Question

Oxidation of ammonia is an exothermic reaction. $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) + 905.6 \text{ kJ}$ then which of the following statement(s) is/are correct?

• A. On increasing temperature, equilibrium mixture will have less [NO]
• B. On increasing temperature, the equilibrium constant remains unchanged
• C. On increasing pressure, equilibrium mixture will have more $[NH_3]$
• D. Addition of He at constant pressure promotes the oxidation of $NH_3$

Answer 1

Answer: (A), (C), (D)

A, C, and D

Answer 2

The reaction is $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) + 905.6 \text{ kJ}$. This is an exothermic reaction.

• A. On increasing temperature, equilibrium mixture will have less [NO]: For an exothermic reaction, increasing temperature shifts the equilibrium to the left (reverse reaction) to absorb heat. The reverse reaction consumes NO, so the concentration of NO will decrease. This statement is correct.

• B. On increasing temperature, the equilibrium constant remains unchanged: The equilibrium constant ($K$) for a reaction is temperature-dependent. For exothermic reactions, $K$ decreases with increasing temperature. This statement is incorrect.

• C. On increasing pressure, equilibrium mixture will have more $[NH_3]$: Reactants have $4 + 5 = 9$ moles of gas. Products have $4 + 6 = 10$ moles of gas. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas (reactants), increasing $[NH_3]$. This statement is correct.

• D. Addition of He at constant pressure promotes the oxidation of $NH_3$: Adding He at constant pressure increases volume, decreasing partial pressures. Equilibrium shifts to produce more moles of gas, favoring the product side and thus promoting $NH_3$ oxidation. This statement is correct.