
\(\frac {4}{5 - α} = \frac {3}{α - 2}\)
\(⇒ 4α - 8 = 15 - 3α\)
\(α = \frac {23}{7}\)
\(A = ( \frac {23}{7},0)\ Q = (5,4)\)
\(R = (\frac {10 + \frac {23}{7}}{3} , \frac 83 )\)
\(= ( \frac {31}{7} , \frac 83 )\)
Bisector of angle PAQ is \(X = \frac {23}{7}\).
\(⇒ M = ( \frac {23}{7} , \frac 83 )\)
So, \(7α + 3β = 31\)
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 : \(A_2 \rightleftharpoons 2A\)
m×n = -1
