Question:medium
Given below are the reactions. Identify the reaction for which \(K_p<K_c\):
Given below are the reactions. Identify the reaction for which \(K_p<K_c\):
Show Hint
If \(\Delta n_g=0\), then \(K_p=K_c\). If \(\Delta n_g<0\), then \(K_p<K_c\).
Updated On: May 28, 2026
- \(\ce{N2(g)+O2(g)<=>2NO(g)}\)
- \(\ce{H2O(g)+CO(g)<=>H2(g)+CO2(g)}\)
- \(\ce{H2(g)+I2(g)<=>2HI(g)}\)
- \(\ce{N2(g)+3H2(g)<=>2NH3(g)}\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
In chemical equilibrium, $K_p$ is the equilibrium constant defined by partial pressures, and $K_c$ is defined by molar concentrations. Depending on the stoichiometry of the gaseous components, these two constants may or may not be numerically equal. The question asks to identify the reaction where they are different.
Step 2: Key Formula or Approach:
The relationship between the two constants is: $K_p = K_c(RT)^{\Delta n_g}$.
$\Delta n_g$ = (Total moles of gaseous products) - (Total moles of gaseous reactants).
If $\Delta n_g = 0$, then $(RT)^0 = 1$, so $K_p = K_c$.
If $\Delta n_g \neq 0$, then $K_p \neq K_c$.
Step 3: Detailed Explanation:
Reaction A: $H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g)$. $\Delta n_g = (1 + 1) - (1 + 1) = 0$. So $K_p = K_c$.
Reaction B: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. $\Delta n_g = (2) - (1 + 3) = 2 - 4 = -2$. Since $\Delta n_g \neq 0$, $K_p = K_c(RT)^{-2}$. Thus, $K_p \neq K_c$.
Reaction C: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$. $\Delta n_g = 2 - (1 + 1) = 0$. So $K_p = K_c$.
Reaction D: $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$. $\Delta n_g = 2 - (1 + 1) = 0$. So $K_p = K_c$.
Reaction B is the only case where the number of gaseous molecules changes during the reaction, leading to the inequality.
Step 4: Final Answer:
The reaction where $K_p \neq K_c$ is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
In chemical equilibrium, $K_p$ is the equilibrium constant defined by partial pressures, and $K_c$ is defined by molar concentrations. Depending on the stoichiometry of the gaseous components, these two constants may or may not be numerically equal. The question asks to identify the reaction where they are different.
Step 2: Key Formula or Approach:
The relationship between the two constants is: $K_p = K_c(RT)^{\Delta n_g}$.
$\Delta n_g$ = (Total moles of gaseous products) - (Total moles of gaseous reactants).
If $\Delta n_g = 0$, then $(RT)^0 = 1$, so $K_p = K_c$.
If $\Delta n_g \neq 0$, then $K_p \neq K_c$.
Step 3: Detailed Explanation:
Reaction A: $H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g)$. $\Delta n_g = (1 + 1) - (1 + 1) = 0$. So $K_p = K_c$.
Reaction B: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. $\Delta n_g = (2) - (1 + 3) = 2 - 4 = -2$. Since $\Delta n_g \neq 0$, $K_p = K_c(RT)^{-2}$. Thus, $K_p \neq K_c$.
Reaction C: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$. $\Delta n_g = 2 - (1 + 1) = 0$. So $K_p = K_c$.
Reaction D: $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$. $\Delta n_g = 2 - (1 + 1) = 0$. So $K_p = K_c$.
Reaction B is the only case where the number of gaseous molecules changes during the reaction, leading to the inequality.
Step 4: Final Answer:
The reaction where $K_p \neq K_c$ is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
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