QJEEM24Ch1e29: JEE Main 2024 Chemistry - Class 11 JEEM24Ch1e29 Solution: The ratio for the reaction : is :

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: $\frac{1}{\sqrt{R T}}$