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Question: Answered & Verified by Expert
$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})$
The equilibrium constant of the above reaction is $6.4$ at $300 \mathrm{~K}$. If $0.25$ mole each of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are added to the system, the equilibrium constant will be
ChemistryChemical EquilibriumKCETKCET 2009
Options:
  • A $6.4$
  • B $0.8$
  • C $3.2$
  • D $1.6$
Solution:
1747 Upvotes Verified Answer
The correct answer is: $6.4$
The value of equilibrium constant remains constant for a given reaction at constant temperature.

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