The equilibrium constant for the reaction `N_(2)+3H_(2)hArr2NH_(3)` is 70 at a certain temperature. Hence, equilibrium constant for the reaction `NH_(3)hArr(1)/(2)N_(2)+(3)/(2)H_(2)` of the same temperature will be approximately

Options

A. `1.4xx10^(-2)`
B. `1.2xx10^(-1)`
C. `2.0xx10^(-4)`
D. `2.9xx10^(-2)`

Answer

To find the equilibrium constant for the reaction \( \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \) given that the equilibrium constant for the reaction \( \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \) is 70, we can follow these steps:

### Step 1: Write the equilibrium expression for the first reaction
The equilibrium constant \( K_c \) for the reaction \( \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \) is given by the expression:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Given that \( K_c = 70 \).

### Step 2: Write the equilibrium expression for the second reaction
For the reaction \( \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \), the equilibrium constant \( K_{c}' \) is given by:
\[ K_{c}' = \frac{[\text{N}_2]^{1/2}[\text{H}_2]^{3/2}}{[\text{NH}_3]} \]

### Step 3: Relate the two equilibrium constants
The second reaction can be seen as the reverse of the first reaction, and it also involves a change in stoichiometry. When you reverse a reaction, the equilibrium constant becomes the reciprocal of the original constant. Additionally, if the coefficients of the reaction are halved, the equilibrium constant is raised to the power of \( \frac{1}{2} \).

Thus, we can express \( K_{c}' \) in terms of \( K_c \):
\[ K_{c}' = \frac{1}{K_c^{1/2}} \]

### Step 4: Substitute the value of \( K_c \)
Substituting the given value of \( K_c = 70 \):
\[ K_{c}' = \frac{1}{70^{1/2}} = \frac{1}{\sqrt{70}} \]

### Step 5: Calculate \( K_{c}' \)
Calculating \( \sqrt{70} \):
\[ \sqrt{70} \approx 8.37 \quad (\text{using a calculator or estimation}) \]
Thus,
\[ K_{c}' \approx \frac{1}{8.37} \approx 0.1197 \]

### Step 6: Final answer
Rounding this value gives:
\[ K_{c}' \approx 0.12 \]

### Conclusion
The equilibrium constant for the reaction \( \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \) is approximately \( 0.12 \) or \( 1.2 \times 10^{-1} \).

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