Solid carbon in the form of graphite reacts with oxygen gas to produce carbon dioxide…

CHEMISTRY · Multi-part · A Level · Chemistry · Energetics · Energetics

Problem

Solid carbon in the form of graphite reacts with oxygen gas to produce carbon dioxide gas according to the overall chemical equation below. C(graphite)+O2(g)→CO2(g)\text{C(graphite)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} Two independent data sets are available: • Data Set 1 (Two-step combustion sequence) Step I: C(graphite)+12 O2(g)→CO(g)\text{C(graphite)} + \frac{1}{2}\,\text{O}_2\text{(g)} \rightarrow \text{CO(g)}, ΔH=−110.5 kJ mol−1\Delta H = -110.5\,\text{kJ mol}^{-1} Step II: CO(g)+12 O2(g)→CO2(g)\text{CO(g)} + \frac{1}{2}\,\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}, ΔH=−283.0 kJ mol−1\Delta H = -283.0\,\text{kJ mol}^{-1} • Data Set 2 (Standard formation enthalpies, 298 K298\,\text{K}, 1 bar1\,\text{bar}) ΔHf∘[C(graphite)]=0 kJ mol−1\Delta H^\circ_f[\text{C(graphite)}] = 0\,\text{kJ mol}^{-1} (reference) ΔHf∘[O2(g)]=0 kJ mol−1\Delta H^\circ_f[\text{O}_2\text{(g)}] = 0\,\text{kJ mol}^{-1} (reference) ΔHf∘[CO2(g)]=−393.5 kJ mol−1\Delta H^\circ_f[\text{CO}_2\text{(g)}] = -393.5\,\text{kJ mol}^{-1} Answer the following questions. (1) Part 1: (a) Calculate the molar enthalpy change for the overall reaction using Data Set 1. (2) Part 2: (b) Determine the molar enthalpy change for the same overall reaction using Data Set 2. (3) Part 3: (c) Compare the numerical results obtained in parts (a) and (b). State whether they agree within the precision of the data and briefly discuss what this reveals about the thermodynamic quantity ΔH\Delta H. (4) Part 4: (d) Suggest one experimental factor that might cause a small difference between the two numerical values if higher-precision calorimetry were performed, and explain your reasoning.

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