$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})$ 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

• Correct answer
  A

  $6.4$
• B

  $0.8$
• C

  $3.2$
• D

  $1.6$

Hence, the correct answer is option: 'A'.