For the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) ightleftharpoons 2…

For the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) ightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ at $721 \mathrm{~K}$, the value of equilibrium constant is 50 , when equilibrium concentration of both is $5 \mathrm{M}$. Value of $\mathrm{K}_{\mathrm{p}}$ under the same conditions will be
  1. $0.02$
  2. $0.2$
  3. 50
  4. $50 \mathrm{RT}$

Solution

$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) ightleftharpoons 2 \mathrm{HI}(\mathrm{g})$
$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}} ;$
$\Delta \mathrm{n}=2-2=0 ; \therefore \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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