Question
Consider the following reaction :
$2A (g) + B (g) \rightarrow 2D (g)$
$\Delta U^\ominus = -10 \text{ kJ mol}^{-1}$ and $\Delta S^\ominus = -44 \text{ J K}^{-1}$ at 298 K.
Identify the correct option with $\Delta G^\ominus$ for the reaction and spontaneity of the reaction at 298 K. (Given : $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$)
This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$+0.63568 \text{ kJ mol}^{-1}$, non-spontaneous
Calculating Gibbs Free Energy and Reaction Spontaneity
This section details the calculation of the standard Gibbs free energy change ($\Delta G^\ominus$) and determination of the reaction's spontaneity.
Reaction Data Summary
- Reaction: $2A (g) + B (g) \rightarrow 2D (g)$
- Standard Internal Energy Change: $\Delta U^\ominus = -10 \text{ kJ mol}^{-1}$
- Standard Entropy Change: $\Delta S^\ominus = -44 \text{ J K}^{-1} \text{ mol}^{-1}$
- Temperature: $T = 298 \text{ K}$
- Gas Constant: $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$
Calculating Enthalpy Change ($\Delta H^\ominus$)
First, find the change in moles of gas ($\Delta n_g$):
- $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
- $\Delta n_g = 2 - (2 + 1) = -1$
Next, calculate $\Delta H^\ominus$ using the relation $\Delta H^\ominus = \Delta U^\ominus + \Delta n_g RT$. Convert $\Delta U^\ominus$ to Joules for calculation:
- $\Delta U^\ominus = -10 \text{ kJ mol}^{-1} = -10000 \text{ J mol}^{-1}$
- $\Delta H^\ominus = -10000 \text{ J mol}^{-1} + (-1) \times (8.31 \text{ J mol}^{-1} \text{K}^{-1}) \times (298 \text{ K})$
- $\Delta H^\ominus = -10000 \text{ J mol}^{-1} - 2476.58 \text{ J mol}^{-1}$
- $\Delta H^\ominus = -12476.58 \text{ J mol}^{-1}$
Calculating Gibbs Free Energy Change ($\Delta G^\ominus$)
Use the Gibbs Free Energy equation: $\Delta G^\ominus = \Delta H^\ominus - T \Delta S^\ominus$. Use values in Joules for calculation consistency:
- $\Delta H^\ominus = -12476.58 \text{ J mol}^{-1}$
- $\Delta G^\ominus = (-12476.58 \text{ J mol}^{-1}) - (298 \text{ K} \times -44 \text{ J K}^{-1} \text{ mol}^{-1})$
- $\Delta G^\ominus = -12476.58 \text{ J mol}^{-1} + 13112 \text{ J mol}^{-1}$
- $\Delta G^\ominus = 635.42 \text{ J mol}^{-1}$
Convert the result back to kilojoules per mole:
- $\Delta G^\ominus = \frac{635.42}{1000} \text{ kJ mol}^{-1}$
- $\Delta G^\ominus \approx 0.635 \text{ kJ mol}^{-1}$
Determining Reaction Spontaneity
The spontaneity of a reaction depends on the sign of $\Delta G^\ominus$:
- If $\Delta G^\ominus > 0$, the reaction is non-spontaneous.
- If $\Delta G^\ominus < 0$, the reaction is spontaneous.
Since the calculated $\Delta G^\ominus$ is positive ($+0.635 \text{ kJ mol}^{-1}$), the reaction is non-spontaneous at 298 K.
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