The equilibrium constant for the reaction `H_(2)O(g)+CO(g)hArrH_(2)(g)+CO_(2)(g)` is 0.44 at 1660 K. The equilibrium constant for the reaction
`2H_(2)(g)+2CO_(2)(g)hArr 2CO(g)+2H_(2)O(g)`
at 1660 K is equal to

Options

A. 0.44
B. 0.88
C. 5.16
D. 126

Answer

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

\[ 2H_2(g) + 2CO_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g) \]

given that the equilibrium constant for the reaction:

\[ H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \]

is \( K_1 = 0.44 \) at 1660 K.

### Step-by-step Solution:

1. **Reverse the Original Reaction**:
The equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. Therefore, for the reaction:

\[ H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g) \]

The equilibrium constant \( K_2 \) is given by:

\[ K_2 = \frac{1}{K_1} = \frac{1}{0.44} \]

2. **Calculate \( K_2 \)**:
Now, we can calculate \( K_2 \):

\[ K_2 = \frac{1}{0.44} \approx 2.27 \]

3. **Multiply the Reaction by 2**:
We need to find the equilibrium constant for the reaction:

\[ 2H_2(g) + 2CO_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g) \]

When a balanced equation is multiplied by a factor, the equilibrium constant is raised to the power of that factor. Therefore, we have:

\[ K_3 = K_2^2 \]

4. **Calculate \( K_3 \)**:
Now, substituting \( K_2 \) into the equation:

\[ K_3 = (2.27)^2 \]

Calculating this gives:

\[ K_3 \approx 5.15 \]

5. **Final Answer**:
Thus, the equilibrium constant for the reaction \( 2H_2(g) + 2CO_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g) \) at 1660 K is approximately:

\[ K_3 \approx 5.15 \]

### Summary:
The equilibrium constant for the reaction \( 2H_2(g) + 2CO_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g) \) at 1660 K is approximately 5.15.