Equilibrium Question 536
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 $ .
BETA
