Equilibrium Question 621

Question: A quantity of $ PCl_5 $ was heated in a 10 litre vessel at $ 250^{o}C $ ; $ PCl_5(g) $ ⇌ $ PCl_3(g)+Cl_2(g) $ . At equilibrium the vessel contains 0.1 mole of $ PCl_5,0.20 $ mole of $ PCl_3 $ and 0.2 mole of $ Cl_2 $ . The equilibrium constant of the reaction is [KCET 1993, 2001; MP PMT 2003]

Options:

A) 0.02

B) 0.05

C) 0.04

D) 0.025

Show Answer

Answer:

Correct Answer: C

Solution:

$ K_{c}=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{\frac{0.2}{10}\times \frac{0.2}{10}}{[ {0.1}/{10}; ]}=0.04 $ .



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