Equilibrium Question 877

Question: In the reaction $ PC{l_{5(g)}} $

$ \rightarrow $ $PC{l_{3(g)}} $

$ +C{l_{2(g)}}. $

The equilibrium concentrations of $ PCl_5 $ and $ PCl_3 $ are 0.4 and 0.2 mole/litre respectively. If the value of $ K_{c} $ is 0.5 what is the concentration of $ Cl_2 $ in moles/litre

Options:

A) 2.0

B) 1.5

C) 1.0

D) 0.5

Show Answer

Answer:

Correct Answer: C

Solution:

$ K_{c}=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.2\times x}{0.4}=0.5 $ , $ x=1 $



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