JEE Main 2024 — Chemical Equilibrium Question with Solution

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:

$$2\times 10^{-2}$$