The value of `K_(p)` for the following reaction `2H_(2)S(g)hArr2H_(2)(g) + S_(2)(g)` is `1.2xx10^(-2)` at `10.6.5^(@)C`. The value of `K_(c)` for this reaction is

Options

A. `1.2xx10^(-2)`
B. `lt1.2xx10^(-2)`
C. 83
D. `gt 1.2xx10^(-2)`

Answer

To find the value of \( K_c \) for the reaction \( 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \) given that \( K_p = 1.2 \times 10^{-2} \) at \( 106.5^\circ C \), we can use the relationship between \( K_p \) and \( K_c \):

### Step-by-Step Solution:

1. **Write the relationship between \( K_p \) and \( K_c \)**:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]
where:
- \( R \) is the gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

2. **Calculate \( \Delta n \)**:
- For the reaction \( 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \):
- Moles of products = 2 (from \( H_2 \)) + 1 (from \( S_2 \)) = 3
- Moles of reactants = 2 (from \( H_2S \))
- Therefore, \( \Delta n = 3 - 2 = 1 \).

3. **Convert the temperature to Kelvin**:
\[ T(K) = 106.5 + 273 = 379.5 \, K \]

4. **Substitute the values into the equation**:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]
Since \( \Delta n = 1 \), we can simplify:
\[ K_p = K_c \cdot R \cdot T \]
Rearranging gives:
\[ K_c = \frac{K_p}{R \cdot T} \]

5. **Substitute the known values**:
\[ K_c = \frac{1.2 \times 10^{-2}}{0.0821 \cdot 379.5} \]

6. **Calculate \( R \cdot T \)**:
\[ R \cdot T = 0.0821 \cdot 379.5 \approx 31.16495 \]

7. **Calculate \( K_c \)**:
\[ K_c = \frac{1.2 \times 10^{-2}}{31.16495} \approx 3.84 \times 10^{-4} \]

### Final Answer:
The value of \( K_c \) for the reaction is approximately \( 3.8 \times 10^{-4} \).

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