10: Thermodynamics

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Energy

Exercise \(\PageIndex{1.1}\)

Express the quantity of 422 J in calories.
Express the quantity of 3.225 kJ in calories.

Answer

101 cal

\(422 J\left ( \frac{1 calorie}{4.184 } \right )=101 calorie\)

770.8 cal

Exercise \(\PageIndex{1.2}\)

Express the quantity 55.69 cal in joules.
Express the quantity 965.3 kcal in joules.
Express the quantity 965.3 Cal in joules.

Answer

233.0 joules

\(55.69 cal\left ( \frac{4.184 J}{1 cal} \right )=233.0 J\)

4.039 x 10⁶ joules
4.039 x 10⁶ joules (Cal is kcal)

Heat Capacity

Exercise \(\PageIndex{2.1}\)

The specific heat capacity of water(liquid) is 4.18 J/g⋅°C. What is the molar specific heat capacity of this substance? The molar mass of water is 18.01 g/mol.

Answer

75.3 J/mol * °C

\(c_{m}=\frac{4.18 J}{\left ( \frac{1}{18.01} mol * °C\right )}=75.2818 J/mol * °C\)

Exercise \(\PageIndex{2.2}\)

Exactly 149.6 J will raise the temperature of 10.0 g of a metal from 25.0 °C to 60.0 °C. What is the specific heat capacity of the metal?

Answer: 0.427 J/(g *°C)

Exercise \(\PageIndex{2.3}\)

A 100 g sample of each of the following metals is heated from 35°C to 45°C. Which metal absorbs the lowest amount of heat energy?

Metal Specific Heat Capacity

Copper 0.385 J/(g*°C)

Magnesium 1.02 J/(g*°C)

Mercury 0.138 J/(g*°C)

Silver 0.237 J/(g*°C)

Lead 0.129 J/(g*°C)

Answer: Lead

Exercise \(\PageIndex{2.4}\)

35.0 kJ raises the temperature of an gold object from 22^oC to its melting point of 1064.18^oC. What is the mass of the gold object?

Answer: \[q=mc\Delta T \] [\ m=\frac{q}{c\left ( T_f-T_i \right )}\] \[=\frac{35kJ\left ( \frac{1000J}{kJ} \right )}{\frac{0.129J}{g\cdot ^{o} C}\left (1064.18^{o}C-22^{o}C \right )}=260. g \nonumber\]

Exercise \(\PageIndex{2.5}\)

What is the specific heat capacity in cal/g^oC of a substance if it takes 5.09 kJ to raise a 30.1 g mass sample of the material from 25.0 to 400.0^oC?

Answer: \[q=mc\Delta T\\c=\frac{q}{m\left ( T_f-T_i \right )}=\frac{5.09kJ\left ( \frac{1000J}{kJ} \right )\left ( \frac{cal}{4.184J} \right )}{30.1g\left (400.0^{o}C-25.0^{o}C \right )}=0.108\left ( \frac{cal}{g\cdot^{o}C} \right ) \nonumber\]

Exercise \(\PageIndex{2.6}\)

What is the final temperature if 20.0kJ is added to a 340 g of gold at 25.0^oC?

Answer: \[q=mc\Delta T \\ T_F = T_i+\frac{q}{mc} \\ = 25.0^o+\frac{20.0kJ}{0.340kg(0.108)( \frac{kcal}{kg \cdot^oC)}}=545^oC\nonumber \]

Exercise \(\PageIndex{2.7}\)

35.0 kJ raises the temperature of a 260.0g object from 22^oC to its melting point of 1064.18^oC. Could the object be gold?

Answer: \[q=mc\Delta T\\c=\frac{q}{m\left ( T_f-T_i \right )}=\frac{35.0kJ\left ( \frac{1000J}{kJ} \right )\left ( \frac{cal}{4.184J} \right )}{260.0g\left (1064.18.0^{o}C-25.0^{o}C \right )}=0.108\left ( \frac{cal}{g\cdot^{o}C} \right ) \\=0.0309 \\ \text{wrong heat capacity, can not be gold}\nonumber\]

Exercise \(\PageIndex{2.8}\)

How much heat is required to be removed from water for it to cool the temperature of 1.00 liter of water (density = 0.9982g/mL) from 25.0^oC to 32.0^oF?

Answer: \[q=mc\Delta T= 1.00L\left ( 0.9982\frac{kg}{L} \right )4.18J\frac{kJ}{kg \cdot^oC}\left ( 0.00^oC-25.0^oC \right )\\q=-104kJ\]

Exercise \(\PageIndex{2.9}\)

How much heat is required to be removed from ice to cool 1.00 liter of water (density = 0.9982g/mL) from its freezing point to -5.00^oC?

Answer: \[q=mc\Delta T= 1.00L\left ( 0.9982\frac{kg}{L} \right )2.06\frac{kJ}{kg \cdot^oC}\left ( -5^oC-0^oC \right )\\q=-10.3kJ\]

Energy and Phase Changes

Exercise \(\PageIndex{1}\)

How much energy is needed to convert 5.88 g of ice at 0.00 °C to liquid water at 85.0 °C?

