PCl_{5} , PCl_{3} and Cl_{2} are at equilibrium at 500 K in a closed container and their concentrations are 0.8\times 10^{-3}\; mol\; L^{-1} , 1.2 \times 10^{-3} mol L^{-1}   and 1.2 \times 10^{-3} mol L^{-1} respectively. The value of Kc for the reaction PCl5 (g) \rightleftharpoonsPCl3 (g) + Cl2 (g) will be

 (i) 1.8\times 10^{3}mol\; L^{-1}   

(ii) 1.8\times 10^{-3} mol L-1

 (iii) 1.8\times 10^{-3}\; Lmol^{-1}  

(iv) 0.55\times 10^{4}

Answers (1)

The answer is the option (ii) 1.8\times 10^{-3} mol L-1

Explanation :

K_{c}=\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}=\frac{1.2\times 10^{-3}\times 1.2\times 10^{-3}}{0.8\times 10^{-3}}

=1.8\times 10^{-3}\; mol\; L^{-1}

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