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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.

$$\displaystyle K_p = \frac{P_{H_2}P_{I_2}}{P_{HI}^2} =\frac {0.08 \times 0.08}{(0.04)^2} = 4.0$$

Chemistry
NCERT
Standard XI

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