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Chemical Equilibrium question

2002 · Shift 0 · Q9
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Chemical Equilibrium question

2002 · Shift 0 · Q9

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
For the reaction COCOCO (g) + (1/2) O2O_2O2​ (g) ⇋\leftrightharpoons⇋ CO2CO_2CO2​ (g), Kp/Kc is :
  1. A
    RT
  2. B
    (RT)-1
  3. C
    (RT)-1/2
  4. D
    (RT)1/2
View written solutionFree

Correct answer: C

  1. For a gaseous equilibrium, Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n}Kp​=Kc​(RT)Δn where Δn=(moles of gaseous products)−(moles of gaseous reactants)\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}Δn=(moles of gaseous products)−(moles of gaseous reactants)

  2. Given reaction: CO(g)+12O2(g)⇌CO2(g)CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g)CO(g)+21​O2​(g)⇌CO2​(g)

  3. Count gaseous moles:

    • Products: 111 mole of CO2CO_2CO2​
    • Reactants: 1+12=321 + \frac{1}{2} = \frac{3}{2}1+21​=23​ moles

    So, Δn=1−32=−12\Delta n = 1 - \frac{3}{2} = -\frac{1}{2}Δn=1−23​=−21​

  4. Substitute into the relation: Kp=Kc(RT)−1/2K_p = K_c (RT)^{-1/2}Kp​=Kc​(RT)−1/2

  5. Therefore, KpKc=(RT)−1/2\frac{K_p}{K_c} = (RT)^{-1/2}Kc​Kp​​=(RT)−1/2

  6. Matching with options:

    • A: RTRTRT ❌
    • B: (RT)−1(RT)^{-1}(RT)−1 ❌
    • C: (RT)−1/2(RT)^{-1/2}(RT)−1/2 ✅
    • D: (RT)1/2(RT)^{1/2}(RT)1/2 ❌

Hence, the correct answer is Option C.

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