Question

Heat of formation of CO gas at 300K is -110KJ at constant pressure. Its heat of formation at the same temperature but at the constant volume is: (Given: R = 8.344 $J\quad K^{ -1 }\quad mol^{ -1 }$)
A- -108.753 KJ
B- -110 KJ
C- -111.247 KJ
D- -112.249 KJ

Answer 1

Hint: Try to figure out what will be the effect on the reaction when it is performed under constant volume conditions. At constant volume, the volume will remain constant. How does the enthalpy of reaction be affected by changing conditions?

Complete step by step answer:
We know that heat of formation is the change of enthalpy of reaction when one mole of given chemical species is formed from its elements. When a given reaction is carried at constant pressure, the pressure is kept constant throughout the reaction. Whereas in constant volume condition, the volume is kept constant throughout the reaction.
But the heat of formation of reaction is not the same for all reactions under these 2 conditions. This is because of the number of gaseous moles changes in the reaction. If the volume is not constant then gas will expand.
The reaction will be:

$C_{ \left( S \right) }\quad +\quad \frac { 1 }{ 2 } { O }_{ 2 }\quad →\quad CO_{ \left( g \right) }$

Change in number of gaseous moles $\Delta n_{ g }\quad =\quad 1-\frac { 1 }{ 2 } \quad =\quad \frac { 1 }{ 2 } $
\[\Delta H\quad =\quad \Delta U\quad +\quad \Delta n_{ g }RT\]
$\Delta H$ is enthalpy change, $\Delta U$ is the internal energy change
\[\Delta H\quad -\quad \Delta U\quad =\quad \Delta n_{ g }RT\]

Given $\Delta H$ is the heat of formation at constant pressure, $\Delta U$ is the heat of formation at constant volume.

\[\Delta U\quad =\quad \Delta H\quad -\quad \Delta n_{ g }RT\]
\[\Delta n_{ g }RT\quad =\quad \frac { 1 }{ 2 } \times 8.344\times 300\quad =\quad 1251.6\quad J\]
\[\Delta U\quad =\quad -110KJ\quad -\quad 1251.6\quad J\quad =\quad -110\quad -\quad 1.2516\quad =\quad -111.2516KJ\]
Hence, the correct answer is option C.

Note:
Heat of formation involves reactants that are in their most stable form. Internal energy change will be equal to the heat of formation at constant volume because there will be no work done in expanding gas so enthalpy at constant volume is the same as internal energy change under constant pressure condition.