right) = -72.0\; cal$

Video Solution

## Q11.49

A 2.0 g sample of Ar (g) at 0.5 atm pressure and 30 ºC is allowed to expand into an empty vessel of 4.0 L. Does the gas do work?

## S11.49

No, the gas does not do work because the number of moles do not change.

$w= -ΔnRT$

$w= (0)(8.3145)(30+273.15) = 0$

## Q11.53

What is the volume when 2.2kJ of work is done on the system at a pressure of 750 mmHg?

## S11.53

$w = P \Delta V$

$2.2kJ = 750mmHg \times \Delta V$

$\dfrac{2.2kJ}{(750mmHg)( \dfrac{1atm}{760mmHg})} = \Delta V$

$\Delta V = \dfrac{2.2kJ}{0.98684atm}$

$\Delta V = 2.229338kJ/atm \times \dfrac{9.86923267 L}{1kJ/atm} = 22.0L$

## Q11.55

Calculate the change in internal energy for each of the following systems?

1. 100 J of energy is added to a system that does 40 J of external work?
2. 75 J of energy is removed from a system that has 50 J of work done on it?
3. 150 J of energy is added to a system that has 75 J of work done on it?

## S11.55

(a.)

$q = 100J; w = -40J$

$\Delta U = q+w$

$\Delta U = 100J - 40J$

$\Delta U = 60J$

(b.)

$q = -75J; w = 50J$

$\Delta U = q+w$

$\Delta U = -75J + 50J$

$\Delta U = -25J$

(c.)

$q = 150J; w = 75J$

$\Delta U = q+w$

$\Delta U = 150J + 75J$

$\Delta U = 225J$

## Q11.55

What happens to the internal energy when

1. The system release 20 J of heat and does 20 J of work
2. Absorbs 1 kJ of heat and has 100 J of work done on it
3. Release 1 kcal of heat and does 1 kJ of work

## S11.55

(a)

$q = -20J; w = -20J$

$\Delta U = q + w$

$\Delta U = -20J + (-20J)$

$\Delta U = -40J$

(b)

$q = 1kJ \times \dfrac{1000J}{1kJ} = 1000J; w = 100J$

$\Delta U = q + w$

$\Delta U = 1000J + 100J$

$\Delta U = 110J$

(c)

$q = -1kcal \times \dfrac{4184J}{1kcal} = 4184J; w = -1000J$

$\Delta U = q + w$

$\Delta U = 4184J + (-1000J)$

$\Delta U = 3184J$

## Q11.57

An isothermal process is a change of a system, however the temperature stays constant. Ideal gasses are allowed to expand at this constant temperature. Considering the ideal gas, answer the following questions.

1. Does the ideal gas do work on its surroundings?
2. Is there a heat exchange between the system and the surroundings?
3. Does the temperature of the gas increase, decrease, or remain constant?
4. What is $$\Delta U$$ for the gas?

## S11.57

1. Yes
2. Yes
3. The temperature remains constant.
4. $$\Delta U$$ = 0

## Q11.57

In an isothermal expansion with the internal energy of gas only dependent on temperature:

1. Is work done by gas?
2. Are the surroundings heated through gas exchange?
3. Does the temperature increase, decrease, or stay the same?
4. What is the $$∆U$$ of the gas?

## S11.57

1. Yes, work is done by gas
2. Yes. By absorbing energy, the gas exchanges energy with the surroundings
3. Since the expansion is isothermal, the temperature of the gas stays the same.
4. The internal energy $$∆U = 0$$ is a state function dependent on temperature. Hence, due to the constant temperature of the isothermal expansion, the internal energy change is zero.

## Q11.57

There is a fixed quantity of an ideal gas that is allowed to expand at a 25℃ and its temperature is held constant, called isothermal expansion. Answer the following questions about the gas:

(a) What is the temperature of the gas before and during expansion?

(b) How does this change the energy of the ideal gas?

(c) How does the system (the gas) interact with its surroundings (work + heat)?

## S11.57

Remember, the internal energy of an ideal gas whose quantity is fixed, is only dependent on its temperature.

(a) This is a trick question; the temperature is constant and does not change, so it will always remain at 25℃.

(b) The energy of an ideal gas, or ∆U, is described by the equation ∆U= 3/2 nRT. R is constant, and the moles do not change as a gas expands. Since there is no change in temperature, there is no change in energy. ∆U=0

(c) Because there is no change in energy, as seen above, we know that the work and heat of the system are balancing each other out. Since this system is expanding, the system is doing work on to its surroundings, and the w value is negative. This means that gas is absorbing energy (heat) from its surroundings. As long as they are balanced, ∆U remains 0.

## Q11.57

For a fixed vessel at constant pressure, the volume of the vessel expands 3.5L

1. Is there any work done by the random object in the vessel?
2. Is there any heat exchange with the surroundings?
3. What happens to the internal energy?
4. What happens to the pressure?

## S11.57

1. No
2. No
3. Nothing
4. Remains Constant

## Q11.59

An ideal gas that expands isothermally does two times more work than the heat being absorbed from the surroundings. True or False? Explain why.

## S11.53

False. When an ideal gas expands, the amount of work and heat absorbed are the exact same with opposite signs. We can symbolize this by saying:

$q = -w$

where q is positive because heat is absorbed and, therefore, the amount of work done is negative, resulting in:

$\Delta U = 0$.

## Q11.59

There is an ideal gas expanding isothermally (at a constant temperature). The gas is absorbing 3 times as much heat from its surroundings as it is performing work. Explain if this is possible or not (see above question for help).

## S11.59

This situation is impossible, assuming the gas is ideal. If the gas is expanding isothermally, its ∆U or change in energy is always 0. This means that work and heat absorbed must always be equal.

## Q11.63

According to the First Law of Thermodynamics, explain why each of the following equations will not always provide a true value for the heat of chemical reactions, regardless of the reaction.

1. Heat energy for a constant volume
2. Heat energy for a constant pressure
3. Change in internal energy
4. Change in enthalpy

## S11.63

1. The heat energy depends on a constant volume.
2. The heat energy depends on a constant pressure
3. Internal energy is heat energy + work. If you subtracted work, this would be correct.
4. Change in enthalpy is the change in internal energy + pressure \times volume. Although this accounts for some other variables, it still involved work and is, therefore, incorrect.

## Q11.63

Which corresponds with the heat of a chemical reaction and why?

1. qp
2. ∆U-w
3. ∆H
4. qv
5. ∆U

## S1163

Both (a)qp and (d)qv represent heat carried out under constant pressure or constant volume. (c)∆H and (e)∆U both have the involvement of work and heat. So the answer is (b) ∆U-w

## Q11.63

Which of the follow is best represented as heat of a chemical equation?

1. $$q_{rxn}$$
2. $$\Delta U$$ + q + w
3. $$\Delta U$$ - w
4. $$\Delta U$$ – q
5. $$\Delta U$$ – q – w

## S11.63

$$\Delta U$$ = q + w is the original equation.

$$\Delta U$$ - w = q + w - w is the original equation removing work from both sides.

$$\Delta U$$ - w = q is the simplified equation.

Therefore, (c) is the correct answer.

## Q11.65

3-ethanol temperature is increased from 290 K to 305 K, determined in a bomb calorimeter, is -45.62 kJ/g. For the combustion of one mole of 3-ethanol determine (a) Cp, (b) ∆H.

## S11.65

(a)

$q_{react} = -q_{cal}$

$q=45.62 kJ/g$

$C_{p}= \dfrac{45.62 kJ/g}{(305K-290K)}$

$C_{p}=3.04 J/g°C$

(b)

$q = n \times ∆H$

$45.62 kJ/g=1mol ∆H$

$∆H= 45.62$

## Q11.65

At 25˚C the combustion of $$C_3H_8O$$ is -33.41 kJ/g. Find:

1. $$∆H$$
2. $$∆U$$

for the combustion of one mole of $$C_3H_8O$$.

