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Chemical Equilibrium question

2004 · Shift 0 · Q5
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  5. /2004 · Shift 0 · Q5

Chemical Equilibrium question

2004 · Shift 0 · Q5

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
For the reaction, COCOCO(g) + Cl2Cl_2Cl2​(g) ⇋\leftrightharpoons⇋ COCl2COCl_2COCl2​(g) the KpKc{{{K_p}} \over {{K_c}}}Kc​Kp​​ is equal to :
  1. A
    RT\sqrt {RT}RT​
  2. B
    RT
  3. C
    1/RT
  4. D
    1.0
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)

  1. Given reaction:

CO(g)+Cl2(g)⇋COCl2(g)CO(g) + Cl_2(g) \leftrightharpoons COCl_2(g)CO(g)+Cl2​(g)⇋COCl2​(g)

  • Product side gaseous moles =1= 1=1
  • Reactant side gaseous moles =2= 2=2

So,

Δn=1−2=−1\Delta n = 1 - 2 = -1Δn=1−2=−1

  1. Substitute into the relation:

Kp=Kc(RT)−1=KcRTK_p = K_c (RT)^{-1} = \frac{K_c}{RT}Kp​=Kc​(RT)−1=RTKc​​

Thus,

KpKc=1RT\frac{K_p}{K_c} = \frac{1}{RT}Kc​Kp​​=RT1​

  1. Compare with options:
  • A: RT\sqrt{RT}RT​
  • B: RTRTRT
  • C: 1RT\dfrac{1}{RT}RT1​
  • D: 111

Hence, the correct option is:

C: 1RT\boxed{\text{C: } \frac{1}{RT}}C: RT1​​

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