First slide

Chemical Equilibrium

Question

In the hypothetical equation $A + B \underset{K_{2}}{\overset{K_{1}}{⇌}} C + D$ if equilibrium constant is $2 \times 10^{4} at 298 K$ . If forward rate constant $(K_{1}) = 4 \times 10^{3} {mole}^{2} l i t^{- 2} at 298 K$ , the value of backward rate constant $(K_{2})$ at same temperature is ___?

Moderate

Solution

$\begin{array}{l} Equilibrium constant = \frac{Forward rate constant}{Backward rate constant} \\ 2 \times 10^{4} = \frac{4 \times 10^{3}}{K_{2}} \\ K_{2} = 0.2 {mole}^{2} {lit}^{- 2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App