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

Options

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

Explanation

To determine when the reaction will proceed to the right side, we need to analyze the reaction quotient (Q) and the equilibrium constant (Kc) for the reaction:

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

### Step 1: Understand the Reaction Quotient (Q)

The reaction quotient (Q) is defined for the reaction as follows:

\[ Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

This expression compares the current concentrations of the products and reactants at any point in time.

### Step 2: Understand the Equilibrium Constant (Kc)

At equilibrium, the reaction quotient (Q) becomes equal to the equilibrium constant (Kc):

\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 3: Determine the Direction of the Reaction

The direction in which the reaction will proceed can be determined by comparing Q and Kc:

- If \( Q < K_c \): The reaction will proceed to the right (towards products).
- If \( Q = K_c \): The system is at equilibrium.
- If \( Q > K_c \): The reaction will proceed to the left (towards reactants).

### Step 4: Conclusion

For the reaction to proceed to the right side, we need:

\[ Q < K_c \]

This means that the concentration of the products (NH3) needs to be less than what it would be at equilibrium, or the concentration of the reactants (N2 and H2) needs to be more than what it would be at equilibrium.

### Final Answer

The reaction will proceed towards the right side if:

\[ Q < K_c \]

---