The reaction quotient `(Q)` for thereaction
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
is given by
`Q = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
The reaction will proceed from right to left if where `K_(C)` is the equilibrium constant.

Options

A. `Q=K_(C)`
B. `Q lt K_(C)`
C. `Q gt K_(C)`
D. `Q = 0`

Explanation

To determine when the reaction will proceed from right to left for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

we need to analyze the relationship between the reaction quotient \( Q \) and the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Understand the Reaction Quotient \( Q \)**:
The reaction quotient \( Q \) is defined for the given reaction as:
\[ Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
This expression gives us a measure of the ratio of the concentrations of products to reactants at any point in time.

2. **Understand the Equilibrium Constant \( K_c \)**:
The equilibrium constant \( K_c \) is a specific value that represents the ratio of the concentrations of products to reactants when the reaction is at equilibrium. For this reaction, it is also given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

3. **Determine the Direction of the Reaction**:
- If \( Q < K_c \): The reaction will proceed in the forward direction (to the right) to reach equilibrium.
- If \( Q = K_c \): The reaction is at equilibrium, and there will be no net change.
- If \( Q > K_c \): The reaction will proceed in the reverse direction (to the left) to reach equilibrium.

4. **Conclusion**:
The reaction will proceed from right to left (backward direction) if the reaction quotient \( Q \) is greater than the equilibrium constant \( K_c \):
\[ Q > K_c \]

### Final Answer:
The reaction will proceed from right to left if \( Q > K_c \).

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