Chemical Equilibrium — NEET UG practice

1. Relating Kp and Kc: For any gaseous equilibrium, the pressure-based equilibrium constant $K_P$ and the concentration-based equilibrium constant $K_C$ are connected by the formula below, where $\Delta n_g$ is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). $K_P = K_C(RT)^{\Delta n_g}$

2. When are they equal? Since $(RT)^0 = 1$, $K_P$ equals $K_C$ only when $\Delta n_g = 0$. If $\Delta n_g \neq 0$, then $K_P \neq K_C$.

3. Checking Option A: $H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g)$ has 2 moles of gas on each side, so $\Delta n_g = 2 - 2 = 0$ Hence $K_P = K_C$ here.

4. Checking Option B: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ has 4 moles of gaseous reactants and 2 moles of gaseous product, so $\Delta n_g = 2 - 4 = -2$ Since $\Delta n_g \neq 0$, this is the reaction where $K_P \neq K_C$.

5. Checking Options C and D: Both $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$ and $N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$ have 2 moles of gas on both sides, so $\Delta n_g = 0$ and $K_P = K_C$ for both.

Hence, the answer is Option B.

1. Relating rate constants to Kc: At equilibrium, the forward rate equals the backward rate, so the equilibrium constant $K_C$ can be written as the ratio of the forward rate constant $k_f$ to the backward rate constant $k_b$. $K_C = \frac{k_f}{k_b}$

2. Using the given data: Since the backward rate constant is 2500 times the forward rate constant, $K_C = \frac{1}{2500}$

3. Converting Kc to Kp: Since the reaction involves gases whose moles change, we need the relation $K_P = K_C(RT)^{\Delta n_g}$ where $\Delta n_g$ is the change in gaseous moles.

4. Finding $\Delta n_g$: For $A(g) \rightleftharpoons 2B(g)$, there is 1 mole of gas on the reactant side and 2 moles on the product side, so $\Delta n_g = 2 - 1 = 1$

5. Substituting values: Using $R = 0.0831\ L\ atm\ mol^{-1}K^{-1}$ and $T = 1000\ K$, $K_P = \frac{1}{2500} \times 0.0831 \times 1000$

6. Calculating the result: $K_P = 0.033$

Hence, the answer is Option A.

1. Understanding the reaction: The reaction $N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$ has $\Delta H = +180.7\ kJ\ mol^{-1}$, which means it is endothermic (heat is absorbed to go forward).

2. Effect of temperature (Le Chatelier's Principle): For an endothermic reaction, increasing the temperature supplies more heat, which the system relieves by favouring the forward (heat-absorbing) direction, so the yield of NO increases. Thus A (higher temperature) favours more NO, and B is incorrect.

3. Effect of reactant concentration: Increasing the concentration of a reactant shifts the equilibrium forward to consume the extra reactant and restore equilibrium.

4. Effect of higher $N_2$: Since $N_2$ is a reactant, increasing its concentration pushes the equilibrium towards more NO, so C increases yield.

5. Effect of higher $O_2$: Similarly, since $O_2$ is a reactant, increasing its concentration also pushes the equilibrium towards more NO, so D increases yield.

6. Combining results: Options A, C, and D all favour higher yield of NO, while B (lower temperature) does not.

Hence, the answer is Option B (A, C, D only).