1. Tardigrade
  2. Question
  3. Chemistry
  4. Consider following reactions in equilibrium concentration 0.01 M of every species (I) PCl5(g) leftharpoons PCl3(g) + Cl2(g) (II) 2HI(g) leftharpoons H2(g) + I2(g) (III) N2(g) + 3H2(g) leftharpoons 2NH3(g) Extent of the reactions taking place is

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Q. Consider following reactions in equilibrium concentration $0.01$ M of every species
$(I) PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g)$
$(II) 2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)$
$(III) N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$
Extent of the reactions taking place is

Equilibrium

Solution:

Extent of reaction can be calculated by the value of $K_{c}$. The higher value of $K_{c}$ the larger the extent of reaction.
$(I)\,\,\,PCl_{5}(g) \rightleftharpoons PCl_{3}(g) +Cl_{2}(g)$
$\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}=\frac{0.01 \times0.01}{0.01}=0.01$
$(II)\,\,\,2HI \rightleftharpoons H_{2} +I_{2},K_{C} =\frac {[H_{2}][I_{2}]}{[HI]^{2}}$
$=\frac{(0.01)(0.01)}{(0.01)^{2}}=1$
$(III)\,\,\, N_{2} (g) +3H_{2}(g) \rightleftharpoons 2NH_{3} (g)$
$K_{c}=\frac{[NH_{3}]^{2}}{[N_{2}[H_{2}]^3}=\frac{(0.01)^{2}}{(0.01)^{4}}\frac{1}{0.01 \times 0.01}=10000$
Extent of reaction $I < II < III$