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 : \(A_2 \rightleftharpoons 2A\)

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

Answer 1

The Correct Option is A

The correct option is(A) : 0.16×10⁻¹¹
\(A_2 \rightleftharpoons 2A\)
Concentration of \(A_{2}\) at equilibrium \(=\frac{0.5}{5}\) 
Concentration of \(A\) at equilibrium \(=\frac{2\times 10^{-6}}{5}\) 
Equilibrium constant, \(K_{c}=\frac{\left[A\right]^{2}}{\left[A_{2}\right]}=\frac{\left(\frac{2\times10^{-6}}{5}\right)^{2}}{\frac{0.5}{5}}\)
\(=\frac{4\times5}{25\times0.5}\times10^{-12}\)
\(=0.16\times10^{-11}\)