Question

For a reaction \[ 2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g), \] \(\Delta_rG^\circ=-128\,\text{kJ}\) at \(300\,\text{K}\). If \(\Delta_rS^\circ\) of the reaction is \(-40\,\text{J K}^{-1}\), calculate \(\Delta_rU\) of the reaction.

• A. \(-137.5\,\text{kJ}\)
• B. \(-128\,\text{kJ}\)
• C. \(-140\,\text{kJ}\)
• D. \(126.2\,\text{kJ}\)

Hint:

Use \[ \Delta G=\Delta H-T\Delta S \] first to calculate \(\Delta H\), then use \[ \Delta H=\Delta U+\Delta n_gRT \] to find \(\Delta U\).

Answer 1

The Correct Option is A

Step 1: Use the relation between Gibbs energy, enthalpy, and entropy.
We know that \[ \Delta G=\Delta H-T\Delta S \] Therefore, \[ \Delta H=\Delta G+T\Delta S \]

Step 2: Substitute the given values.
Given, \[ \Delta G=-128\,\text{kJ} \] \[ T=300\,\text{K} \] \[ \Delta S=-40\,\text{J K}^{-1} \] Convert entropy into \(\text{kJ K}^{-1}\): \[ -40\,\text{J K}^{-1}=-0.040\,\text{kJ K}^{-1} \] Thus, \[ T\Delta S=300\times(-0.040) \] \[ T\Delta S=-12\,\text{kJ} \] So, \[ \Delta H=-128-12 \] \[ \Delta H=-140\,\text{kJ} \]

Step 3: Use relation between enthalpy and internal energy.
For gaseous reactions, \[ \Delta H=\Delta U+\Delta n_gRT \] Therefore, \[ \Delta U=\Delta H-\Delta n_gRT \]

Step 4: Find \(\Delta n_g\).
For the reaction, \[ 2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g) \] Moles of gaseous products: \[ 2 \] Moles of gaseous reactants: \[ 2+1=3 \] Hence, \[ \Delta n_g=2-3=-1 \]

Step 5: Calculate \(\Delta U\).
\[ \Delta U=\Delta H-\Delta n_gRT \] \[ \Delta U=-140-(-1)(8.314\times300)\,\text{J} \] \[ \Delta U=-140\,\text{kJ}+2494.2\,\text{J} \] \[ \Delta U=-140\,\text{kJ}+2.494\,\text{kJ} \] \[ \Delta U=-137.506\,\text{kJ} \] \[ \Delta U\approx -137.5\,\text{kJ} \]

Step 6: Final conclusion.
Hence, \[ \boxed{-137.5\,\text{kJ}} \]