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Standard XII

Chemistry

Characteristics of Equilibrium Constant

2 moles of ...

Question

$2$ moles of $N_{2} O_{4}$ is heated in a $5$ lit vessel till the equilibrium reached. If $0.5$ moles of $N_{2} O_{4}$ is present at equilibrium, the equilibrium constant for the reaction is:

3.6

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Solution

The correct option is A $3.6$
The given reaction is :-
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
Initial moles : $2$ $0$
At eqm : $(2 - x)$ $2 x$ (let)
Given, $(2 - x) = 0.5$
$\Rightarrow x = 1.5$
Now, No. of moles of ${N O}_{2}$ at eqm $= 2 \times 1.5 = 3$
At eqm, $[N_{2} O_{4}] = \frac{0.5}{5} = 0.1 M$
$[{N O}_{2}] = \frac{3}{5} = 0.6 M$
Now, $K_{C} = \frac{{[{N O}_{2}]}^{2}}{[N_{2} O_{4}]} = \frac{0.6 \times 0.6}{0.1} = 3.6 M$

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2 moles of N2O4 is heated in a 5 lit vessel till the equilibrium reached. If 0.5 moles of N2O4 is present at equilibrium, the equilibrium constant for the reaction is:

The correct option is A 3.6The given reaction is :- N2O4(g)⇌2NO2(g)Initial moles : 2 0At eqm : (2−x) 2x (let)Given, (2−x)=0.5 ⇒x=1.5Now, No. of moles of NO2 at eqm =2×1.5=3At eqm, [N2O4]=0.55=0.1M [NO2]=35=0.6MNow, KC=[NO2]2[N2O4]=0.6×0.60.1=3.6M

$2$ moles of $N_{2} O_{4}$ is heated in a $5$ lit vessel till the equilibrium reached. If $0.5$ moles of $N_{2} O_{4}$ is present at equilibrium, the equilibrium constant for the reaction is: