As per law of mass action, for `NH_(4) `HS(s) `hArr NH_(3)(g) + H_(2) S(g)` ratio of rate constants of forward `(K_(f))` & backword `(K_(b))` reactions at equilibrium equals ot

Options

A. `[NH_(4) HS]`
B. `P_(NH_(3)) + P_(H_(2)S)`
C. `[H_(2)S] + [NH_(3)]`
D. `[NH_(3)][H_(2)S]`

Answer

To solve the problem regarding the law of mass action for the reaction:

\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

we need to determine the ratio of the rate constants of the forward reaction (\(K_f\)) and the backward reaction (\(K_b\)) at equilibrium.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

2. **Write the Expression for the Equilibrium Constant (\(K_c\))**:
According to the law of mass action, the equilibrium constant expression for this reaction can be written as:
\[ K_c = \frac{[\text{NH}_3][\text{H}_2\text{S}]}{[\text{NH}_4\text{HS}]} \]

3. **Consider the State of the Reactants and Products**:
In this case, NH₄HS is a solid. When writing the equilibrium constant expression, the concentration of solids is not included because it is considered to be constant (unity). Therefore, we can simplify the expression:
\[ K_c = [\text{NH}_3][\text{H}_2\text{S}] \]

4. **Relate \(K_c\) to the Rate Constants**:
At equilibrium, the relationship between the equilibrium constant and the rate constants for the forward and backward reactions is given by:
\[ K_c = \frac{K_f}{K_b} \]

5. **Conclusion**:
Therefore, the ratio of the rate constants of the forward and backward reactions at equilibrium is:
\[ K_f = K_c \cdot K_b \]
This indicates that the equilibrium constant \(K_c\) is equal to the ratio of the forward rate constant to the backward rate constant.

### Final Answer:
The ratio of the rate constants of the forward (\(K_f\)) and backward (\(K_b\)) reactions at equilibrium equals \(K_c\).

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