## S11. 65

$C_{3}H_{8}O_{(l)} + \dfrac{9}{2} O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_{2}O_{(l)}$

$3mol_{product(g)} - \dfrac{9}{2}mol_{reactant(g)}$

$\Delta n_{gas}=-1.5 mol$

(a)

$∆U = -33.41 kJ/g \times \dfrac{60.096g}{mol}= -2008 kJ/mol$

$∆H = ∆U -w= ∆U - (-P∆V) = ∆U - (-∆n_{gas}RT) = ∆U + ∆n_{gas}RT$

$∆H = -2008 kJ/mol + (-1.5 mol) \dfrac{8.3145 \times 10-3 kJ}{molK})(298.15 K) = -2012 kJ/mol$

## Q11.65

The heat of combustion of 2-butonol at 298 K is -36.1 kJ/g (which was determined by a bomb calorimeter). Determine the internal energy $$∆U$$ and enthalpy $$∆H$$ for one mole of 2-butonol.

## S11.65

First, write out the balanced reaction:

$C_4 H_{10}O_{(l)}+ 6O_{2(g)} \rightarrow 4CO_2{(g)} + 5H_2O_{(l)}$

Identify the change in moles of gases for reaction:

$$∆n_{gas}= -2\; mol$$

$∆U= (-36.1\; kJ/g) \times (\dfrac{74.12\; g C_4H_{10}O}{1\; mol C_4H_{10}O}) = -2,676 \;kJ⁄mol$

$∆H= ∆U-w$

$∆H= ∆U-(-P∆V)= ∆U-(-∆n_{gas}RT)$

$∆H=∆U+ ∆n_{gas}RT$

$∆H=-2676 kJ⁄mol + (-2 mol) \times \dfrac{8.3145 \times 10^{-3} kJ}{molK} \times 298 K=-2,681\; kJ/mol$

## Q11.65

The heat of combustion of 2-propanol at 303.15 K (determined in a bomb calorimeter) is -45.50 kJ/g. Determine ∆U and ∆H for the combustion of one mole of 2-propanal.

## S11.65

First, convert grams to moles and multiply by heat of combustion:

1 mol 2-proponal = 0.0166 \times -45.50 = -0.757 kJ/g

$∆U = ∆H – P∆V$

$C_3H_8O + O_2 \rightarrow H_2O + CO_2$

(balance combustion equation)

$C_3H_8O_{(l)} + 9/2 O_{2(g)} \rightarrow 4 H_2O_{(l)} + 3 CO_{2(g)}$

$P∆V = RT∆(n_f – n_i)$

$P∆V = (0.0083145)(303.15 K)(9/2 – 3)$

(only use moles of gas in products and reactants)

$P∆V =3.78 kJ$

So ∆U = -45.50 kJ – 3.78 kJ

$∆U = -49.28\; kJ$

$$∆H$$ is equal to the initial energy, which is -0.757 kJ/g

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Standard_Enthalpy_Of_Formation

## Q11.67

∆H of the reaction below is 197.87 kJ. What is the standard enthalpy of formation of SO2(g). (Answer should be per mole SO2(g))

## S11.67

⅔SO2(g) ⇄ S(s) + ⅔O2(g) $$\Delta H^o_f =197.87 kJ$$

S(s) + ⅔O2(g) ⇄ ⅔SO2(g) $$\Delta H^o_f= -197.87 kJ$$

$$\Delta H^o_f$$ of ⅔SO2(g)= -197.87 kJ \times (1/( ⅔ mole SO2)) = -296.8kJ/mol of SO2

## Q11.67

The standard enthalpy of formation of $$CO_2$$ is -35.9 kJ/mol $$CO_2$$.

$2CO_{2\;(g)} \rightarrow 2CO_{(g)} + O_{2\;(g)}$

What is the enthalpy of this reaction?

## S11.67

$2CO_{2(g)} \rightarrow 2CO_{(g)} + O_{2(g)}$

$\Delta H^o_f = -'x'kJ/mol$

The enthalpy in the problem is concerned with half the amount of reactant, therefore producing half the products. Thus, doubling all factors involved will double the enthalpy of formation. We can apply this to solve the enthalpy of this reaction.

$x = -35.9kJ/mol \times 2 = -71.8kJ/mol$

## Q11.67

The standard enthalpy of formation of $$HCl_{(g)}$$ is -52.4 kJ/mol HCl. What is $$\Delta H^o_{rxn}$$ for the following reaction?

$2 HCl_{(g)} \rightarrow Cl_{2(g)} + H_{2(g)}$

with $$\Delta H^o_f=52.4\; kJ/mol$$

## S11.67

Consider the original:

$2 HCl_{(g)} \rightarrow Cl_{2(g)} + H_{2(g)}$

Now consider the converse:

$Cl_{2(g)} + H_{2(g)} \rightarrow 2 HCl_{(g)}$

Since it is merely the inverse of the original equation, including all the standard states of the elements involved, it is the inverse of $$\Delta H^o_f$$ of $$HCl_{(g)}$$. So $$\Delta H^o_{rxn}$$ is -52.4 kJ/mol.

## Q11.67

What is $$∆H^{o}_{rxn}$$ of

$C_2H_4 + 6F_2 \rightarrow 2CF_4 + 4HF$

given knowledge of the thermodynamics of the following reactions?

$H_{2(g)} + F_{2(g)} \rightarrow 2HF_{(g)} \;\;\;\;\; \Delta H^o_f = -537kJ$

$C_{(s)} + 2F_{2(g)} \rightarrow CF_{4(g)} \;\;\;\;\; \Delta H^o_f = -680kJ$

$2C_{(s)} + 2H_{2(g)} \rightarrow C_{2}H_{4(g)} \;\;\;\;\; \Delta H^o_f = +52.3kJ$

## S11.67

$2 \times (H_{2(g)} + F_{2(g)} \rightarrow 2HF_{(g)})$

$2 \times (C_{(s)} + 2F_{(g)} \rightarrow CF_{4(g)})$

$-1 \times (2C_{(s)} + 2H_{2(g)} \rightarrow C_{2}H_{4(g)})$

$\Delta {H^{o}_{rxn}}=\sum_i \Delta {H_i} = (2 \times -537\; kJ) + (2 \times -680kJ) + (-1 \times 52.3\; kJ) = -2486\; kJ$

## Q11.67

The standard enthalpy of formation of $$B_2O_{3\;(s)}$$ is -1273 kJ/mol $$B_2 O_3$$. What is the ∆H° for the following reaction?

$4B_{(s)}+3O_{2\; (g)} \rightarrow 2B_2O_{3\;(s)}$

## S11.67

(-1273 kJ) –(1273 kJ) × 1/2=636.5 kJ

## Q11.69

Calculate $$∆H˚$$ of the full reaction of

$2NO(g) + O_3 \rightleftharpoons 2NO_2(g)$

given knowledge of the thermodynamics of the following reactions?

$2O_3(g) \rightleftharpoons 3O_2(g) \;\;\;\;\; \Delta H^o_f= -420 kJ$

$O_2(g) \rightleftharpoons 2O(g) \;\;\;\;\; \Delta H^o_f= 622 kJ$

$NO(g) + O_3(g) \rightleftharpoons NO_2(g) + O_2(g) \;\;\;\;\; \Delta H^o_f= -137 kJ$

## S11.69

$-1 \times( 2O_3(g) \rightleftharpoons 3O_2(g))$

$0 \times (O_2(g) \rightleftharpoons 2O(g))$

$2 \times (NO(g) + O_3(g) \rightleftharpoons NO_2(g) + O_2(g))$

$\Delta {H^{o}_{rxn}}=\sum_i \Delta {H_i} = -1(-420\; kJ) + 0 + 2(-137\; kJ) = 146\; kJ$

## Q11.69

Use Hess’s Law to determine the enthalpy of $$3C_{(Graphite)} + 4H_{2(g)} \rightarrow C_3H_{8(g)}$$

1. $$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H2O_{(l)} \;\;\;\;\; \Delta H^o_f = -4567.1kJ$$
2. $$C_{(graphite)} + O_{2(g)} \rightarrow CO2(g) \;\;\;\;\; \Delta H^o_f = -432.5kJ$$
3. $$H_{2(g)} + 1/2O_{2(g)} \rightarrow H_2O_{(l)} \;\;\;\;\; \Delta H^o_f = -123.4kJ$$

## S11.69

$3C_{(Graphite)} + 4H_{2(g)} \rightarrow C_3H_{8(g)}$

$-1 \times (C_{3}H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_{2}O_{(l)})$

$3 \times C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}$

$4 \times H_{2(g)} + 1/2O_{2(g)} \rightarrow H_{2}O_{(l)}$

$\Delta H^{o}_{rxn} = \sum_i \Delta {H_i} = (-1 \times -4567.1kJ) + (3 \times -432.5kJ) + (4 \times -123.4kJ)$

