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For a gas-phase reaction at temperature $T$, the relationship between the equilibrium constants $K_p$ (in partial pressures) and $K_c$ (in molar concentrations) is

A$K_p = K_c\,(RT)^{\Delta n}$
B$K_p = K_c\,(RT)^{-\Delta n}$
C$K_p = K_c / (\Delta n)$
D$K_p$ and $K_c$ are always equal
Answer & Solution
Correct answer: A. $K_p = K_c\,(RT)^{\Delta n}$
1. NCERT §6.4.1 derives the relationship for an ideal-gas reaction $aA + bB \rightleftharpoons cC + dD$. 2. Use the ideal-gas law $p = (n/V)RT = [X]RT$ to substitute partial pressure for concentration. 3. The result (Eq. 6.15): $K_p = K_c\,(RT)^{\Delta n}$, where $\Delta n = (c + d) - (a + b)$ is the change in moles of gaseous species (products minus reactants). 4. When $\Delta n = 0$ (e.g. $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$), $K_p = K_c$. Otherwise they differ by a power of $RT$. 5. Option B inverts the exponent. Option C is dimensionally wrong. Option D is only true for $\Delta n = 0$. _Source: NCERT Class 11 Chemistry, Ch 6, §6.4.1 (Eq. 6.15), p. 10–11._
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