AIPMT PRELIMS 2004 Chemistry Gibbs Free Energy MCQ Question

To find the standard Gibbs energy change ($\Delta G^\circ$) for the oxidation of ammonia at 298 K, we can use the following thermodynamic relationship:

$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$

Where:

• $\Delta G^\circ$ is the standard Gibbs energy change,
• $\Delta H^\circ$ is the standard enthalpy change,
• $T$ is the temperature in Kelvin,
• $\Delta S^\circ$ is the standard entropy change.

Step-by-Step Calculation

1. Identify values from the question:

   • Standard enthalpy change, $\Delta H^\circ = -382.64 \, \text{kJ mol}^{-1}$
   • Standard entropy change, $\Delta S^\circ = -145.6 \, \text{JK}^{-1} \text{mol}^{-1}$
   • Temperature, $T = 298 \, \text{K}$

2. Convert units if necessary:

   • Convert $\Delta H^\circ$ to Joules for consistency with $\Delta S^\circ$: $\Delta H^\circ = -382.64 \, \text{kJ mol}^{-1} \times 1000 \, \text{J kJ}^{-1} = -382640 \, \text{J mol}^{-1}$

3. Substitute values into the Gibbs equation: Using the equation from step 1, we can substitute the values: $\Delta G^\circ = -382640 \, \text{J mol}^{-1} - (298 \, \text{K} \times -145.6 \, \text{JK}^{-1} \text{mol}^{-1})$

4. Calculate the $T \Delta S^\circ$ term: $T \Delta S^\circ = 298 \, \text{K} \times -145.6 \, \text{JK}^{-1} \text{mol}^{-1} = -43329.6 \, \text{J mol}^{-1}$

5. Combine the results: Now substitute this back into the Gibbs energy equation: $\Delta G^\circ = -382640 \, \text{J mol}^{-1} - (-43329.6 \, \text{J mol}^{-1})$ $\Delta G^\circ = -382640 \, \text{J mol}^{-1} + 43329.6 \, \text{J mol}^{-1}$ $\Delta G^\circ = -339310.4 \, \text{J mol}^{-1}$

6. Convert back to kJ: $\Delta G^\circ = -339.31 \, \text{kJ mol}^{-1}$

Conclusion

The standard Gibbs energy change for the oxidation of ammonia at 298 K is approximately $-339.3 \, \text{kJ mol}^{-1}$. Thus, the correct answer is Option A: −339.3 kJ mol⁻¹.

Clarification of Incorrect Options

• Option B: −439.3 kJ mol⁻¹: This value does not match the calculated Gibbs energy change and likely results from an incorrect calculation or misunderstanding of the enthalpy and entropy contributions.
• Option C: −523.2 kJ mol⁻¹: This is also inconsistent with our calculation. It suggests an overestimation of the Gibbs energy change.
• Option D: −221.1 kJ mol⁻¹: This value is too low and indicates a failure to account for the significant negative enthalpy change properly.

In summary, by applying the Gibbs energy equation correctly and performing the necessary calculations, we arrive at the conclusion that the standard Gibbs energy change for the reaction is $-339.3 \, \text{kJ mol}^{-1}$, confirming Option A as the correct answer.