JEE MAIN - Chemistry (2024 - 6th April Evening Shift - No. 17)
The ratio $$\frac{K_P}{K_C}$$ for the reaction :
$$\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}$$ is :
Explanation
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: $$\frac{1}{\sqrt{R T}}$$
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