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 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 \( K = 0.44 \) at 1660 K, we can follow these steps:

### Step 1: Identify the relationship between the two reactions

The first reaction can be written as:

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

The second reaction is:

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

### Step 2: Reverse the first reaction

If we reverse the first reaction, we get:

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

When we reverse a reaction, the equilibrium constant for the reversed reaction \( K' \) is the reciprocal of the original equilibrium constant \( K \):

\[ K' = \frac{1}{K} = \frac{1}{0.44} \]

### Step 3: Multiply the reversed reaction by 2

Now, if we multiply the entire reversed reaction by 2, we get:

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

When we multiply a reaction by a coefficient, the equilibrium constant is raised to the power of that coefficient. Therefore, the equilibrium constant for the new reaction \( K'' \) is:

\[ K'' = (K')^2 = \left(\frac{1}{0.44}\right)^2 \]

### Step 4: Calculate \( K'' \)

Now, we can calculate \( K'' \):

\[ K'' = \left(\frac{1}{0.44}\right)^2 = \frac{1}{0.1936} \approx 5.1529 \]

### Step 5: Round the answer

Rounding \( 5.1529 \) gives us approximately \( 5.16 \).

### 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 \( 5.16 \).

---