Question: If the value of equilibrium constant $1 \times 10^{- 12}$ for the reaction, \(3.8 \times 10^{- 3}\...

Question

If the value of equilibrium constant $1 \times 10^{- 12}$ for the reaction, $3.8 \times 10^{- 3}$ + $3.8$ $5.04$ is 7. The equilibrium constant

for the reaction $2.42$ $9.2$ will be

Answer 1

: If reaction is multiplied by 2, the equilibrium constant becomes square of the previous value $K = 7^{2} = 49$

Question: If the value of equilibrium constant \(1 \times 10^{- 12}\) for the reaction, \(3.8 \times 10^{- 3}\...