Relation Between Kp And Kc
At 500 K, the equilibrium constant K_c for the reaction PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) is 1.8 \times 10^{-2}. What is the value of K_p at the same temperature? (R = 0.0831 L bar mol^{-1} K^{-1})
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- relation between kp and kc
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
7.47×10−1
The relationship between K_p and K_c is given by the expression K_p = K_c \times (RT)^{\Delta n_g}, where \Delta n_g is the change in moles of gaseous species (moles of gaseous products minus moles of gaseous reactants). For PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g), \Delta n_g = (1 + 1) - 1 = +1. Therefore, K_p = K_c \times (RT)^1 = 1.8 \times 10^{-2} \times (0.0831 \times 500) = 1.8 \times 10^{-2} \times 41.55 = 0.7479 \approx 7.47 \times 10^{-1}. Option 4.33 \times 10^{-4} is wrong because it would correspond to dividing by (RT) instead of multiplying, implying \Delta n_g = -1 which is incorrect. Option 1.8 \times 10^{-2} is wrong because K_p equals K_c only when \Delta n_g = 0, which is not the case here. Option 9.0 \times 10^{-1} uses an incorrect value of RT. This derivation is rooted in NCERT's thermodynamic treatment of gaseous equilibria and is a standard JEE Main numerical. Plausibility check: since \Delta n_g > 0, K_p > K_c, which is confirmed as 0.747 > 0.018.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of relation between kp and kc. It appeared in the 2025 exam.
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