Question

What will be the equilibrium constant of the given reaction carried out in a $ 5 \,L $ vessel and having equilibrium amounts of $ A_2 $ and $ A $ as $ 0.5 $ mole and $ 2 \times 10^{-6} $ mole respectively ? The reaction : $ \ce{A2 <=> 2A} $

• A. $ 0.16 \times 10^{-11} $
• B. $ 0.25 \times 10^{5} $
• C. $ 0.4 \times 10^{-5} $
• D. $ 0.2 \times 10^{-11} $

Hint:

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Answer 1

The Correct Option is A

$A_{2}\ce{<=>}2A$ Concentration of $A_{2}$ at equilibrium $=\frac{0.5}{5}$ Concentration of $A$ at equilibrium $=\frac{2\times 10^{-6}}{5}$ Equilibrium constant, $K_{c}=\frac{\left[A\right]^{2}}{\left[A_{2}\right]}=\frac{\left(\frac{2\times10^{-6}}{5}\right)^{2}}{\frac{0.5}{5}}$ $=\frac{4\times5}{25\times0.5}\times10^{-12}$ $=0.16\times10^{-11}$