Question: On a given condition, the equilibrium concentration of $K_{p} > K_{c}$ and $K_{p} < K_{c}$ are 0...

Question

On a given condition, the equilibrium concentration of $K_{p} > K_{c}$ and $K_{p} < K_{c}$ are 0.80, 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction $K_{p} = K_{c}$ ⇌ $K_{p} = 1/K_{c}$ will be.

Answer 1

Solution

$PCl_{5}(g)$ ⇌ 2HI; [HI] = 0.80, $250^{o}C$ , $K_{c}$

$K_{p}$

Question: On a given condition, the equilibrium concentration of \(K_{p} > K_{c}\) and \(K_{p} < K_{c}\) are 0...

Solution