Specific heat capacity (ice) = 2.10 J/g*°C
Specific heat capacity (liquid water) = 4.18 J/g*°C
Heat of fusion = 333 J/g

Answer

4.05 * 10³ J

\(q_{1} = m\Delta H_{f} = \left ( 5.88 g \right )\left ( 333 J/g \right ) = 1958.04 J\)

\(q_{2} = mc\Delta T = \left ( 5.88 g \right )\left ( 4.18 J/g*°C \right )\left ( 85 °C - 0 °C \right ) = 2089.164 J\)
q_Total = q₁ + q₂ = 1958.04 J + 2089.164 J = 4047.204 J

Exercise \(\PageIndex{2}\)

Calculate the energy in the form of heat (in kJ) required to convert 125 grams of liquid water at 20.0 °C to steam at 115 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: liquid water = 4.184 J/g*°C, steam = 1.92 J/g*°C)

Answer: 3.28 * 10² kJ

Exercise \(\PageIndex{3}\)

Calculate the energy in the form of heat (in kJ) required to change 81.5 g of liquid water at 30.6 °C to ice at –13.5 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g*°C, liquid water = 4.184 J/g*°C)

Answer: -39.8 kJ

Exercise \(\PageIndex{4}\)

What is the minimum mass of ice at 0.0 °C that must be added to 2.50 kg of water to cool the water from 28.0 °C to 9.0 °C? (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/g*°C, liquid water = 4.184 J/g*°C)

Answer

m = 536 g

\(mc\Delta T = mc\Delta T + m\Delta H_{f}\)

\((2500g)(4.184 J/g*°C)(28 °C - 9 °C) = x(4.184 J/g*°C)(9 °C - 0 °C)+ x(333 J/g)\)

\(198740 = x370.656\)

\(x = \frac{198740}{370.656}\)

x = 536.184 g

Exercise \(\PageIndex{5}\)

How much heat is required to be removed from 1.00L water to freeze it at it's freezing point

Answer: \[\Delta H=-n\Delta H_f=1.00L\left ( 998.2\frac{g}{L} \right )\frac{mol_{H_2O}}{18.015g}\left ( \frac{6.01kJ}{mol_{H_2O}} \right )= 332kJ\]

First Law of Thermodynamics

Exercise \(\PageIndex{1}\)

If a 55.8 g piece of silver at 75.8°C is placed in 150.0 g of water at 35.2°C, what is the final temperature? Assume that no heat is transferred to the surroundings.

Answer

36.0 °C

\((mc_{s}(T_{final}-T_{initial}))_{Ag} + (mc_{s}(T_{final}-T_{initial}))_{H_{2}O} = 0\)

\((55.8 g)(0.240 J/g*°C)(T_{final} - 75.8°C) + (150 g)(4.184 J/g*°C)(T_{final} - 35.2 °C) = 0\)
\(T_{final}(13.392 J/g*°C) -1015.1136 J + T_{final}(627.6 J/g*°C) - (22091.52 J) = 0\)

\(T_{final}(640.992 J/g*°C) = 23106.6336 J\)

\(T_{final} = 36.048 °C\)

Exercise \(\PageIndex{2}\)

If a 105.3 g piece of copper pipe at 63.8°C is placed in 94.7 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings.

Answer: 30.4 °C

Enthalpy of Reaction

Exercise \(\PageIndex{1}\)

Decide whether each of these reactions is exothermic or endothermic:

When two chemicals mix their temperature rises
A solid burns brightly and releases heat, light and sound
When two chemicals are mixed their temperatures drops
Two chemicals will only react if you heat them continually
Plants take in light energy for photosynthesis

Answer

exothermic
exothermic
endothermic
endothermic
endothermic

Exercise \(\PageIndex{2}\)

What quantity, in moles, of hydrogen is consumed when 185.6 kJ of energy is evolved from the combustion of a mixture of H₂(g) and O₂(g)?

H₂(g) + O₂(g) → H₂O(l); Δ_rH° = –285.8 kJ/mol-rxn

Answer

0.6494 moles

\(\frac{185.6 kJ}{285.8 kJ/mol} = 0.649405 moles\)

Exercise \(\PageIndex{3}\)

The following reaction of iron oxide with aluminum is an exothermic reaction.

Fe₂O₃(s) + 2 Al(s) → 2 Fe(s) + Al₂O₃(s)

The reaction of 7.50 g of Fe₂O₃ with excess Al(s) evolves 31.5 kJ of energy in the form of heat. Calculate the enthalpy change per mole of Fe₂O₃.

Answer: -6.71*10² kJ*mol^-1 (heat evolved, negative sign)

Exercise \(\PageIndex{4}\)

How much heat is liberated at constant pressure if 0.675 g of calcium carbonate reacts with 45.1 mL of 0.498 M hydrochloric acid?

