Question

# A sample of $$\displaystyle HI\left( g \right)$$ is placed in a flask at a pressure of $$0.2\ atm$$. At equilibrium the partial pressure of $$\displaystyle HI\left( g \right)$$ is $$0.04\ atm$$. What is $$\displaystyle { K }_{ p }$$ for the given equilibrium?$$\displaystyle 2HI\left( g \right) \rightleftharpoons { H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right)$$

Solution

## Change in the pressure of HI is $$\displaystyle 0.2-0.04 = 0.16$$ atm.Equilibrium prtial pressure of hydrogen is $$\displaystyle \frac {0.16}{2}=0.08\ atm$$.Equilibrium partial pressure of iodine is 0.08 atm.$$\displaystyle K_p = \frac{P_{H_2}P_{I_2}}{P_{HI}^2} =\frac {0.08 \times 0.08}{(0.04)^2} = 4.0$$ChemistryNCERTStandard XI

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