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Chemical Equilibrium question

2024 · 6 Apr · Shift 2 · Q17

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Chemical Equilibrium
2024 · 6 Apr · Shift 2 · Q17

Chemical Equilibrium question

2024 · 6 Apr · Shift 2 · Q17

JEE MainChemistryChemical EquilibriumMCQ+4 / −1

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 :

A
1
B
$(\mathrm{RT})^{1 / 2}$
C
RT
D
$\mathrm{ \frac{1}{\sqrt{R T}}}$

View written solutionFree

Correct answer: D

For gaseous equilibria, the relation between $K_P$ and $K_C$ is:

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

where

$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

For the reaction

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

Product side gaseous moles $= 1$
Reactant side gaseous moles $= 1 + \frac{1}{2} = \frac{3}{2}$

So,

$\Delta n = 1 - \frac{3}{2} = -\frac{1}{2}$

Substitute into the formula:

$K_P = K_C (RT)^{-1/2}$

Therefore,

$\frac{K_P}{K_C} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}$

Hence the correct option is:

$\boxed{\text{D}}$

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