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Chemistry LibreTexts

Thermochemistry Solutions

  • Page ID
    42458
  • S1

    \[ q=m\, C_p \, \Delta T\]

    1. raise the temperature of 10 L of water from 23 ˚C to 30 ˚C;
    • \(\Delta T= (30˚C - 23˚C) = 7˚\)

    \[q = (10\; \cancel{L} ) \left(\dfrac{1000\; \cancel{cm^3}}{1\; \cancel{L}} \right) \left(\dfrac{1\; g}{1\; \cancel{cm^3}}\right) (4.18\; \cancel{J}/(g˚C)) \left( \dfrac{ 1 \;kJ}{1000\; \cancel{J}} \right) \Delta T\]

    \[= +1,254 kJ\]

    1. associated with a 33 ˚C decrease in temperature in a 6.80 kg aluminum bar (specific heat of aluminum = 0.903 J/(g ˚C)).
    • \(\Delta T= 33˚C\)

    \[q = (6.8\;\cancel{kg}) \left( \dfrac{1000\;g}{1\;\cancel{kg}} \right) (0.903J/(g˚C)) (-33˚C) \left( \dfrac{1\;kJ}{1000\;J} \right) = -203\; J\]

    Video Solution

    S2

    a) For this we will use the equation from the previous question, but rearrange to find the final temperature.

    \[ 13.8\; g \times 4.18\; J/(g°C) \times (T_f - 21.7\; °C) = 679\; J\]

    \[(T_f-21.7\;°C)= \dfrac{679\;J}{(13.8g \times 4.18\; J/(g°C)}=11.8\; °C\]

    \(T_f=33.5°C\)

    b) For this we will use the previous equation, but first we must convert kilograms to grams, and kilocalories to joules.

    Grams of aluminum = \((2.43\;kg ) \left( \dfrac{1000\;g}{1\;kg}\right) =2,430\;g\)

    Joules of heat = \(2.04\; kcal×\left(\dfrac{1000\; cal}{1\;kcal} \right) \left(\dfrac{4.18\;J}{1\;cal} \right) =8,527\; J\)

    \[2430g \times 0.902 J/(g°C) \times (T_f-69.5°C)= 8527J\]

    \((T_f - 69.5°C)=\dfrac{8,527\;J}{2430\; g \times 0.902J/(g°C)}=3.89 °C\)

    \(T_f=73.4°C\)

    S3

    Solve for the q of water:

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

    conservation of energy (assume work is negligible in this system since no gases are generated)

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

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

    S4

    a)

    \[ C_p=\dfrac{q}{m∆T}\]

    \[C_p(zinc) = \dfrac{(-1.02\; \cancel{kJ}) \left(\dfrac{1000\; J}{1\; \cancel{kJ}}\right)}{(119.5\; g)(43.5 °C – 65.7 °C)} = 0.384\; \frac{J}{g °C}\]

    b)

    \[C_p = \dfrac{q}{m∆T}\]

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

    S5

    \[w = -P_{ext} \Delta V\]

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

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

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

    S6

    \[ \Delta U=q+w \]

    \[∆U = (-239 \;J)+(489\; J)=250\; J\]

    S29

    First identify how much heat is available

    \[ q= mc_{sp}\Delta T\]

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

    and equate to the heat of fusion equation

    \[q_{heat}= q_{fus} = m \Delta H_{fus}\]

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

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

    \[m= \dfrac{11.9 \times 10^3\; J}{333.6\; J/g} = 35.7\; g\; H_2O\]

    S33

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

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

    S35

    \[C_v = \dfrac{q}{\Delta T}\]

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

    S47

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

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

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

    Video Solution

    S67

    It is just the reverse since the products are all the standard states of the elements involved in \(\Delta_f^o\) of \HCl_{(g)}\). So \(\Delta H_{rx}°\) is -52.4 kJ/mol.

    S69

    \[ -1 \times( 2O_3(g) \rightleftharpoons 3O_2(g)) \tag{-1 x ∆H1= -1(-420 kJ)}\]

    \[0 \times (O_2(g) \rightleftharpoons 2O(g)) \tag{0 x ∆H2= 0 kJ}\]

    \[2 \times (NO(g) + O_3(g) \rightleftharpoons NO_2(g) + O_2(g)) \tag{ 2 x ∆H3= 2(-137 kJ)}\]

    \[\Delta{H}=\sum_i \Delta{H_i} = -1(-420\; kJ) + 0 + 2(-137\; kJ) = 146\; kJ\]

    S73

    \[2 \times (N_2 (g) + O_2 (g) \rightarrow 2 NO (g)) \tag{2 x ΔH = 2(-210.2 kJ)}\]

    \[ -2 \times (N_2 (g) + 3 H_2 (g) \rightarrow 2 NH_3 (g)) \tag{-2 x ΔH = -2(-80.8 kJ)}\]

    \[ 3 \times (2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g)) \tag{3 x ΔH = 3(-376.2 kJ)}\]

    \[\Delta{H}=\sum_i \Delta{H_i} = 2 (-210.2\; kJ) + -2 (-80.8\; kJ) + 3 (-376.2\; kJ) = -1387.4\; kJ\]

    Video Solution

    S75

    \[ \dfrac{-1}{2} \times [N_2(g) + 3 H_2 (g) \rightarrow 2 NH_3 (g)] \tag{-1/2 x ΔH = -1/2(-82.6 kJ)}\]

    \[ -1 \times [C (s) + 2 H_2 (g) \rightarrow CH_4 (g)] \tag{-1 x ΔH = -1(-90.3 kJ)}\]

    \[ \dfrac{1}{2} \times [ H_2 (g) + 2 C (s) + N_2 (g) \rightarrow 2 HCN (g)] \tag{1/2 x ΔH = 1/2 (293.3 kJ)}\]

    \[\Delta{H}=\sum_i \Delta{H_i} = -\dfrac{-1}{2}(-82.6 \; kJ) + -1(-90.3\; kJ) + \dfrac{1}{2} (293.3\; kJ) = 278.25 \; kJ\]

    S81

    \[∆H˚= 1 \times ∆H_f˚[CdSO_4] - 1 \times ∆H_f˚[Cd^{2+}] - 1 \times ∆H_f˚[SO_4^{2-}]\]

    \[= (-1473\; kJ - (-75.9 \; kJ - (-909.3 \; kJ \]

    \[= -487.8\;kJ\]