Question

For the reaction, $PCl _{5(g)} \rightleftharpoons PCl _{3(g)}+ Cl _{2(g)}$ in a 3 litre vessel at $250^{\circ} C , K_{c}$ is $0.04$ Calculate the initial number of moles of $PCl _{5}$ if equilibrium concentration of $Cl _{2}$ is $0.15\,M$

• A. 2.1
• B. 0.56
• C. 0.71
• D. 0.24

Answer 1

Answer: (A)

$K_{c}=\frac{\left[ PCl _{3}\right]\left[ Cl _{2}\right]}{\left[ PCl _{5}\right]} $

$\Rightarrow 0.04=\frac{0.15 \times 0.15}{(C-x)}$

$ \Rightarrow C-x=0.5625 \,M$

Initial concentration of $PCl _{5}(C)=0.5625+0.15=0.7125 M$

Initial number of moles of $PCl _{5}=$ Concentration $\times$ Volume (L)

$=0.7125 \times 3=2.1375$