\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} + \frac {65}{2} - \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
\(\frac {(169π)}{4} + \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
The correct option is (A): \(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\).
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 : \(A_2 \rightleftharpoons 2A\)
Read More: Area under the curve formula