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 $
