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 \( \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \) given that the equilibrium constant \( K_c \) for the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \) is 64, we can follow these steps:

### Step 1: Write the expression for the original reaction
For the reaction:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Given that \( K_c = 64 \).

### Step 2: Write the expression for the new reaction
For the new reaction:
\[ \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \]
The equilibrium constant \( K'_c \) can be expressed as:
\[ K'_c = \frac{[\text{N}_2]^{1/2}[\text{H}_2]^{3/2}}{[\text{NH}_3]} \]

### Step 3: Relate the new equilibrium constant to the original
To relate \( K'_c \) to \( K_c \), we can manipulate the original reaction. The new reaction is essentially the reverse of the original reaction, and it is also halved.

When a reaction is reversed, the equilibrium constant becomes the reciprocal:
\[ K' = \frac{1}{K_c} \]
When the coefficients of the reaction are halved, the equilibrium constant is taken to the power of \( \frac{1}{2} \):
\[ K' = \left(\frac{1}{K_c}\right)^{1/2} \]

### Step 4: Calculate the new equilibrium constant
Substituting the value of \( K_c \):
\[ K' = \left(\frac{1}{64}\right)^{1/2} = \frac{1}{8} \]

### Conclusion
Thus, the equilibrium constant for the reaction \( \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \) is:
\[ K'_c = \frac{1}{8} \]