Equilibrium Question 767
Question: $ K_{c} $ for $ PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g) $ is 0.04 at $ 250{}^\circ C $ . How many moles of $ PCl_5 $ must be added to a 3 L flask to obtain a $ Cl_2 $ concentration of 0.15M
Options:
A) 4.2 moles
B) 2.1 moles
C) 5.5 moles
D) 6.3 moles
Show Answer
Answer:
Correct Answer: B
Solution:
At equilibrium the moles of $ Cl_2 $ must be $ =0.15\times 3=0.45 $
$ \begin{aligned} & \begin{matrix} {} & {} & {} & ,PCl_5\rightarrow PCl_3+Cl_2 \\ \end{matrix} \\ & \begin{matrix} Eqm\text{.}Conc\text{.} & \frac{x-0.45}{3} & \frac{0.45}{3} & \frac{0.45}{3} \\ \end{matrix} \\ \end{aligned} $
$ K_{c}=\frac{[ PCl_3 ][ Cl_2 ]}{[ PCl_5 ]} $
$ \therefore 0.04=\frac{0.15\times 0.15}{(x-0.45)/3} $
$ \therefore x=2.1moles $
