Question

For the equilibrium, $\underset{\mathrm{Nitrosyl}\mathrm{chloride}}{2\mathrm{NOCl}\left(\mathrm{g}\right)}\rightleftharpoons \underset{\mathrm{Nitric}\mathrm{oxide}}{2\mathrm{NO}\left(\mathrm{g}\right)}+\underset{\mathrm{chlorine}}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}$. The value of the equilibrium constant $\mathrm{Kc}$ is $3.75\times {10}^{-4}$ at $1069\mathrm{K}$. Calculate $\mathrm{Kp}$ for the reaction at this temperature.

Answer 1

Step 1 Given data:

Reaction: $2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

Value of $\mathrm{Kc}$$=3.75\times {10}^{-4}$

Temperature $=1069\mathrm{K}$

Step 2 Applying formula:

$\mathrm{Kp}=\mathrm{Kc}\times {\left(\mathrm{RT}\right)}^{∆\mathrm{ng}}$

$∆\mathrm{ng}=\mathrm{mole}\mathrm{of}\mathrm{product}-\mathrm{mole}\mathrm{of}\mathrm{reactant}$

$∆\mathrm{ng}=3-2=1$

$$\begin{gathered} \mathrm{Kp}=3.75\times {10}^{-4}{(8.314\times 1069)}^1 \\ \mathrm{Kp}=3.33 \end{gathered}$$

Therefore, $\mathrm{Kp}$ for the reaction at $1069\mathrm{K}$ is$3.33$.