If \( \alpha>\beta>\gamma>0 \), then the expression \[ \cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right) \] is equal to:
The given expression involves inverse cotangents and some algebraic manipulation. By applying trigonometric identities and simplifying, we can show that the expression simplifies to \( \pi \).
Final Answer: \( \pi \).
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 : \(A_2 \rightleftharpoons 2A\)