JEE Main 2024 — Chemical Equilibrium Question with Solution

To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure, $$K_P$$, and the equilibrium constant in terms of concentration, $$K_C$$. The relationship between these two constants for a general reaction is given by:

$$K_P=K_C(RT{)}^{\Delta n}$$

where:

$$\Delta n$$ is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),

$$R$$ is the ideal gas constant, and

$$T$$ is the temperature in Kelvin.

For the given reaction:

$${\mathrm{CO}}_{(\mathrm{g})}+\frac{1}{2}{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons {\mathrm{CO}}_{2(\mathrm{g})}$$

we need to calculate $$\Delta n$$.

On the reactants side, we have:

• 1 mole of $$\mathrm{CO}$$ (g)
• 0.5 moles of $${\mathrm{O}}_2$$ (g)

So, the total number of moles of reactants is:

$$1+\frac{1}{2}=\frac{3}{2}=1.5$$

On the products side, we have:

• 1 mole of $${\mathrm{CO}}_2$$ (g)

The total number of moles of products is:

$$1$$

Therefore, $$\Delta n$$ is:

$$\Delta n=\text{moles of products}-\text{moles of reactants}=1-1.5=-0.5$$

Now, we can use the relationship between $$K_P$$ and $$K_C$$:

$$K_P=K_C(RT{)}^{\Delta n}$$

Substituting $$\Delta n=-0.5$$, we get:

$$K_P=K_C(RT{)}^{-0.5}$$

or equivalently:

$$\frac{K_P}{K_C}=(RT{)}^{-0.5}$$

Therefore, the correct option is:

Option D: $$\frac{1}{\sqrt{RT}}$$