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Chemical Equilibrium question

2008 · Q116
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Chemical Equilibrium question

2008 · Q116

NEETChemistryChemical EquilibriumMCQ+4 / −1
The value of equilibrium constant of the reaction
HI(g) ⇌\rightleftharpoons⇌ 12{1 \over 2}21​H2(g) + 12{1 \over 2}21​I2(g)
is 8.0. The The equilibrium constant of the reaction
H2(g) + I2(g) ⇌\rightleftharpoons⇌ 2HI(g) will be
  1. A
    16
  2. B
    1/8
  3. C
    1/16
  4. D
    1/64
View written solutionFree

Correct answer: D

HI(g) ⇌\rightleftharpoons⇌ 12{1 \over 2}21​H2(g) + 12{1 \over 2}21​I2(g),

For this reaction, K = 8.0

Reversing the equation,

\therefore $${1 \over 2}H2(g) + 12{1 \over 2}21​I2(g) ⇌\rightleftharpoons⇌ HI(g) ......(1)

For this reaction, K1 = 18{1 \over 8}81​

H2(g) + I2(g) ⇌\rightleftharpoons⇌ 2HI(g)

We get this equation by multiplying equation (1) by 2,

∴\therefore∴ For this reaction, K2 = (18)2{\left( {{1 \over 8}} \right)^2}(81​)2 = 164{1 \over {64}}641​

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