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: Identify the backward reaction.
The forward reaction is N\(_2\) + 3H\(_2\) \( ightleftharpoons\) 2NH\(_3\). The backward reaction is 2NH\(_3\) \( ightleftharpoons\) N\(_2\) + 3H\(_2\).

Step 2: Determine the Q vs K\(_c\) relationship.
If the backward reaction is to proceed spontaneously, the products of the backward reaction (N\(_2\), H\(_2\)) must increase, meaning the current mixture has excess NH\(_3\). This means Q (written for the backward reaction) > K\(_c\) for the backward reaction, which drives it toward the right (forward direction of the backward reaction). \[ oxed{Q \gt K_c} \]