$\Delta H^{o}_{rxn} = 2776kJ$

## Q11.73

Use Hess’s Law to determine the enthalpy of $$C_{2}H_{4(g)} + Cl_{2(g)} \rightarrow C_{2}H_{4}Cl_{2(l)}$$

1. $$4HCl(g) + O_{2(g)} \rightarrow 2Cl_{2(g)} + 2H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -345.6kJ$$
2. $$2HCl(g) + C_{2}H_{4(g)} + 1/2O_{2(g)} \rightarrow C_{2}H_{4}Cl_{2(l)} + H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -111.1kJ$$

## S11.73

$C_{2}H_{4(g)} + Cl_{2(g)} \rightarrow C_{2}H_{4}Cl_{2(l)}$

$\dfrac{-1}{2} \times(4HCl(g) + O_{2(g)} \rightarrow 2Cl_{2(g)} + 2H_{2}O_{(l)})$

$1 \times 2HCl(g) + C_{2}H_{4(g)} + 1/2O_{2(g)} \rightarrow C_{2}H_{4}Cl_{2(l)} + H_{2}O_{(l)}$

$\Delta H^{o}_{rxn} = \sum_i \Delta {H_i} = (\dfrac{-1}{2} \times -345.6kJ) + (1 \times -111.1kJ)$

$\Delta H^{o}_{rxn} = 61.7kJ$

## Q11.73

Find $$ΔH^{o}_{rxn}$$:

$4 NH_3 (g) + 5 O_2 (g) \rightarrow 4 NO (g) + 6 H_2O (g)$

by using Hess's law and the following equations with known thermodynamic properties

$N_2 (g) + O_2 (g) \rightarrow 2 NO (g) \;\;\;\;\; \Delta H^o_f= -210.2 kJ$

$N_2 (g) + 3 H_2 (g) \rightarrow 2 NH_3 (g) \;\;\;\;\; \Delta H^o_f = -80.8 kJ$

$2H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \;\;\;\;\; \Delta H^o_f = -376.2 kJ$

## S11.73

Video Solution

$2 \times (N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)})$

$-2 \times (N_2 (g) + 3 H_2 (g) \rightarrow 2 NH_3 (g))$

$3 \times (2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g))$

$\Delta H^{o}_{rxn} =\sum_i \Delta {H_i} = 2 (-210.2\; kJ) + -2 (-80.8\; kJ) + 3 (-376.2\; kJ) = -1387.4\; kJ$

## Q11.73

Find the $$\Delta H^o_{rxn}$$:

$Fe_{(s)} + \dfrac{3}{2}Cl_{2(g)} \rightarrow FeCl_{3(s)}$

Use the following reactions to calculate $$\Delta H^o_{rxn}$$.

$Fe (g) + Cl_2 (g) \rightarrow FeCl_2 (s) \;\;\;\;\; \Delta H^o_f = -342 kJ$

$2 FeCl_3 (s) \rightarrow 2 FeCl_2 (s) + Cl_2 (g) \;\;\;\;\; \Delta H^o_f = 114 kJ$

## S11.73

Need the equation:

$Fe_{(s)} + 3/2 Cl_{2(g)} \rightarrow FeCl_{3(s)}$

Multiply the following equations to get rid of the intermediates.

$1 \times (Fe_{(s)} + Cl_{2(g)} \rightarrow FeCl_{2(s)})$

$-\dfrac{1}{2} \times (2FeCl_{3(s)} \rightarrow 2FeCl_{2(s)} + Cl_{2(g)})$

$\Delta H^{o}_{rxn} = (1 \times -343\; kJ) + (-\dfrac{1}{2} \times 114\; kJ) = -399\; kJ$

## Q11.75

Find the $$ΔH^{o}_{rxn}$$:

$CH_4(g) + NH_3 (g) \rightarrow HCN(g) + 3 H_2(g)$

by using Hess's law.

$N_2(g) + 3 H_2 (g) \rightarrow 2 NH_3 (g) \;\;\;\;\; \Delta H^o_f = -82.6 kJ$

$C (s) + 2 H_2 (g) \rightarrow CH_4 (g) \;\;\;\;\; \Delta H^o_f = -90.3 kJ$

$H_2 (g) + 2 C (s) + N_2 (g) \rightarrow 2 HCN (g) \;\;\;\;\; \Delta H^o_f = 293.3 kJ$

## S11.75

$-\dfrac{1}{2} \times [N_2(g) + 3 H_2 (g) \rightarrow 2 NH_3 (g)]$

$-1 \times [C (s) + 2 H_2 (g) \rightarrow CH_4 (g)]$

$\dfrac{1}{2} \times [ H_2 (g) + 2 C (s) + N_2 (g) \rightarrow 2 HCN (g)]$

$\Delta H^{o}_{rxn} = \sum_i \Delta {H_i} = -\dfrac{1}{2}(-82.6 \; kJ) + -1(-90.3\; kJ) + \dfrac{1}{2} (293.3\; kJ) = 278.25 \; kJ$

## Q11.75

Use Hess’s Law to determine the enthalpy of $$4CO_{(g)} + 8H_{2(g)} \rightarrow 3CH_{4(g)} + CO_{2(g)} + 2H_{2}O_{(l)}$$

$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)}\;\;\;\;\; \Delta H^o_f = -444.5kJ$

$H_{2(g)} + 1/2O_{2(g)} \rightarrow H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -112.2kJ$

$C_{(Graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)} \;\;\;\;\; \Delta H^o_f = -223.3kJ$

$C_{(Graphite)} + 1/2O_{2(g)} \rightarrow CO_{(g)} \;\;\;\;\; \Delta H^o_f = -334.4kJ$

$CO_{(g)} + 1/2O_{2(g)} \rightarrow CO_{2(g)} \;\;\;\;\; \Delta H^o_f = -555.5kJ$

## S11.75

$-2 \times (CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)})$

$6 \times (H_{2(g)} + 1/2O_{2(g)} \rightarrow H_{2}O_{(l)})$

$1 \times (C_{(Graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)})$

$-1 \times (C_{(Graphite)} + 1/2O_{2(g)} \rightarrow CO_{(g)})$

$3 \times (CO_{(g)} + 1/2O_{2(g)} \rightarrow CO_{2(g)})$

$\Delta H^{o}_{rxn} = \sum_i \Delta {H_i}=(-444.5kJ \times -2)+(-112.2kJ \times 6)+(-223.3kJ \times 1)+(-334.4kJ \times -1)+(-555.5kJ \times 3)$

$∆H^{o}_{rxn} = -1339.6kJ$

## Q11.77

Use Hess’s Law to determine the enthalpy of $$N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)}$$

$1/2N_{2(g)} + 3/2H_{2(g)} \rightarrow NH_{3(g)} \;\;\;\;\; \Delta H^o_f = -999.9kJ$

$H_{2(g)} + 1/2O_{2(g)} \rightarrow H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -454.8kJ$

$NH_{3(g)} + 5/4O_{2(g)} \rightarrow NO_{(g)} + 3/2H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -222.2kJ$

## S11.77

$1/2N_{2(g)} + 3/2H_{2(g)} \rightarrow NH_{3(g)}$

$H_{2(g)} + 1/2O_{2(g)} \rightarrow H_{2}O_{(l)}$

$NH_{3(g)} + 5/4O_{2(g)} \rightarrow NO_{(g)} + 3/2H_{2}O_{(l)}$

$\Delta H^{o}_{rxn} =\sum_i \Delta {H_i}= (2 \times -999.9kJ) + (-3 \times -454.8kJ) + (2 \times -222.2kJ)$

$\Delta H^{o}_{rxn} = -1079.8kJ$

## Q11.79

Calculate $$\Delta H^{o}_{rxn}$$ for the thermite reaction

$2Al_{(s)}+Fe_2O_{3(s)} → Al_2O_{3(s)} + 2Fe_{(s)}$

$$\Delta H^o_f[Al]~=~0$$

$$\Delta H^o_f[Fe_{2}O_{3}]~=~-826~kJ/mol$$

$$\Delta H^o_f[Al_{2}O_{3}] ~=~ -1675.7~kJ/mol$$

$$\Delta H^o_f[Fe]~=~0$$

## S11.79

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (1mol_{Al_{2}O_{3}} \times -1675.7kJ/mol + 2mol_{Fe} \times 0kJ/mol) - (2mol_{Al} \times 0kJ'mol + 1mol_{Fe_{2}O_{3}} \times -826kJ/mol)$

$\Delta H^{o}_{rxn} = (-1675.7kJ) - (-826kJ)$

$\Delta H^{o}_{rxn} = -849.7kJ$

## Q11.79

Calculate the enthalpy changes in the following reactions using the standard enthalpies of formation.

