The equilibrium constant `(K_(c))` for the reaction
`2HCl(g)hArrH_(2)(g)+Cl_(2)(g)`
is `4xx10^(-34)`at `25^(@)C` .what is the equilibrium constant for the reaction ?
`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)hArrHCl(g)`

Options

A. (a) `2xx10^(-17)`
B. (b) `2.5xx10^(33)`
C. (c) `5xx10^(6)`
D. (d) none of these

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction:

\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightleftharpoons HCl(g) \]

given that the equilibrium constant \( K_c \) for the reaction:

\[ 2 HCl(g) \rightleftharpoons H_2(g) + Cl_2(g) \]

is \( 4 \times 10^{-34} \) at \( 25^\circ C \).

### Step-by-Step Solution:

1. **Identify the given reaction and its equilibrium constant**:
The original reaction is:
\[ 2 HCl(g) \rightleftharpoons H_2(g) + Cl_2(g) \]
with \( K_c = 4 \times 10^{-34} \).

2. **Reverse the reaction**:
When we reverse the reaction, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. Therefore, for the reaction:
\[ H_2(g) + Cl_2(g) \rightleftharpoons 2 HCl(g) \]
the equilibrium constant \( K_c' \) is:
\[ K_c' = \frac{1}{K_c} = \frac{1}{4 \times 10^{-34}} = 2.5 \times 10^{33} \]

3. **Halve the coefficients of the reversed reaction**:
Now, we need to consider the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightleftharpoons HCl(g) \]
When the coefficients of a balanced reaction are halved, the equilibrium constant for this new reaction is the square root of the equilibrium constant of the reversed reaction. Thus:
\[ K_c'' = \sqrt{K_c'} = \sqrt{2.5 \times 10^{33}} \]

4. **Calculate the square root**:
To find \( K_c'' \):
\[ K_c'' = \sqrt{2.5} \times \sqrt{10^{33}} = \sqrt{2.5} \times 10^{16.5} \]
Approximating \( \sqrt{2.5} \approx 1.58 \):
\[ K_c'' \approx 1.58 \times 10^{16.5} \approx 5 \times 10^{16} \]

5. **Final answer**:
Thus, the equilibrium constant for the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightleftharpoons HCl(g) \]
is approximately:
\[ K_c'' \approx 5 \times 10^{16} \]

### Conclusion:
The equilibrium constant for the reaction \( \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightleftharpoons HCl(g) \) is \( 5 \times 10^{16} \).

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