At 500 K, equlibrium constant, `K_(c)` for the following reaction is 5.
`1//2 H_(2)(g)+ 1//2(g)hArr HI (g)`
What would be the equilibrium constant `K_(c)` for the reaction `2HI(g)hArrH_(2)(g)+l_(2)(g)`

Options

A. 0.04
B. 0.4
C. 25
D. 2.5

Explanation

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

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

\[ \frac{1}{2} \text{H}_2(g) + \frac{1}{2} \text{I}_2(g) \rightleftharpoons \text{HI}(g) \]

is 5 at 500 K.

### Step-by-Step Solution:

1. **Write the given reaction and its equilibrium constant:**
\[ \frac{1}{2} \text{H}_2(g) + \frac{1}{2} \text{I}_2(g) \rightleftharpoons \text{HI}(g) \quad K_c = 5 \]

2. **Reverse the reaction:**
When we reverse a reaction, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant.
\[ \text{HI}(g) \rightleftharpoons \frac{1}{2} \text{H}_2(g) + \frac{1}{2} \text{I}_2(g) \quad K_c' = \frac{1}{5} \]

3. **Multiply the reversed reaction by 2:**
We now multiply the entire reversed reaction by 2 to obtain the desired reaction.
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]
When we multiply a reaction by a coefficient, we raise the equilibrium constant to the power of that coefficient.
\[ K_c'' = \left( K_c' \right)^2 = \left( \frac{1}{5} \right)^2 \]

4. **Calculate the new equilibrium constant:**
\[ K_c'' = \frac{1}{25} = 0.04 \]

5. **Final answer:**
The equilibrium constant \( K_c \) for the reaction \( 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \) is \( 0.04 \).