Ncert Exercises & Solutions

7.19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be
0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5 (g) PCl3 (g) + Cl2(g)

Let the concentrations of both ${PCl}_{3} and {Cl}_{2}$ at equilibrium be x mol $L^{- 1}$ . The given reaction is:

${PCl}_{5 (g)} \leftrightarrow {PCl}_{3 (g)} + {Cl}_{2 (g)}$
$At equilibrium 0.5 \times 10^{- 1} mol L^{- 1} x mol L^{- 1} x mol L^{- 1}$

It is given that of the value of the equilibrium constant, $K_{c} is 8.3 \times 10^{- 3}$ .

Now we can write the expression for equilibrium as:

$\frac{[{PCl}_{2}] [{Cl}_{2}]}{[{PCl}_{5}]} = K_{c}$
$\Rightarrow \frac{x \times x}{0.5 \times 10^{- 1}} = 8.3 \times 10^{- 3}$
$\Rightarrow x^{2} = 4.15 \times 10^{- 4}$
$\Rightarrow x = 2.04 \times 10^{- 2}$
$= 0.0204$
$= 0.02 (approx .)$

Therefore, at equilibrium,

$[{PCl}_{3}] = [{Cl}_{2}] = 0.02 mol L^{- 1} .$

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