$CO_{(g)} +CH_{4\;(g)} \rightarrow C_2H_{4\;(g)} +H_2O_{(g)}$

$SO_{3(g)} + NO_{(g)} \rightarrow SO_{2(g)} + NO_{2(g)}$

## S11.79

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (1mol_{C_{2}H_{4}} \times 52.26kJ/mol + 1mol_{H_{2}O} \times -241.818kJ/mol) - (1mol_{CO} \times -110.525kJ/mol + 1mol_{CH_{4}} \times -74.81kJ/mol)$

$\Delta H^{o}_{rxn} = (-233.5kJ) - (-185.335kJ)$

$\Delta H^{o}_{rxn} = -48.17kJ$

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (1mol_{SO_{2}} \times -296.83kJ/mol + 1mol_{NO_{2}} \times 33.18kJ/mol) - (1mol_{SO_{3}} \times -395.72kJ/mol + 1mol_{NO} \times 90.25kJ/mol)$

$\Delta H^{o}_{rxn} = (-263.65kJ) - (-305.47kJ)$

$\Delta H^{o}_{rxn} = 41.82kJ$

## Q11.79

Determine the enthalpy changes based on the enthalpy of formation from table 11.2 in the chemistry book.

$\dfrac{2}{3}NH_{3(g)} \rightarrow \dfrac{1}{3}N_{2(g)} + H_{2(g)}$

$4CO_{(g)} + 8H_{2(g)} \rightarrow 3CH_{4(g)} + CO_{2(g)} + 2H_{2}O_{(l)}$

## S11.79

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (\dfrac{1}{3}mol_{N_{2}} \times 0kJ/mol + 1mol_{H_{2}} \times 0kJ/mol) - (\dfrac{2}{3}mol_{NH_{3}} \times -45.9kJ/mol)$

$\Delta H^{o}_{rxn} = (0kJ) - (-30.6kJ)$

$\Delta H^{o}_{rxn} = 30.6kJ$

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (3mol_{CH_{4}} \times -74.6kJ/mol + 1mol_{CO_{2}} \times -393.5kJ/mol + 2mol_{H_{2}O} \times -285.8kJ/mol) - (4mol_{CO} \times -110.5kJ/mol + 8mol_{H_{2}} \times 0kJ/mol)$

$\Delta H^{o}_{rxn} = (-1188.9kJ) - (-442kJ)$

$\Delta H^{o}_{rxn} = -746.9kJ$

## Q11.81

Use Hess’s Law to calculate the standard enthalpy formation of acetone.

$C_{3}H_{6}O_{(l)} + 4O_{2(g)} \rightarrow 3CO_{2(g)} + 3H_{2}O_{(l)} \;\;\;\;\; \Delta H^o_f = -1790kJ$

## S11.81

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$-1790kJ = (3mol_{CO_{2}} \times -393.5kJ/mol + 3mol_{H_{2}O} \times -285.8kJ/mol) - (1mol_{C_{3}H_{6}O} \times kJ/mol + 4mol_{O_{2}} \times 0kJ/mol)$

$-1790kJ = (-2037.9kJ) - (\Delta H^o_{f_{acetone}})$

$\Delta H^o_{f_{acetone}} = -247.9kJ/mol$

## Q11.81

Calculate the $$\Delta H^o_f$$ for the reaction

$Cd^{2+} (aq) + SO_4^{2-}(aq) \rightarrow CdSO_4(s)$

given the following Enthalpies of formation:

• $$∆H^o_f[Cd^{2+}_{(aq)}]=-75.9\; kJ/mol$$,
• $$∆H^o_f[SO^{2-}_{4(aq)}]=-909.3 \; kJ/mol$$, and
• $$∆H^o_f[CdSO_{4(s)}]= -1473\; kJ/mol$$.

## S11.81

$\Delta H^{o}_{rxn} = \sum \Delta H^o_{f_{products}} - \sum \Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (1mol_{CdSO_{4}} \times -1473kJ/mol) - (1mol_{Cd}^{2+} \times -75.9kJ/mol + 1mol_{(SO_{4})^{2+}} \times -909.3kJ/mol)$

$\Delta H^{o}_{rxn} = (-1473kJ) - (-985.2kJ)$

$\Delta H^{o}_{rxn} = -487.8kJ$

## Q11.81

Use the equation below to calculate the standard enthalpy for the reaction.

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

## S11.81

$\Delta H^{o}_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (3mol_{CO_{2}} \times -393.5kJ/mol + 4mol_{H_{2}O} \times -285.8kJ/mol) - (1mol_{C_{3}H_{8}} \times -103.8kJ/mol + 5mol_{O_{2}} \times 0kJ/mol)$

$\Delta H^{o}_{rxn} = (-2323.7kJ) - (-103.8kJ)$

$\Delta H^{o}_{rxn} = -2219.9kJ$

## Q11.83

Calculate the enthalpy change for the following reactions at 298.15 K.

$\Delta H^o_{f}~C_{2}H_{6(g)} = -84.68kJ/mol$

$\Delta H^o_{f}~CO_{2(g)} = -393.5kJ/mol$

$\Delta H^o_{f}~H_{2}O_{(l)} = -285.8kJ/mol$

## S11.83

$2C_{2}H_{6(g)} \rightarrow 4C_{graphite} + 6H_{2} \tag{\Delta H^{o}_{rxn} = -2(-84.68) = 169.36kJ/mol}$

$7O_{2} \rightarrow 14O \tag{\Delta H^{o}_{rxn} = 0kJ/mol}$

$4C_{graphite} + 8O \rightarrow 4CO_{2(g)} \tag{\Delta H^{o}_{rxn} = 4(-393.5) = -1574kJ/mol}$

$6H_{2} + 6O \rightarrow 6H_{2}O_{(l)} \tag{\Delta H^{o}_{rxn} = 6(-285.8) = -1714.8kJ/mol}$

$2C_{2}H_{6(g)} + 7O_{2(g)} \rightarrow 4CO_{2(g)} + 6H_{2}O_{((l)}$

$\Delta H^{o}_{rxn} = \sum \Delta H^o_{f_{products}} - \sum \Delta H^o_{f_{reactants}}$

$\Delta H^{o}_{rxn} = (4mol_{CO_{2(g)}} \times -393.5kJ/mol + 6mol_{H_{2}O} \times -285.8kJ/mol) - (2mol_{C_{2}H_{6}} \times -84kJ/mol + 7mol_{O_{2}} \times 0kJ/mol)$

$\Delta H^{o}_{rxn} = (-3288.8kJ) - (-168kJ)$

$\Delta H^{o}_{rxn} = -3120.8kJ$

## Q11.83

Calculate the enthalpy change in the following reaction using the standard enthalpies of formation from table 11.2. (Hint: Use equation 1.21)

$C_{3}H_{8(g)} + CH_{4(g)} \rightarrow 2C_{2}H_{6(g)}$

## S11.83

$\Delta H^o_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^o_{rxn} = (2mol_{C_{2}H_{6}} \times -84kJ/mol) - (1mol_{C_{3}H_{8}} \times -103.8kJ/mol + 1mol_{CH_{4}} \times -74.81kJ/mol)$

$\Delta H^o_{rxn} = (-168kJ) - (-178.61kJ)$

$\Delta H^o_{rxn} = 10.61kJ$

## Q11.83

Determine the enthalpy changes based on the enthalpy of formation from table 11.2 in the chemistry book.

