Question

For the formation of \(\mathrm{NH_3(g)}\) from its constituent elements, which of the relation between the reaction quotient \((Q)\) and equilibrium constant \((K_c)\) is correct for the backward reaction?

• A. \(Q=K_c\)
• B. \(Q\gt K_c\)
• C. \(Q\lt K_c\)
• D. \(Q=K=1\)

Hint:

Remember the relation between \(Q\) and \(K_c\): \[ Q\lt K_c \Rightarrow \text{Forward reaction proceeds} \] \[ Q\gt K_c \Rightarrow \text{Backward reaction proceeds} \] \[ Q=K_c \Rightarrow \text{System is at equilibrium} \]

Answer 1

The Correct Option is B

Step 1: Write the reaction for formation of ammonia.
The formation of ammonia from its constituent elements is \[ N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) \] This is the forward reaction.
The backward reaction is \[ 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) \]

Step 2: Recall the significance of \(Q\) and \(K_c\).
The reaction quotient \(Q\) predicts the direction in which a reaction will proceed to attain equilibrium.
\[ Q=K_c \] indicates equilibrium.
\[ Q\lt K_c \] indicates the forward reaction is favored.
\[ Q\gt K_c \] indicates the reverse (backward) reaction is favored.

Step 3: Apply the condition for the backward reaction.
The question asks for the condition under which the backward reaction proceeds.
For the reaction to move in the reverse direction and produce more reactants, \[ Q\gt K_c \] must hold.
Under this condition, the system contains more products than required at equilibrium, so the reaction shifts backward to attain equilibrium.

Step 4: Final conclusion.
Therefore, for the backward reaction, \[ \boxed{Q\gt K_c} \] Hence, the correct option is \[ \boxed{(2)} \]