Question

For the reversible reaction at 300 K, \( \Delta H^o = 28.4 \, \text{KJ/mole} \) and equilibrium constant \( K = 1.8 \times 10^{-7} \), then calculate the magnitude of \( \Delta S^o \) in Joule/K-mole.
[Given: log 2 = 0.3, \( \ln 10 = 2.3 \), \( R = 8.314 \, \text{J/K.mole} \), log 3 = 0.47]

Answer 1

Correct Answer: 34

Step 1: Understanding the Question:
We need to calculate the standard entropy change (\(\Delta S^\circ\)) using the relationship between enthalpy, temperature, and the equilibrium constant.
Step 2: Key Formula or Approach:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -2.303 RT \log K \]
\[ \Delta S^\circ = \frac{\Delta H^\circ + 2.303 RT \log K}{T} \]
Step 3: Detailed Explanation:
Given: \(T = 300 \text{ K}, \Delta H^\circ = 28400 \text{ J/mol}, K = 1.8 \times 10^{-7}\).
Calculate \(\log K\):
\[ \log K = \log(1.8 \times 10^{-7}) = \log(1.8) - 7 \]
\[ \log(1.8) = \log(2 \times 9 / 10) = \log 2 + 2\log 3 - \log 10 \]
\[ \log(1.8) = 0.3 + 2(0.47) - 1 = 0.3 + 0.94 - 1 = 0.24 \]
\[ \log K = 0.24 - 7 = -6.76 \]
Now substitute into the entropy equation:
\[ \Delta S^\circ = \frac{28400 + (2.303 \times 8.314 \times 300 \times (-6.76))}{300} \]
\[ \Delta S^\circ = \frac{28400 - 38741.2}{300} = \frac{-10341.2}{300} \approx -34.47 \text{ J/K}\cdot\text{mol} \]
The magnitude of \(\Delta S^\circ\) is approximately 34 J/K-mol.
Step 4: Final Answer:
The magnitude of \(\Delta S^\circ\) is 34 J/K-mole.