Use the thermochemical equations shown below to determine the enthalpy of the final reaction

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Use the thermochemical equations shown below to determine the enthalpy of the final reaction: (1) CO2 (g) ? C (s) + O2 (g) ?? = 551 kJ (2) H2(g) + 1/2 O2 (g) ? H2O (l) ?? = -400 kJ (3) CH4 (g) + 2O2 (g) ? CO2 (g) + 2H2O (l) ?? = -1247 kJ CH4 (g) ? C (s) + 2H2 (g) ?? =? 2) The heat released on combustion of benzoic acid (C6H5COOH) is 26.38 kJ/g. When 1.000 g of benzoic acid was combusted in a bomb calorimeter, the temperature of the calorimeter rose from 24.32 °C to 27.05 °C. What is the heat capacity, C, of the calorimeter (in kJ/°C)? NOTE: Marks are deducted if units and significant figures are ignored.

The Correct Answer and Explanation is:

Problem 1: Enthalpy of the Final Reaction

We are given the following thermochemical equations:

  1. CO2(g)→C(s)+O2(g)ΔH1=551 kJCO_2 (g) \rightarrow C (s) + O_2 (g) \quad \Delta H_1 = 551 \, \text{kJ}CO2​(g)→C(s)+O2​(g)ΔH1​=551kJ
  2. H2(g)+12O2(g)→H2O(l)ΔH2=−400 kJH_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O (l) \quad \Delta H_2 = -400 \, \text{kJ}H2​(g)+21​O2​(g)→H2​O(l)ΔH2​=−400kJ
  3. CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH3=−1247 kJCH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2O (l) \quad \Delta H_3 = -1247 \, \text{kJ}CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l)ΔH3​=−1247kJ

We need to determine the enthalpy of the reaction:
CH4(g)→C(s)+2H2(g)CH_4 (g) \rightarrow C (s) + 2H_2 (g)CH4​(g)→C(s)+2H2​(g).

Solution Strategy:

We can manipulate the given reactions to derive the desired reaction. Here’s the breakdown of steps:

  1. Start with the combustion of methane (reaction 3) and reverse it so that methane (CH4) is on the left: CO2(g)+2H2O(l)→CH4(g)+2O2(g)CO_2 (g) + 2H_2O (l) \rightarrow CH_4 (g) + 2O_2 (g)CO2​(g)+2H2​O(l)→CH4​(g)+2O2​(g) The enthalpy change for this reversed reaction is +1247 kJ+1247 \, \text{kJ}+1247kJ.
  2. Add reaction (1) to remove CO2(g)CO_2 (g)CO2​(g) from the equation: CO2(g)→C(s)+O2(g)ΔH1=551 kJCO_2 (g) \rightarrow C (s) + O_2 (g) \quad \Delta H_1 = 551 \, \text{kJ}CO2​(g)→C(s)+O2​(g)ΔH1​=551kJ This results in the cancellation of CO2(g)CO_2 (g)CO2​(g).
  3. Add reaction (2) to remove H2O(l)H_2O (l)H2​O(l) and form H2(g)H_2 (g)H2​(g): H2(g)+12O2(g)→H2O(l)ΔH2=−400 kJH_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O (l) \quad \Delta H_2 = -400 \, \text{kJ}H2​(g)+21​O2​(g)→H2​O(l)ΔH2​=−400kJ

By adding all the reactions, we can now derive the following: CH4(g)→C(s)+2H2(g)ΔH=1247+551−400=1398 kJCH_4 (g) \rightarrow C (s) + 2H_2 (g) \quad \Delta H = 1247 + 551 – 400 = 1398 \, \text{kJ}CH4​(g)→C(s)+2H2​(g)ΔH=1247+551−400=1398kJ

Thus, the enthalpy of the reaction CH4(g)→C(s)+2H2(g)CH_4 (g) \rightarrow C (s) + 2H_2 (g)CH4​(g)→C(s)+2H2​(g) is ΔH=1398 kJ\Delta H = 1398 \, \text{kJ}ΔH=1398kJ.

Problem 2: Heat Capacity of the Calorimeter

We are told the heat released by the combustion of benzoic acid is 26.38 kJ/g. When 1.000 g of benzoic acid is combusted, the temperature of the calorimeter rises from 24.32 °C to 27.05 °C.

Solution Strategy:

  1. Determine the heat released during combustion:
    The heat released by the combustion of 1.000 g of benzoic acid is: Q=26.38 kJ/g×1.000 g=26.38 kJQ = 26.38 \, \text{kJ/g} \times 1.000 \, \text{g} = 26.38 \, \text{kJ}Q=26.38kJ/g×1.000g=26.38kJ
  2. Calculate the temperature change:
    The temperature change of the calorimeter is: ΔT=27.05 °C−24.32 °C=2.73 °C\Delta T = 27.05 \, \text{°C} – 24.32 \, \text{°C} = 2.73 \, \text{°C}ΔT=27.05°C−24.32°C=2.73°C
  3. Use the heat capacity formula:
    The formula for the heat absorbed by the calorimeter is: Q=C×ΔTQ = C \times \Delta TQ=C×ΔT Where:
    • Q=26.38 kJQ = 26.38 \, \text{kJ}Q=26.38kJ
    • ΔT=2.73 °C\Delta T = 2.73 \, \text{°C}ΔT=2.73°C
    • CCC is the heat capacity of the calorimeter.
    Rearranging the formula to solve for CCC: C=QΔT=26.38 kJ2.73 °C=9.66 kJ/°CC = \frac{Q}{\Delta T} = \frac{26.38 \, \text{kJ}}{2.73 \, \text{°C}} = 9.66 \, \text{kJ/°C}C=ΔTQ​=2.73°C26.38kJ​=9.66kJ/°C

Thus, the heat capacity of the calorimeter is C=9.66 kJ/°CC = 9.66 \, \text{kJ/°C}C=9.66kJ/°C.

Final Answers:

  1. The enthalpy of the final reaction CH4(g)→C(s)+2H2(g)CH_4 (g) \rightarrow C (s) + 2H_2 (g)CH4​(g)→C(s)+2H2​(g) is 1398 kJ.
  2. The heat capacity of the calorimeter is 9.66 kJ/°C.

Both results are rounded to the correct number of significant figures based on the given data.

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