For an equlibrium reaction `N_2(g) + 3H_2(g) harr 2NH_3(g) , K_c` = 64. what is the equilibriu constant for the reaction `NH_3(g) harr 1/2N_2(g) + 3/2 H_2(g)`

Options

A. 1/8
B. 1/4
C. 1/68
D. 1/2

Answer

To find the equilibrium constant for the reaction \( NH_3(g) \rightleftharpoons \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \) given that the equilibrium constant \( K_c \) for the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) is 64, we can follow these steps:

### Step 1: Write down the original reaction and its equilibrium constant.
The original reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
with \( K_c = 64 \).

### Step 2: Reverse the original reaction.
When we reverse the reaction, we get:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]
The equilibrium constant for the reverse reaction is the inverse of the original \( K_c \):
\[ K' = \frac{1}{K_c} = \frac{1}{64} \]

### Step 3: Adjust the stoichiometry of the reversed reaction.
Now, we need to divide the entire equation by 2 to match the target reaction:
\[ NH_3(g) \rightleftharpoons \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \]
When we divide the reaction by 2, we also need to raise the equilibrium constant to the power of the coefficient (which is 1/2 in this case):
\[ K'' = (K')^{1/2} = \left(\frac{1}{64}\right)^{1/2} \]

### Step 4: Calculate the new equilibrium constant.
Calculating \( K'' \):
\[ K'' = \frac{1}{\sqrt{64}} = \frac{1}{8} \]

### Conclusion
Thus, the equilibrium constant for the reaction \( NH_3(g) \rightleftharpoons \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \) is:
\[ \boxed{\frac{1}{8}} \]