The equilibrium constant for, `2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g) "is" 0.0118 "at" 1300 K` while the heat of dissociation is `597.4KJ`. The standard equilibrium constant of the reaction at `1200K` is:

Options

A. `1.180xx10^(-4)`
B. `11.80`
C. `118.0`
D. cannot be calculated from given data

Explanation

To solve the problem, we will use the van 't Hoff equation to relate the equilibrium constants at two different temperatures. The equation is given by:

\[ \log \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{2.303 R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

### Step-by-Step Solution:

1. **Identify the Known Values:**
- \( K_1 = 0.0118 \) (equilibrium constant at \( T_1 = 1300 \, K \))
- \( \Delta H^\circ = 597.4 \, \text{kJ} = 597400 \, \text{J} \) (heat of dissociation)
- \( R = 8.314 \, \text{J/(mol K)} \) (universal gas constant)
- \( T_1 = 1300 \, K \)
- \( T_2 = 1200 \, K \)

2. **Substitute the Known Values into the Equation:**
\[ \log \left( \frac{K_2}{0.0118} \right) = -\frac{597400}{2.303 \times 8.314} \left( \frac{1}{1200} - \frac{1}{1300} \right) \]

3. **Calculate the Right Side of the Equation:**
- First, calculate \( \frac{1}{1200} - \frac{1}{1300} \):
\[ \frac{1}{1200} - \frac{1}{1300} = \frac{1300 - 1200}{1200 \times 1300} = \frac{100}{1560000} \approx 6.41 \times 10^{-5} \]

- Now calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.1 \]

- Now substitute these values back into the equation:
\[ \log \left( \frac{K_2}{0.0118} \right) = -\frac{597400}{19.1} \times 6.41 \times 10^{-5} \]

- Calculate \( -\frac{597400}{19.1} \):
\[ -\frac{597400}{19.1} \approx -31200.48 \]

- Now multiply by \( 6.41 \times 10^{-5} \):
\[ -31200.48 \times 6.41 \times 10^{-5} \approx -1.999 \]

4. **Final Calculation for \( K_2 \):**
\[ \log \left( \frac{K_2}{0.0118} \right) \approx -2 \]
\[ \log K_2 - \log 0.0118 = -2 \]
\[ \log K_2 = -2 + \log 0.0118 \]

- Calculate \( \log 0.0118 \):
\[ \log 0.0118 \approx -1.928 \]
- Substitute back:
\[ \log K_2 = -2 - 1.928 = -3.928 \]
- Convert from logarithmic to exponential form:
\[ K_2 = 10^{-3.928} \approx 1.18 \times 10^{-4} \]

### Final Answer:
The standard equilibrium constant \( K_2 \) at \( 1200 \, K \) is approximately \( 1.18 \times 10^{-4} \).

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