QJEEM24Ch553b: JEE Main 2024 Chemistry - Class 11 JEEM24Ch553b Solution: For the reaction at for the reaction at same temperature is…

For the reaction:

$\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})}$

we need to find the equilibrium constant $\mathrm{K}_{\mathrm{c}}$ at 300 K given that $\mathrm{K}_{\mathrm{p}} = 0.492 \text{ atm}$ .

The relationship between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ is given by the following equation:

$\mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{c}} (RT)^{\Delta n}$

where:

$R = 0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$ is the gas constant,

$T = 300 \, \mathrm{K}$ is the temperature,

$\Delta n$ is the change in moles of gas from reactants to products.

For the reaction given:

The number of moles of gaseous products is 2 (for $2 \mathrm{NO}_2$ ).

The number of moles of gaseous reactants is 1 (for $\mathrm{N}_2 \mathrm{O}_4$ ).

Thus, $\Delta n = 2 - 1 = 1$ .

Substitute the known values into the equation:

$0.492 = \mathrm{K}_{\mathrm{c}} (0.082 \times 300)^{1}$

Solve for $\mathrm{K}_{\mathrm{c}}$ :

$0.492 = \mathrm{K}_{\mathrm{c}} \times 24.6$

$\mathrm{K}_{\mathrm{c}} = \frac{0.492}{24.6}$

Calculate $\mathrm{K}_{\mathrm{c}}$ :

$\mathrm{K}_{\mathrm{c}} = 0.02$

Thus, $\mathrm{K}_{\mathrm{c}}$ for the reaction at 300 K is:

$2 \times 10^{-2}$