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g); Δ_rH° = –15.2 kJ/mol-rxn

Answer: ΔH = -0.103 kJ

Enthalpy Calculations

Hess's Law And Enthalpy of Reaction

Exercise \(\PageIndex{1}\)

What is the overall chemical equation that results from the sum of the given steps?

2 C(s) + 2 H₂O(g) → 2 CO(g) + 2 H₂(g)

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

CO(g) + 3 H₂(g) → CH₄(g) + H₂O(g)

Answer: 2 C(s) + 2 H₂O(g) → CO₂(g) + CH₄(g)

Exercise \(\PageIndex{2}\)

Determine the heat of evaporation of carbon disulfide,
CS₂(l) → CS₂(g)
given the enthalpies of reaction below.

C(s) + 2 S(s) → CS₂(l) Δ_rH° = +89.4 kJ/mol-rxn

C(s) + 2 S(s) → CS₂(g) Δ_rH° = +116.7 kJ/mol-rxn

Answer

+27.3 kJ

Reverse (1) and add to (2) ΔH_rxn = -ΔH₁ + ΔH₂

(1) CS₂(l) → ~~C(s)~~ + ~~2 S(s)~~ = -89.4 kJ/mol + 116.7 kJ/mol = 27.3 kJ/mol

(2) ~~C(s)~~ + ~~2 S(s)~~ → CS₂(g)

CS₂ (l) → CS₂ (g)

Enthalpy of Formation

Exercise \(\PageIndex{3}\)

Determine the standard enthalpy of formation of Fe₂O₃(s) given the thermochemical equations below.

Fe(s) + 3 H₂O(l) → Fe(OH)₃(s) + 3/2 H₂(g) Δ_rH° = +160.9 kJ/mol-rxn

H₂(g) + 1/2 O₂(g) → H₂O(l) Δ_rH° = –285.8 kJ/mol-rxn

Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s) Δ_rH° = +288.6 kJ/mol-rxn

Answer

–824.2 kJ/mol-rxn

(Fe(s) + 3 H₂O(l) → Fe(OH)₃(s) + 3/2 H₂(g) Δ_rH° = +160.9 kJ/mol-rxn) * 2

2 Fe(s) + 6 H₂O(l) → 2 Fe(OH)₃(s) + 3 H₂(g) Δ_rH° = +321.8 kJ/mol-rxn

(H₂(g) + 1/2 O₂(g) → H₂O(l) Δ_rH° = –285.8 kJ/mol-rxn) * 3

3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l) Δ_rH° = –857.4kJ/mol-rxn

(Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s) Δ_rH° = +288.6 kJ/mol-rxn) Reverse Equation or ( * -1)

2 Fe(OH)₃(s) → Fe₂O₃(s) + 3 H₂O(l) Δ_rH° = -288.6 kJ/mol-rxn

Add the equations and Δ_rH°.

2 Fe(s) + ~~6 H₂O(l)~~ → ~~2 Fe(OH)₃(s)~~ + ~~3 H₂(g)~~ Δ_rH° = +321.8 kJ/mol-rxn

~~3 H₂(g)~~ + 3/2 O₂(g) → ~~3 H₂O(l)~~ Δ_rH° = –857.4kJ/mol-rxn

~~2 Fe(OH)₃(s)~~ → Fe₂O₃(s) + ~~3 H₂O(l)~~ Δ_rH° = -288.6 kJ/mol-rxn

2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH_f = -824.2 kJ/mol

Exercise \(\PageIndex{4}\)

Determine the standard enthalpy of formation of calcium carbonate from the thermochemical equations given below.

Ca(OH)2(s) → CaO(s) + H2O(l) Δ_rH° = 65.2 kJ/mol-rxn

Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l) Δ_rH° = −113.8 kJ/mol-rxn

C(s) + O2(g) → CO2(g) Δ_rH° = −393.5 kJ/mol-rxn

2 Ca(s) + O2(g) → 2 CaO(s) Δ_rH° = −1270.2 kJ/mol-rxn

Answer: −1207.6 kJ/mol-rxn

Enthalpy of Reaction from Standard Enthalpies of Formation

Exercise \(\PageIndex{5}\)

Calculate ΔrH° for the combustion of ammonia, using standard molar enthalpies of formation:

4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(l)

Molecule ΔfH° (kJ/mol-rxn)

NH₃(g) –45.9

NO₂(g) +33.1

H₂O(l) –285.8

Answer

–1398.8 kJ/mol-rxn

ΔH°rxn = Σ ΔH°_f(products) - Σ ΔH°_f(reactants)

= [(4 * ΔH°_f(NO₂)) + (6 * ΔH°_f(H₂O))] - [(4 * ΔH°_f(NH₃)) + (7 * ΔH°_f(O₂))]

= [(4 * 33.1) + (6 * -285.8)] - [(4 * -45.9) + (7 * 0)]

= -1398.8 kJ/mol-rxn

Exercise \(\PageIndex{6}\)

What is ΔrH° for the following phase change?

LiF(s) → LiF(l)

Substance ΔH°f (kJ/mol-rxn)

LiF(s) –616.93

LiF(l) –598.65

Answer: 18.28 kJ/mol-rxn