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When 3 moles of A and 1 mole of B are mixed in 1 litre vessel, the following reaction takes place $A_{(g)} + B_{(g)} ⇋ 2 C_{(g)} .$ 1.5 moles of C are formed. The equilibrium constant for the reaction is

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0.12

0.25

0.50

4.0

answer is D.

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Detailed Solution

$\underset{(3 - 0.75)}{A} + \underset{(1 - 0.75)}{B} \to \underset{1.5}{2 C}$

$K = \frac{{[C]}^{2}}{[A] [B]} = \frac{{(1.5)}^{2}}{2.25 \times 0.25} = \frac{2.25}{2.25 \times 0.25} = 4.0$

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