$N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)}$

## S11.83

$\Delta H^o_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^o_{rxn} = (2mol_{NO} \times 90.25kJ/mol) - (1mol_{N_{2}} \times 0kJ/mol + mol_{O_{2}} \times 0kJ/mol)$

$\Delta H^o_{rxn} = (90.25kJ) - (0kJ)$

$\Delta H^o_{rxn} = 90.25kJ$

## Q11.83

Calculate the $$\Delta H^o_{rxn}$$ of

$$2I_{2(g)} + 2H_{2}O_{(l)} \rightarrow 4HI_{(g)} + O_{2(g)}$$ given that:

• $$∆H^o_f[I_2(g)]=62.44 \; kJ/mol$$,
• $$∆H^o_f[H_2O(l)]=-285.8 \; kJ/mol$$, and
• $$∆H^o_f[HI(g)]= 26.48 \; kJ/mol$$

## S11.83

$\Delta H^o_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^o_{rxn} = (4mol_{HI} \times 26.5kJ/mol + 1mol_{O_{2}} \times 0kJ/mol) - (2mol_{I_{2}} \times 62.4kJ/mol + 2mol_{H_{2}O} \times -285.8kJ/mol)$

$\Delta H^o_{rxn} = (106kJ) - (-446.8kJ)$

$\Delta H^o_{rxn} = 552.8kJ$

## Q11.83

Determine the enthalpy change at 25°C for the following reaction:

$2 Br_{2(g)}+ H_2O_{(l)} \rightarrow 4HBr_{(g)}+ O_{2(g)}$

given the standard enthalpy of formation of $$HBr_{(g)}$$ is -36.40 kJ and the standard enthalpy of formation for $$H_2O$$ is -285.8.

## S11.83

$\Delta H^o_{rxn} = \sum\;\Delta H^o_{f_{products}} - \sum\;\Delta H^o_{f_{reactants}}$

$\Delta H^o_{rxn} = (4mol_{HBr} \times -36.4kJ/mol + 1mol_{O_{2}} \times 0kJ/mol) - (2mol_{Br_{2}} \times 0kJ/mol + 1mol_{H_{2}O} \times -285.8kJ/mol)$

$\Delta H^o_{rxn} = (-145.6kJ) - (-285.8kJ)$

$\Delta H^o_{rxn} = 140.2kJ$

## Q11.85

Find the heat of combustion of

$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)}$

given that

• ∆Hf˚[H2O(l)]=-285.8kJ/mol, and
• ∆Hf˚[CO2(g)]= 26.48 kJ/mol.

## S11.85

$\Delta H_{rxn} = \sum\;\Delta H_{f_{products}} - \sum\;\Delta H_{f_{reactants}}$

$\Delta H_{rxn} = (1mol_{CO_{2}} \times -393.5kJ/mol + 2mol_{H_{2}O} \times -285.8kJ/mol) - (1mol_{CH_{4}} \times -74.81kJ/mol + 2mol_{O_{2}} \times 0kJ/mol)$

$\Delta H_{rxn} = (-965.1kJ) - (-299.24kJ)$

$\Delta H_{rxn} = -665.86kJ$

## Q11.85

Determine the standard heat of combustion of $$CH_3 OH_{(l)}$$ if the reactants and products are maintained at standard conditions. Given the standard molar enthalpy of formation for $$CH_3OH_{(l)}$$ is -238.7 kJ, for $$CO_{2(g)}$$ is -393.5 kJ and for $$H_{2}O_{(l)}$$ is -285.8 kJ.

$2CH_{3}OH_{(g)} + 3O_{2(g)} \rightarrow 2 CO_{2(g)} + 4H_{2}O_{(l)}$

## S11.85

$\Delta H_{rxn} = \sum\;\Delta H_{f_{products}} - \sum\;\Delta H_{f_{reactants}}$

$\Delta H_{rxn} = (2mol_{CO_{2}} \times -393.5kJ/mol + 4mol_{H_{2}O} \times -285.8kJ/mol) - (2mol_{CH_{3}OH} \times -238.7kJ/mol + 3mol_{O_{2}} \times 0kJ/mol)$

$\Delta H_{rxn} = (-1930.2kJ) - (-477.4kJ)$

$\Delta H_{rxn} = -1452.8kJ$

## Q11.85

At 50°C and 1 bar, the standard heat of combustion for C2H5OH is -317.2 kJ/mol. What is the standard heat of combustion of two moles C2H5OH at 50°C?

## S11.85

The standard heat of combustion is not dependent on moles, but temperature.. Therefore, considering the temperature is consistent, it is still -317.2 kJ/mol

## Q11.87

Calculate the standard energy of combustion at 25 °C and 1 bar for the following reaction:

$C_{6}H_{12}O_{6(s)} +6O_{2(g)} \rightarrow 6CO_{2(g)} + 6H_{2}O_{(l)}$

Given these values:

• $$\Delta H^o_f = -1275.0\; kJ/mol$$ for $$C_{6}H_{12}O_{6(s)}$$
• $$\Delta H^o_f = -393.5\; kJ/mol$$ for $$CO_{2(g)}$$
• $$\Delta H^o_f = -285.8\; kJ/mol$$ for $$H_{2}O_{(l)}$$

## S11.87

$\Delta H_{rxn} = \sum\;\Delta H_{f_{products}} - \sum\;\Delta H_{f_{reactants}}$

$\Delta H_{rxn} = (1mol_{C_{6}H_{12}O_{6}} \times -1275kJ/mol + 6mol_{O_{2}} \times 0kJ/mol) - (6mol_{CO_{2}} \times -393.5kJ/mol + 6mol_{H_{2}O} \times -285.8kJ/mol)$

$\Delta H_{rxn} = (-1275kJ) - (kJ)$

$\Delta H_{rxn} = 2800.8kJ$

## Q11.87

Calculate the $$∆H˚$$ of

$SO_{3(g)} + H_{2}O_{(l)} \rightarrow H_{2}SO_{4(l)}$

given that

• $$∆H_f˚[H_2SO_4(l)] = -814.8 \; kJ/mol$$,
• $$∆H_f˚[H_2O(l)] = -285.8 \; kJ/mol$$, and
• $$∆H_f˚[SO_3(g)] = -395.7\; kJ/mol$$

## S11.87

$\Delta H_{rxn} = \sum\;\Delta H_{f_{products}} - \sum\;\Delta H_{f_{reactants}}$

$\Delta H_{rxn} = (1mol_{H_{2}SO_{4}} \times -814.8kJ/mol) - (1mol_{SO_{3}} \times -395.7kJ/mol + 1mol_{H_{2}O} \times -285.8kJ/mol)$

$\Delta H_{rxn} = (-814.8kJ) - (-681.5kJ)$

$\Delta H_{rxn} = -133.3kJ$

## Q11.87

Determine the standard enthalpy of formation of $$C_{2}H_{2(g)}$$ at 25°C and 1 bar given $$∆H°_f$$ of $$CO_{2(g)}$$ is -393.5 kJ/mol, and the $$∆H°_f$$ of $$H_2O_{(l)}$$ is -285.8. ∆H°= -2599 kJ.

## S11.87

$2C_{2}H_{2(g)} + 5O_{2(g)} \rightarrow 4CO_{2(g)} + 2H_{2}O_{(l)}$

$\Delta H_{rxn} = \sum\;\Delta H_{f_{products}} - \sum\;\Delta H_{f_{reactants}}$

$-2599kJ = (4mol_{CO_{2}} \times -393.5kJ/mol + 2mol_{H_{2}O} \times -285.8kJ/mol) - (2mol_{C_{2}H_{2}} \times kJ/mol + 5mol_{O_{2}} \times 0kJ/mol)$

$-2599kJ = (-2145.6kJ) - (2mol_{C_{2}H_{2}} \times kJ/mol)$

$\Delta H^o_{f_{(C_{2}H_{2})}} = \dfrac{2599kJ-2145.6kJ}{2mol_{C_{2}H_{2}}}$

$\Delta H_{f_{C_{2}H_{2}}} = 226.7kJ/mol$

## Q11.89

Use a table of Enthalpies of Formation to figure out the change of enthalpy in the following reaction

$PO^{3-}_4 + 3Li^+ \rightarrow PO_4Li_3$

## S11.89

$\Delta H^o_{rxn} = (1mol_{PO^{3-}_{4}} \times -1277.4kJ/mol) + (3mol_{Li^+} \times -278.5kJ/mol)$

$\Delta H^o_{rxn} = -2113kJ$

## Q11.89

Determine the standard enthalpy change in the following reaction given:

$\Delta H^o_{f}~~ Ba^{2+}_{aq} = -537.6 kJ/mol$

$\Delta H^o_{f} ~~OH^{-}_{aq} = -230kJ/mol$

$\Delta H^o_{f} ~~Ba(OH)_{2(s)} = -944.7kJ/mol$

## S11.89

$Ba^{2+}_{aq} + 2OH^{-}_{aq} \rightarrow Ba(OH)_{2(s)}$

$\Delta H^o = \sum \Delta H^o_{products} - \sum \Delta H^o_{reactants}$

$\Delta H^o = (-944.7kJ/mol) - [(-537.6kJ/mol) + (2 \times -230kJ/mol)] = 52.9kJ/mol$

## Q11.89

Determine how much heat is required to decompose $$2.15 \times 10^{3} kg.~of~Ca( OH)_{2(s)}~into~CaO_{(s)}~and~H_{2}O_{(l)}$$, given that:

$\Delta H^o_{f} ~of~ Ca(OH)_{2(s)} = -986.1kJ/mol$

$\Delta H^o_{f} ~of~ CaO_{(s)} = -635.1 kJ/mol$

$\Delta H^o_{f} ~of~ H_{2}O_{(l)} = -285.8 kJ/mol$

## S11.89

$Ca(OH)_{2(s)} \rightarrow CaO_{(s)} + H_{2}O_{(l)}$

$\Delta H^o = \sum \Delta H^o_{products} - \sum \Delta H^o_{reactants}$

$\Delta H^o = [(-635.1 kJ/mol) + (-285.8 kJ/mol)] – (-986.1 kJ/mol) = 65.2 kJ/mol$

$q = 2.15 \times 10^3 kg ~ Ca(OH)_{2} \times \frac{1000g}{1kg} \times \frac{1mol_{Ca(OH)_{2}}}{74.093g_{Ca(OH)_{2}}} \times \frac{65.2kJ}{1mol_{Ca(OH)_{2}}}$

$q = 1.89 \times 10^{6} kJ$

## Q11.89

Calculate the $$∆H˚$$ of

$Ba^{2+} (aq) + 2 OH^-(aq) \rightarrow Ba(OH)_2(s)$

given that

• $$∆H_f˚[Ba^{2+}(aq)]=-537.6 \; kJ/mol$$,
• $$∆H_f˚[OH^-(aq)]= -230.0 \; kJ/mol$$, and
• $$∆H_f˚[Ba(OH)_2(s)]= -944.6 \; kJ/mol$$.

## S11.89

$\Delta H^o ~=~ \sum \Delta H^o_{products} ~-~ \sum \Delta H^o_{reactants}$

$\Delta H^o ~=~ (-944.6 kJ/mol) ~–~ [(2 \times -230 kJ/mol) ~+~ (-537.6 kJ/mol)] ~=~ 53 kJ/mol$

## Q11.91

A 3.0 gram sample of Hydrogen gas is burned in excess nitrogen in a bomb calorimeter. The heat capacity is equal to 14.6 kJ/C. The temperature increases from 13.0 C to 23.0 C. What is the energy of combustion per kilogram?

## S11.91

1. $q=m c_{s} \Delta T$

$q = (3g)( 14.6kJ/{^{\circ}}C)( -10{^{\circ}}C)$

$q = - kJ/kg$

q=-C∆T = -(14.6 kJ/ C) (-10 C)=-146 kJ

-146 kJ/3.0 grams x 100 grams/ 1 kg = -4866.67 kJ/kg.

## Q11.91

The decomposition of limestone, CaCO3 (s), into quicklime, CaO(s), and CO2(g) is carried out in a gas-fired kiln. Use data from Appendix D to determine how much heat is required to decompose 4.6 x 10^4 kg CaCO3(s). (Assume that heats of reaction are the same as at 25°C and 1 bar.)

## S11.91

$\Delta H^o = \sum \Delta H^o_{products} - \sum \Delta H^o_{reactants}$

$\Delta H^o = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] – (-1207 kJ/mol) = 178.4 kJ/mol$

$q = 4.6 \times 10^4 kg ~ CaCO_{3} \times \frac{1000g}{1kg} \times \frac{1mol_{CaCO_{3}}}{100.0869g_{CaCO_{3}}} \times \frac{178.4kJ}{1mol_{CaCO_{3}}}$

$q = 8.199 \times 10^{7} kJ$

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Thermodynamisc_Cycles/Hess'_Law

## Q11.91

Calculate the $$∆H˚$$ needed to decompose $$MgCO_3(s)$$

$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$

given that

• $$∆H_f˚[MgO(s)]=-601.7 \; kJ/mol$$,
• $$∆H_f˚[CO_2(g)]= -393.5 \; kJ/mol$$,
• $$∆H_f˚[MgCO_3(s)]= -1096 \; kJ/mol$$, and

## S11.91

$\Delta H^o = \sum \Delta H^o_{products} - \sum \Delta H^o_{reactants}$

$\Delta H^o = [(-601.7 kJ/mol) + (-393.5 kJ/mol)] – (-1096 kJ/mol) = 100.8 kJ/mol$

## Q11.95

One British thermal unit= quantity of heat needed to change the temperature of 1 lb of water by one degree F. Specific heat of water is independent of temperature. How much heat is needed to raise 20 lbs of water from 50 to 100 degrees F in a) Btu b) joules and c) calories.

## S11.95

$1 BTU ~=~ amount of energy for \Delta T_{H_{2}O} per 1~ ^{o}F$

$20lb. \times 50^{o} ~=~ 1000~BTU$

$20lbs. ~=~ 9.07185kg_{H_{2}O} \times \frac{1000g}{1kg} ~=~ 9071.85g{H_{2}O}$

$50^{o}F ~=~ 27.778^{o}C; q ~=~ mc \DeltaT$

$q ~=~ 9071.85g \times \frac{4.184J}{1~g^{o}C} \times 27.778^{o}C ~=~ 1054359J$

$\frac{1~cal}{4.1858J} \times 1054359J = 251889.4835cal$

## Q11.95

A water heater that holds 50 gal must raise the temperature of the water from 45 to 138˚F. How much heat, in (a)kcal; (b)kJ; (c)Btu, is needed to reach this temperature. (Btu = amount of heat required to raise the temperature of 1lb of water by 1˚F)

## S11.95

1. heat (kcal) = 1.5x105g x 1cal/(g˚C) x (138˚F - 45˚F) x (5˚C/9˚F) x 1kcal/1000cal = 7750kcal
2. heat (kJ) = 7.75x103kcal x 4.184kJ/1kcal = 3.24x104 kJ
3. mass H2O = 50gal x 4qt/1gal x 1L/1.06qt x 1000mL/1L x 1g/1mL x 1kg/1000g x 2.205lb/1kg = 416 lb.heat (Btu) = 416lb x 1Btu/(lb˚F) x (138˚F - 45˚F) = 3.9x104 Btu

## Q11.95

How much heat is required to lower the temperature of 20 kg of sulfuric acid from 98 °C to 23 ° C in a) kJ b) kcal c) Btu? (specific heat of sulfuric acid = 1.34 kJ / kg • K)

## S11.95

(1 kJ / kg • K = 0.2389 kcal / kg °C = 0.430 Btu / pound °C)

∆T = 23 – 98 °C = -75 °C

∆Kevlin = -75 °C = -75 K

a) q = mC∆T

q = (20 kg)(1.34 kJ / kg • K)(-75 K) = -2010 kJ

b) q = (20 kg)(1.34 kJ / kg • K)(.2389 kcal / kg °C / 1 kJ / kg • K)(-75 °C) = -480 kcal

c) q = (20 kg)(2.20 pounds / 1 kg)(1.34 kJ / kg • K)(.430 Btu / pound °C / 1 kJ / kg • K)(-75°C) = -1901 Btu

## Q11.99

Why does Hydrogen have two values for the heat of combustion if liquid water and liquid vapor is formed?

## S11.99

Hydrogen has two different values because the difference in enthalpy of vaporization of water. When water evaporates less heat is evolved.

## Q11.116

There is one possible reaction for the formation of $$MgCO_3(s)$$. What is the heat of reaction?

$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$

## S11.116

∆HfMgCO3(s) = -1096 kJ mol-1

∆Hf MgO(s) = -601.7 kJ mol-1

∆Hf CO2(g) = -393.5 kJ mol-1

∆H˚rxn = [(-393.5 kJ mol-1) + (-601.7 kJ mol-1)] - (-1096 kJ mol-1)

= 494.3 kJ mol-1

## Q11.124

Find the $$∆H˚_{rxn}$$ of

$Mg + 2H_2O \rightarrow Mg(OH)_2 + H_2$

## S11.124

$$∆H_f$$ of $$Mg(OH)_{2\;(s)} is -924.5 kJ \(mol^{-1}$$

$$∆H_f$$ of $$Mg_{(aq)}$$ = -466.9 kJ mol-1

$$∆H_f$$ of $$H_2O_{(g)}$$ = -285.8 kJ mol-1

∆H˚rxn = [(-924.5 kJ mol-1) + (-0 kJ mol-1)] - [(-466.9 kJ mol-1) + (-285.8 kJ mol-1)]

= -171.8 kJ mol-1

## Q11.2

Calculate the quantity of heat, in kJ required to:

1. Raise the temperature of a 32.2 g bar of pure gold from 22.0°C to 100°C (Note: $$C_p$$ of gold is 0.129 J/g \times C)
2. Decrease the temperature of 45.5 L of water 35.0°C

## S11.2

-> q = (m) x (c) x (∆T)

-> = (32.2g)(0.129J/g \times C)(78 \times C)

-> = 3.24 x 102 J

a. q = 3.24 x 102 J

-> q = (m) x (c) x (∆T)

-> = (45,500g)(4.18J/g \times C)(-35 \times C)

-> = -6.66 x 106 J

b. q = -6.66 x 106 J

## Q11.3

A 96.5 g chunk of an unknown metal heated to 98.0°C is dropped into a coffee cup calorimeter holding 45.0 g of pure water. Before the addition of the metal, the temperature of the water is measured at 25.0°C. After the addition of the metal, the temperature of the water raises to 33.3°C.

1. What is the specific heat of the given metal?
2. Given that the specific heats of gold, aluminum, and silver are 0.129 J/g \times C, 0.900 J/g \times C, and 0.240 J/g \times C respectively, what metal most likely comprises a majority of the unknown metal?

## S11.3

-> q = (m) x (c) x (∆T)

= (45g.)(4.18J/g \times C)(8.3 \times C)

= 1561.23 J absorbed by the water

-> 1561.23 J lost by the chunk of metal

-> 1561.23 J / 64.7 \times C

= 24.13 J/ \times C is the heat capacity of the metal

-> 24.13 J/ \times C divided by 96.5g

= 0.25 J/g \times C is the specific heat of the unknown metal

a. Cunknown metal = 0.25 J/g \times C

b. Comparing 0.240 J/g \times C and 0.25 J/g \times C, this chunk is most likely comprised a majority of Silver.

## Q11.4

When 13.6 g of sucrose is burned in a bomb calorimeter, the temperature increases by 5.6 C. The heat capacity of the calorimeter is calculated at 3.40 kJ/C. Calculate the heat of combustion of the sugar per mole.

## S11.4

C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)

q = (c) x (∆T)

= 3.4kJ/ \times C x 5.6 \times C

= 19.04kJ

n of sucrose = 0.03973 moles

∆E = -19.04kJ / 0.03973 molsucrose

= -479.2348 kJ/molsucrose

Since we have same moles of gas in reactants and products, ∆H = ∆E

∆H = 479.2348 kJ/molsucrose

## Q11.5

Calculate the work, in joules, associated with an unknown gas’ expansion from 3.04 L to 5.60 L with an external pressure of 1.4 atm.

## S11.5

W = (-p) x (∆V)

= (-1.4atm) x (2.56L)

= -3.584L atm

## Q11.6

In compressing a gas, 489 J of work is done on the system. Simultaneously, 239 J escape the system. Calculate ∆U for the system.

∆U = w + q

∆U = 489J - 239J

∆U = 250J

## Q11.7

Which of the following is NOT true?

1. When heat is absorbed by a system, q<0
2. If work is done on a system, w<0
3. When heat is released by a system, q<0
4. ∆U of an isolated system is 0

## S11.7

a. When heat is absorbed by a system, q < 0

Q is a positive value when heat is absorbed by the system.

## Q11.8

The heat of combustion of propane is -2220.1 kJ/mol. Calculate the heat of formation given that the ∆H of H20 is -285.8 kJ/mol and the ∆H of CO2 is -393.5 kJ/mol.

## S11.8

\ C_3H_8\;(g) + 5O_2\;(g) \rightarrow 3CO_2\;(g) + 4H_2O\;(l)\

-2220.1kJ/mol = [(3 mol x -393.5kJ/mol) + (4 mol x -285.8kJ/mol)] - [(1 mol x ∆Hf C3H8) + (5 mol x 0kJ/mol)]

-2220.1kJ/mol = (-2321.7kJ/mol) - (∆Hf C3H8)

∆Hf C3H8 = -103.6kJ/mol

## Q11.9

Calculate the ∆H for the following reaction (see standard enthalpy table):

## Q11.10

Given the following equations and the ∆H values:

$BCl_3\;(g) + 3H_2O\;(l) \rightarrow H_3BO_3\;(g) + 3HCl\;(g)\; \Delta H = -112.5\; kJ$

$B_2H_6\;(g) + 6H_2O\;(l) \rightarrow 2H_3BO_3\;(s) + 6H_2\;(g)\; \Delta H = -493.4\; kJ$

$1/2H_2\;(g) + 1/2Cl_2(g) \rightarrow HCl\;(g)\; \Delta H = -92.3\; kJ$

determine the enthalpy of reaction in kJ at 298 K:

$B_2H_6\;(g) + 6Cl_2\;(g) \rightarrow 2BCl_3\;(g) + 6HCl\;(g)$

## Q11.11

Given the following equations and the ∆H values:

$N_2H_4\;(l) + O_2\;(g) \rightarrow N_2\;(g) + 2H_2O\;(l)\; \;\;\; \Delta H = -622.3\; kJ$

$H_2\;(g) + 1/2O_2\;(g)\rightarrow H_2O\;(l)\;\;\;\; \Delta H = -285.8\; kJ$

$H_2\;(g) + O_2\;(g)\rightarrow H_2O_2\;(l)\;\;\;\; \Delta H = -187.8\; kJ$

determine the ∆H for the following reaction using the listed enthalpies:

$N_2H_4\;(l) + 2H_2O_2\;(g) \rightarrow N_2\;(g) + 4H_2O\;(l)$

## Q11.12

Given the following reactions and $$\Delta {H}$$ values:

$$SOCl_2(l)+H_2O(l) \rightarrow SO_2(g)+2HCl(g)$$ $$\Delta {H}$$=10.3 kJ

$$PCl_3(l)+1/2O_2(g) \rightarrow POCl_3(l)$$ $$\Delta {H}$$=-325.7 kJ

$$P(s)+3/2Cl_2(g) \rightarrow PCl_3(l)$$ $$\Delta {H}$$=-306.7 kJ

$$4HCl(g)+O_2(g) \rightarrow 2Cl_2(g)+2H_2O(l)$$ $$\Delta {H}$$=-202.6 kJ

determine the heat of reaction at 298 K for the reaction:

$2SO_2(g)+2P(s)+5Cl_2(g) \rightarrow 2SOCl_2(l)+2POCl_3(l)$

## Q11.13

Calculate enthalpy for the following reaction (see Table T1):

$2C_2H_6 (g)+7O_2(g) \rightarrow 6H_2O(g)+4CO_2(g)$

## Q11.14

Calculate enthalpy for the following reaction (see Table T1):

$4NH_3(g)+5O_2(g) \rightarrow 4NO(g)+6H_2O(g)$

## Q15

Calculate the enthalpy change for the following ionic reaction (see Table T1):

$S^{2-}(aq)+4I_2(s)+8OH^-(aq) \rightarrow SO4^{2-}(aq)+8I^-(aq)+H_2O(l)$

## Q16

The value for the heat of formation of $$Br_2(l)$$ is 0 kJ/mol, while the value for the heat of formation of $$Br_2(g)$$ is 30.91 kJ/mol. Why is there a difference?

## Q17

Explain why water has a significantly higher specific heat than simple elements like lead.

## Q18

A clay pot containing water at 30C is placed in the shade on a day in which the temperature is 40 degrees C. The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot a) increase b) decrease c) stay the same?

## Q11.19

True or False: Combustion reactions are endothermic.

