The equilibrium constant $k_{c}$ of the reaction $A_{2}$ (g) + $B_{2}$ (g) $⇌$ 2AB (g) is 50. If 1 mol of $A_{2}$ and 2 mol of $B_{2}$ are mixed, the amount of AB at equilibrium would be:

.934 mol

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.467 mol

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1.4 mol

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1.867 mol

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Solution

The correct option is D
1.867 mol

$\begin{matrix} A_{2} (g) & + & B_{2} (g) & ⇋ & 2 A B (g) A t s t a r t & 1 & 2 & 0 A t e q u . & 1 - x & 2 - x & 2 x \end{matrix}$

Let us take the volume of the vessel (at equilibrium) to be V litres
$K_{c} = \frac{4 x^{2}}{(1 - x) (2 - x)} = 50$
There won't be a volume term since $Δ$ n = 0
Solving the above quadratic, we get x = .943

But we need (2x) since in the question, the amount of AB formed is required. Hence d) 1.867 mol

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The equilibrium constant $k_{c}$ of the reaction $A_{2}$ (g) + $B_{2}$ (g) $⇌$ 2AB (g) is 50. If 1 mol of $A_{2}$ and 2 mol of $B_{2}$ are mixed, the amount of AB at equilibrium would be: