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Chemical Equilibrium question

2012 · Shift 0 · Q5
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  5. /2012 · Shift 0 · Q5

Chemical Equilibrium question

2012 · Shift 0 · Q5

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The equilibrium constant (KC) for the reaction N2N_2N2​(g) + O2O_2O2​(g) →\to→ 2NONONO(g) at temperature T is 4 ×\times× 10–4. The value of KC for the reaction, NONONO(g) →\to→ 1/2N2N_2N2​(g) + 1/2O2O_2O2​(g) at the same temperature is :
  1. A
    0.02
  2. B
    2.5 ×\times× 102
  3. C
    4 ×\times× 10-4
  4. D
    50.0
View written solutionFree

Correct answer: D

  1. Given reaction and equilibrium constant

The given reaction is:

N2(g)+O2(g)⇌2NO(g)N_2(g) + O_2(g) \rightleftharpoons 2NO(g)N2​(g)+O2​(g)⇌2NO(g)

with

KC=4×10−4K_C = 4 \times 10^{-4}KC​=4×10−4

  1. Required reaction

We need the equilibrium constant for:

NO(g)⇌12N2(g)+12O2(g)NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)NO(g)⇌21​N2​(g)+21​O2​(g)

  1. Relate the required reaction to the given reaction

First, reverse the given reaction:

2NO(g)⇌N2(g)+O2(g)2NO(g) \rightleftharpoons N_2(g) + O_2(g)2NO(g)⇌N2​(g)+O2​(g)

When a reaction is reversed, the equilibrium constant becomes reciprocal:

K′=1KC=14×10−4=2500K' = \frac{1}{K_C} = \frac{1}{4 \times 10^{-4}} = 2500K′=KC​1​=4×10−41​=2500

So for

2NO(g)⇌N2(g)+O2(g),K′=25002NO(g) \rightleftharpoons N_2(g) + O_2(g), \quad K' = 25002NO(g)⇌N2​(g)+O2​(g),K′=2500

  1. Now divide the entire reaction by 2

Required reaction is exactly half of the reversed reaction:

NO(g)⇌12N2(g)+12O2(g)NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)NO(g)⇌21​N2​(g)+21​O2​(g)

When all stoichiometric coefficients are divided by 2, the new equilibrium constant becomes:

K′′=(K′)1/2K'' = (K')^{1/2}K′′=(K′)1/2

Thus,

K′′=2500=50K'' = \sqrt{2500} = 50K′′=2500​=50

  1. Final answer

KC=50.0K_C = 50.0KC​=50.0

So the correct option is D.

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