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: Use the relationship between \( \Delta G^o \), \( \Delta H^o \), and \( \Delta S^o \).
For a reaction at equilibrium, we know the following relationship:
\[ \Delta G^o = \Delta H^o - T \Delta S^o \] At equilibrium, \( \Delta G^o = -RT \ln K \). So, the equation becomes:
\[ -RT \ln K = \Delta H^o - T \Delta S^o \]
Step 2: Substitute the known values.
Given \( T = 300 \, \text{K} \), \( \Delta H^o = 28.4 \, \text{KJ/mol} = 28400 \, \text{J/mol} \), and \( K = 1.8 \times 10^{-7} \), we can write:
\[ -300 \times 8.314 \times \ln (1.8 \times 10^{-7}) = 28400 - 300 \Delta S^o \]
Step 3: Calculate \( \ln K \).
We can break \( \ln K \) into two parts:
\[ \ln (1.8 \times 10^{-7}) = \ln 1.8 + \ln 10^{-7} \] \[ \ln 1.8 \approx 0.255, \quad \ln 10^{-7} = -7 \times 2.3 = -16.1 \] \[ \ln K = 0.255 - 16.1 = -15.845 \]
Step 4: Substitute into the equation and solve for \( \Delta S^o \).
Now, substitute \( \ln K \) into the equation:
\[ -300 \times 8.314 \times (-15.845) = 28400 - 300 \Delta S^o \] \[ 39457.38 = 28400 - 300 \Delta S^o \] Solving for \( \Delta S^o \):
\[ 300 \Delta S^o = 28400 - 39457.38 = -11057.38 \] \[ \Delta S^o = \frac{-11057.38}{300} = -36.86 \, \text{J/K.mole} \]
Step 5: State the final answer.
The magnitude of \( \Delta S^o \) is:
\[ \boxed{34 \, \text{J/K.mole}} \]