\(6+\frac{1}{32}-2\sqrt2\)
\(2+\frac{1}{96}-\frac{2\sqrt2}{3}\)
\(\frac{2\sqrt2}{3}\)
96
The correct answer is (A) : \(\frac{2\sqrt2}{3}\)
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 : \(A_2 \rightleftharpoons 2A\)