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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Equilibrium

BITSAT 2024 · Paper 1 · Chemistry · Chemical Equilibrium

The ratio (K_P)/(K_C) for the reaction CO_ (g) +(1)/(2) O_ 2( g)...

The ratio KPKC\frac{K_P}{K_C} for the reaction CO(g)+12O2(g)⇌CO2(g)\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{ g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{ g})} is:

Options

  1. A(RT)1/2(\mathrm{RT})^{1 / 2}
  2. BRTRT
  3. C11
  4. D1RT\frac{1}{\sqrt{\mathrm{RT}}}

Correct answer

D. 1RT\frac{1}{\sqrt{\mathrm{RT}}}

Step-by-step solution

CO(g)+12O2(g)⇌CO2(g)KP\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{ g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{ g})\mathrm{K}_{\mathrm{P}} and KC\mathrm{K}_{\mathrm{C}} are related as KP=KC(RT)Δng⇒Δng=nP−nR=1−112=−12Δng\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}\Rightarrow \quad \Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{P}}-\mathrm{n}_{\mathrm{R}}=1-1 \frac{1}{2}=-\frac{1}{2}\Delta \mathrm{n}_{\mathrm{g}} for the given reaction =−12∴KPKC=1RT=-\frac{1}{2}\therefore \frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{C}}}=\frac{1}{\sqrt{\mathrm{RT}}}
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