Question

A sample of pure $PCl_{5}$ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of $PCl_{5}$ was found to be 0.5 × $10^{-1}$ mol $L^{-1}$. If value of Kc is 8.3 × $10^{-1}$, what are the concentrations of $PCl_{3}$ and $Cl_{2}$ at equilibrium?

Answer 1

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Answer 2

Consider at equilibrium the concentrations of $PCl_3$ and $Cl_2$ be x

The equation is $PCl_5\rightleftharpoons PCl_3 + Cl_2$

At equilibrium,

$PCl_5(0.5\times10^{-1})\rightleftharpoons PCl_3(x) + Cl_2(x)$

According to law of chemical equilibrium,

Kc = [PCl3][Cl2]/[PCl5]

Substitute the values,

$K_c = x^2/0.5\times 10^{-1} = 8.3\times 10^{-3}$

$x^2 = (8.3\times 10^{-3})\times( 0.5 \times10^{-1})$

$x^2 = 4.15\times 10^{-4}\\x = 2.04\times 10^{-2}\\x = 0.02M$

Therefore, the concentration of $[PCl_3] = [Cl_2] = 0.02M$