Question

For the reversible reaction \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) + \text{Heat}\), the forward reaction is favored by:

• A. High temperature and high pressure
• B. Low temperature and high pressure
• C. High temperature and low pressure
• D. Low temperature and low pressure

Hint:

For any exothermic gaseous reaction where \(\Delta n_g < 0\) (moles of product < moles of reactant), maximum yield is theoretically obtained at low temperature and high pressure.

Answer 1

The Correct Option is B

Step 1: Read the reaction.
The reaction is $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) + \text{Heat}$, the Haber process. We want the temperature and pressure that push it forward.
Step 2: State the guiding principle.
Le Chatelier's Principle says a system at equilibrium shifts so as to oppose any change forced upon it.
Step 3: Decide the temperature.
Heat appears on the product side, so the forward reaction is exothermic. Lowering the temperature makes the system try to replace the lost heat by moving forward. So a low temperature favours product.
Step 4: Count the gas moles.
Left side has $1 + 3 = 4$ moles of gas; right side has $2$ moles of gas. The forward direction shrinks the gas count.
Step 5: Decide the pressure.
Raising the pressure pushes the equilibrium toward the side with fewer gas moles, which is the product side here. So a high pressure favours product.
Step 6: Combine the two.
Forward yield is maximised by low temperature together with high pressure, which is option (B).
\[ \boxed{\text{Low temperature and high pressure}} \]