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Difficulty: Very hardStandard Enthalpy Changes and Hess's Law

The standard enthalpy of formation (ΔHf∘\Delta H_f^\circ) of carbon dioxide gas (CO2(g)\text{CO}_2(g)) is numerically identical to the standard enthalpy of combustion (ΔHc∘\Delta H_c^\circ) of diamond under standard thermochemical conditions (298 K298\text{ K} and 1 atm1\text{ atm}).

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Answer

False
The standard enthalpy of formation (ΔHf∘\Delta H_f^\circ) of a compound is defined as the enthalpy change when one mole of the substance is formed from its constituent elements in their reference standard states. For carbon at 298 K298\text{ K} and 1 atm1\text{ atm}, the reference standard state is graphite. Consequently, ΔHf∘[CO2(g)]\Delta H_f^\circ[\text{CO}_2(g)] represents the reaction of graphite with oxygen. In contrast, the combustion of diamond involves C(s,diamond)\text{C}(s, \text{diamond}). Because conversion of graphite to diamond is endothermic (ΔHf∘[diamond]≈+1.9 kJ mol−1\Delta H_f^\circ[\text{diamond}] \approx +1.9\text{ kJ mol}^{-1}), the combustion of diamond produces more heat than the formation of CO2(g)\text{CO}_2(g) from graphite. Thus, the two values are not numerically identical.

Step-by-Step Solution

1
Define the reaction equation for the standard enthalpy of formation (ΔHf∘\Delta H_f^\circ) of CO2(g)\text{CO}_2(g).
C(s,graphite)+O2(g)→CO2(g)ΔH1∘=ΔHf∘[CO2(g)]\text{C}(s, \text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1^\circ = \Delta H_f^\circ[\text{CO}_2(g)]
By international convention (IUPAC), the standard enthalpy of formation requires elements to be in their most stable allotropic form at 298 K298\text{ K} and 1 atm1\text{ atm}, which is graphite for carbon.
2
Define the reaction equation for the standard enthalpy of combustion (ΔHc∘\Delta H_c^\circ) of diamond.
C(s,diamond)+O2(g)→CO2(g)ΔH2∘=ΔHc∘[diamond]\text{C}(s, \text{diamond}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2^\circ = \Delta H_c^\circ[\text{diamond}]
The standard enthalpy of combustion of diamond measures the enthalpy change when 1 mole of diamond completely reacts with oxygen.
3
Apply Hess's law to relate ΔH1∘\Delta H_1^\circ and ΔH2∘\Delta H_2^\circ.
ΔHc∘[diamond]=ΔHf∘[CO2(g)]−ΔHf∘[diamond]\Delta H_c^\circ[\text{diamond}] = \Delta H_f^\circ[\text{CO}_2(g)] - \Delta H_f^\circ[\text{diamond}]
Since diamond is less stable than graphite (ΔHf∘[diamond]≈+1.9 kJ mol−1\Delta H_f^\circ[\text{diamond}] \approx +1.9\text{ kJ mol}^{-1}), ΔH2∘≠ΔH1∘\Delta H_2^\circ \neq \Delta H_1^\circ.

Key Concept

Standard State Conventions and Allotropy in Hess's Law
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Topics

Standard Enthalpy Changes and Hess's Law