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Question

Chemistry Question on Equilibrium

What will be the equilibrium constant of the given reaction carried out in a 5L5 \,L vessel and having equilibrium amounts of A2A_2 and AA as 0.50.5 mole and 2×1062 \times 10^{-6} mole respectively?
The reaction : A22AA_2 \rightleftharpoons 2A

A

0.16×10110.16 \times 10^{-11}

B

0.25×1050.25 \times 10^{5}

C

0.4×1050.4 \times 10^{-5}

D

0.2×10110.2 \times 10^{-11}

Answer

0.16×10110.16 \times 10^{-11}

Explanation

Solution

The correct option is(A) : 0.16×10−11
A22AA_2 \rightleftharpoons 2A
Concentration of A2A_{2} at equilibrium =0.55=\frac{0.5}{5}
Concentration of AA at equilibrium =2×1065=\frac{2\times 10^{-6}}{5}
Equilibrium constant, Kc=[A]2[A2]=(2×1065)20.55K_{c}=\frac{\left[A\right]^{2}}{\left[A_{2}\right]}=\frac{\left(\frac{2\times10^{-6}}{5}\right)^{2}}{\frac{0.5}{5}}
=4×525×0.5×1012=\frac{4\times5}{25\times0.5}\times10^{-12}
=0.16×1011=0.16\times10^{-11}