Question:
For the reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, the relation between $K_p$ and $K_c$ is}
For the reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, the relation between $K_p$ and $K_c$ is}
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Remember, when the change in moles of gas ($\Delta n_g$) is negative, the exponent on $RT$ will also be negative. Conversely, if $\Delta n_g$ is positive, the exponent will be positive.
Updated On: May 31, 2026
- $K_p = K_c (RT)^{-2}$
- $K_p = K_c (RT)^2$
- $K_p = K_c (RT)^{-1}$
- $K_p = K_c (RT)$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The equilibrium constant $K_p$ is expressed in terms of partial pressures, while $K_c$ is expressed in terms of molar concentrations. The relationship between these two constants depends on the change in the number of moles of gas during the reaction.
Step 2: Meaning
For a general reaction: \[aA + bB \rightleftharpoons cC + dD\] The equilibrium constant $K_c$ is given by: \[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] And the equilibrium constant $K_p$ in terms of partial pressures is: \[K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\] where $P_i$ represents the partial pressure of species $i$. The relationship between $K_p$ and $K_c$ can be derived using the ideal gas law: \[PV = nRT\]
Step 3: Analysis
For the given reaction: \[\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\] The change in moles of gas ($\Delta n_g$) is calculated as follows: \[\Delta n_g = 2 - (1 + 3) = -2\] Using the relationship between $K_p$ and $K_c$: \[K_p = K_c (RT)^{\Delta n_g}\] Substituting $\Delta n_g = -2$: \[K_p = K_c (RT)^{-2}\] This confirms that option A is correct.
Step 4: Conclusion
The relationship between the equilibrium constants $K_p$ and $K_c$ for the given reaction is: \[K_p = K_c (RT)^{-2}\]
Final Answer: (A)
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