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An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine -

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Question

An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide.

\[\ce{H2 + I2 <=> 2HI}\]

The equilibrium constant for the reaction is:

Options

  • 36.0

  • 15.0

  • 0.067

  • 0.028

MCQ

Solution

36.0

Explanation:

  \[\ce{H2}\] \[\ce{+}\] \[\ce{I2}\] \[\ce{<=>}\] \[\ce{2HI}\]
At equ. `0.4/5`   `0.4/5`   `2.4/5`

Kc = `(2.4/5)^2/((0.4/5)(0.4/5))`

Kc = `(2.4 xx 2.4)/(0.4 xx 0.4)`

Kc = `5.76/0.16`

Kc = 36

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Law of Chemical Equilibrium and Equilibrium Constant
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