Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A sample of $H{I_{\left( g \right)}}$ is placed in a flask at a pressure of $0.2$ atm . At equilibrium the partial pressure of $H{I_{\left( g \right)}}$ is $0.04$ atm . What is ${K_p}$ for the given equilibrium ?$2H{I_{\left( g \right)}} \rightleftharpoons {H_2}_{\left( g \right)} + {I_{2\left( g \right)}}$ :A. 6B. 16C. 4D. 2

Last updated date: 20th Jun 2024
Total views: 403.8k
Views today: 9.03k
Verified
403.8k+ views
Hint: ${K_p}$ is the equilibrium constant calculated from the partial pressure of a reaction equation . It is used to express the relationship between product pressure and reactant pressure.

Complete step by step answer: At the initial concentration , $HI$ has a pressure of $0.2$ atm. At equilibrium , it has a partial pressure of $0.04$ atm. Therefore , a decrease in the pressure of $HI$ is $0.2 - 0.04 = 0.16$ atm.
The given reaction is :-
Initial Conc. $2H{I_{\left( g \right)}} \rightleftharpoons {H_{2\left( g \right)}} + {I_{2\left( g \right)}}$
$0.2$atm- $0$ $0$ (for both the species in the product)
At equilibrium $0.04$ atm- $\dfrac{{0.16}}{2}$ $\dfrac{{0.16}}{2}$ (for both the species in the product)
Acc to the formula of ${K_p}$
${K_p} = \dfrac{{{P_{{H_2}}} \times {P_{{I_2}}}}}{{{p_{HI}}}}$
Here ${P_{{H_2}}}$ - partial pressure of ${H_2}$ - $0.08$ atm.
${P_{{I_2}}}$ - partial pressure of ${I_2}$ - $0.08$atm.
${P_{HI}}$ - partial pressure of $HI$ - $0.04$ atm.
On substituting the values in the formula , we get
$\Rightarrow$ ${K_p} = \dfrac{{0.08 \times 0.08}}{{{{\left( {0.04} \right)}^2}}} = \dfrac{{0.0064}}{{0.0016}} = 4$
So , the correct answer is (C).