Question

For the following reaction: \[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g) \] Given that \(\Delta H^\circ = -29.8 \, \text{kJ}\) \text{and} \(\Delta S^\circ = 15 \, \text{J/K}\), what is the value of \(\Delta S_{\text{total}}\) at 298 K?

• A. 29.8 J
• B. 100.0 J
• C. 298.0 J
• D. 115.0 J

Hint:

To find the total entropy change (\(\Delta S_{\text{total}}\)), use the equation \(\Delta S_{\text{total}} = \frac{\Delta H}{T} - \Delta S\), and ensure all units are consistent.

Answer 1

The Correct Option is D

Step 1: Using the Gibbs Free Energy equation.
The total change in entropy (\(\Delta S_{\text{total}}\)) at a given temperature \(T\) can be calculated using the following equation: \[ \Delta G = \Delta H - T \Delta S \] For equilibrium at standard conditions, \(\Delta G = -T \Delta S_{\text{total}}\), so: \[ \Delta S_{\text{total}} = \frac{\Delta H}{T} - \Delta S \]
Step 2: Substituting values.
Given that \(\Delta H^\circ = -29.8 \, \text{kJ}\) and \(\Delta S^\circ = 15 \, \text{J/K}\), and \(T = 298 \, \text{K}\), we first convert \(\Delta H^\circ\) to J: \[ \Delta H^\circ = -29.8 \, \text{kJ} = -29800 \, \text{J} \] Now calculate \(\Delta S_{\text{total}}\): \[ \Delta S_{\text{total}} = \frac{-29800 \, \text{J}}{298 \, \text{K}} + 15 \, \text{J/K} = -100 \, \text{J/K} + 15 \, \text{J/K} = 115.0 \, \text{J/K} \]
Step 3: Conclusion.
The total change in entropy is 115.0 J/K, corresponding to option (D).