## S11.19

False. Combustion reactions release heat (energy) and are thus exothermic.

## Q20

Given the heat of formation of $$BaCl_2(s)$$ is -858.6 kJ/mol, what is the standard enthalpy change for the precipitation of barium chloride?

## Q21

Which of the following is NOT a significant possible effect of global warming?

1. Rise in sea level due to melting of the ice caps
2. Gasoline prices go down
3. The migration of plant and animal species
4. Local temperature changes

## Q22

Cold water and a piece of hot metal come into contact in an isolated container. When the final temperature of the metal and water are identical, is the total energy change in this process:

1. zero
2. negative
3. positive
4. not enough information

## Q8

The heat of combustion of propane ($$C_3H_8$$ is -2220.1 kJ/mol. Calculate the heat of formation of propane given:

• $$∆H$$ of $$H_2O$$ is -285.3 kJ/mol and
• $$∆H$$ of $$CO_2$$ is -393.5 kJ/mol.

## S8

$C_3H_8\;(g)+5O_2\;(g) \rightarrow 3CO_2\; (g)+4H_2O\;(l)$

∆H = -2220.1 kJ/mol

∆H = (heat of formation)products - (heat of formation)reactants

-2220.1 kJ/mol = [3 \times (-393.5 kJ/mol) + 4 \times (-285.3 kJ/mol)] - [(5 \times (0 kJ/mol) + (Hf, C3H8)

(Hf, C3H8) = -101.6 kJ/mol

## Q9

Calculate the ∆H for the following reaction (see standard enthalpy table):

## S9

$8Al\;(s)+3Fe_3O_4\;(s) \rightarrow 4Al_2O_3\;(s)+9Fe\;(s)$

∆H = [(4 \times 1669.8 kJ) - 0 kJ)] - [(3 \times -1120.9 kJ) - 0 kJ)]

∆H = -3316.5 kJ

## Q10

Given the following equations and the ∆H values, determine the enthalpy of reaction in kJ at 298 K:

$B_2H_6\;(g) + 6Cl_2\;(g) \rightarrow 2BCl_3\;(g) + 6HCl\;(g)$

$BCl_3\;(g) + 3H_2O\;(l) \rightarrow H_3BO_3\;(g) + 3HCl\;(g)\; \Delta H = -112.5\; kJ$

$B_2H_6\;(g) + 6H_2O\;(l) \rightarrow 2H_3BO_3\;(s) + 6H_2\;(g)\; \Delta H = -493.4\; kJ$

$1/2H_2\;(g) + 1/2Cl_2(g) \rightarrow HCl\;(g)\; \Delta H = -92.3\; kJ$

## S10

$B_2H_6\;(g) + 6H_20\;(l) \rightarrow 2H_3BO_3\;(s) + 6H_2\;(g)$

$[BCl_3\;(g) + 3H_2O\;(l) \rightarrow H_3BO_3\;(g) + 3HCl\;(g)] \times -2$

$[1/2H_2\;(g) + 1/2Cl_2\;(g) \rightarrow HCl\;(g)] \times 12$

$\Delta H = -493.4\; kJ + (-2 \times -112.5\; kJ) + (12 \times -92.3\; kJ)$

$\Delta H = -1376\; kJ$

## Q11

Determine the ∆H for the following reaction using the listed enthalpies:

$N_2H_4\;(l) + 2H_2O_2\;(g) \rightarrow N_2\;(g) + 4H_2O\;(l)$

$N_2H_4\;(l) + O_2\;(g) \rightarrow N_2\;(g) + 2H_2O\;(l)\; \Delta H = -622.3\; kJ$

$H_2\;(g) + 1/2O_2\;(g)\rightarrow H_2O_2\;(l)\; \Delta H = -285.8\; kJ$

$H_2\;(g) + O_2\;(g)\rightarrow H_2O_2\;(l)\; \Delta H = -187.8\; kJ$

## S11

N2H4(l)+O2(g)-->N2(g)+2H2O(l) ∆H=-622.3 kJ

[2H2O2(l)-->2H2(g)+2O2(g)]X-2 ∆H=(-187.8 kJ)X-2

[2H2(g)+O2(g)-->2H2O(l)]X2 ∆H=(-285.8 kJ)X2

∆H=(-622.3 kJ)+(-187.8 kJ)+(-285.8 kJ \times 2)=-818.3 kJ

## Q12

Given the following equations and ∆H values, determine the heat of reaction (kJ) at 298 K for the reaction:

2SO2(g)+2P(s)+5Cl2(g)-->2SOCl2(l)+2POCl3(l)

SOCl2(l)+H2O(l)-->SO2(g)+2HCl(g) ∆H=10.3 kJ

PCl3(l)+1/2O2(g)-->POCl3(l) ∆H=-325.7 kJ

P(s)+3/2Cl2(g)-->PCl3(l) ∆H=-306.7 kJ

4HCl(g)+O2(g)-->2Cl2(g)+2H2O(l) ∆H=-202.6 kJ

## S12

[SOCl2(l)+H2O(l)-->SO2(g)+2HCl(g)]X2

[PCl3(l)+1/2O2(g)-->POCl3(l)]X2

[P(s)+3/2Cl2(g)-->PCl3(l)]X2

[4HCl(g)+O2(g)-->2Cl2(g)+2H20(l)]X-1

∆H=(-2 \times 10.3 kJ)+(2 \times -325.7 kJ)+(2 \times -306.7 kJ)+(-1 \times -202.6 kJ)

∆H=-1082.8 kJ

## Q13

Calculate enthalpy for the following reaction (see enthalpies table pg. 477):

2C2H6(g)+7O2(g)-->6H20(g)+4CO2(g)

## S13

∆H=[Sum of Vp ∆H(products)]-[Sum of Vp ∆H(reactants)]

∆H=[6 \times (-241.8 kJ)+4 \times (-393.5 kJ)]-[2 \times (-84.68 kJ)+7 \times (0 kJ)

∆H=-2855.44 kJ

## Q14

Calculate enthalpy for the following reaction (see enthalpies table pg. 477):

4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

## S14

∆H=[Sum of Vp ∆H(products)]-[Sum of Vp ∆H(reactants)]

∆H=[6 \times (-241.8 kJ)+4(90.25 kJ)]-[4 \times (-46.11 kJ)+5 \times (0 kJ)]

∆H=-905.36 kJ

## Q15

Calculate the enthalpy for the following ionic reaction (see enthalpies pg. 482):

S^2-(aq)+4I2(s)+8OH-(aq)-->SO4^2-(aq)+8I-(aq)+H2O(l)

## S15

∆H=[Sum of Vp ∆H(products)]-[Sum of Vp ∆H(reactants)]

∆H=[(-909.3 kJ)+(8 \times -55.19 kJ)+(-285.8 kJ)]-[(33.05 kJ)+(4 \times 0 kJ)+(8 \times -230.0 kJ)

∆H=170.33 kJ

## Q16

The value for the heat of formation of $$Br_{2\;(l)}$$ is 0 kJ/mol while the value for the heat of formation of $$Br_{2\;(g)}$$ is 30.91 kJ/mol. What is the origin of the difference?

## S16

Bromine is in its most stable form when it is a liquid. As a gas, it is less stable and requires more energy to form.

## Q17

Explain why water has a significantly higher specific heat than simple elements like lead.

## S17

Water and compounds in general have greater complexity at their molecular level and more ways of storing internal energy.

## Q18

A clay pot containing water at 30C is placed in the shade on a day in which the temperature is 40 C. The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot a) increase b) decrease c) stay the same?

## S18

B) Moisture on the outside of the pot removes heat from the pot as it evaporates.

## Q11.20

Given the $$∆H_{f}˚$$ of $$BaCl_{2(s)}$$ = -858.6 kJ/mol, what is the standard enthalpy change for the precipitation of barium chloride?

## S20

$Ba^{2+}_{(aq)}+Cl^{-}_{(aq)} \rightarrow BaCl_{2(s)}$

$∆H_f˚ = (-858.6 kJ/mol) -[(-537.6kJ/mol)+(-167.2kJ/mol)$

$∆H_f˚ = -153.8kJ/mol$

## Q11.23

Is it possible for a chemical reaction to have a ∆U>0 and a ∆H<0? Explain.

## S11.23

A system that loses heat and has work done on it by the surroundings would result in positive change in the internal energy if there was enough work done on it.