Question

What is the value of \(\Delta S_{\text{total}}\) for the following reaction at 300 K?
\[ \text{Fe}_2\text{O}_3(s) + 3 \text{CO}(g) \rightarrow 2 \text{Fe}(s) + 3 \text{CO}_2(g) \quad \Delta H = -25 \, \text{kJ}, \, \Delta S = 15 \, \text{J K}^{-1} \]

• A. 68.2 \, \text{JK}^{-1}
• B. 98.3 \, \text{JK}^{-1}
• C. 8.32 \, \text{JK}^{-1}
• D. -10.0 \, \text{JK}^{-1}

Hint:

Remember to use the units consistently when calculating entropy changes. Ensure temperature is in Kelvin.

Answer 1

The Correct Option is B

Step 1: Understanding the relation between enthalpy, entropy, and Gibbs free energy.
The total change in entropy (\(\Delta S_{\text{total}}\)) is given by the equation: \[ \Delta S_{\text{total}} = \frac{\Delta H}{T} + \Delta S \] Where: - \(\Delta H = -25 \, \text{kJ} = -25000 \, \text{J}\) - \(T = 300 \, \text{K}\) - \(\Delta S = 15 \, \text{J K}^{-1}\) Step 2: Calculating \(\Delta S_{\text{total}}\).
\[ \Delta S_{\text{total}} = \frac{-25000}{300} + 15 = -83.33 + 15 = 98.3 \, \text{J K}^{-1} \] Step 3: Conclusion.
The correct answer is (B) 98.3 JK^{-1}.