A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI(g) leftharpoons H2(g) +I2(g)

Question

A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI(g) leftharpoons H2(g) +I2(g)  (A)  0.04  (B)  0.4  (C)  40  (D)  4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/47262fc90d9fd47535c37639b3c0c639-.png" /> 0.2-2x=0.04 ⇒ 2x=0.2-0.04 x=0.08 atm Kp=((0.08)(0.08)/(0.04)2)=4

Q. A sample of $H I_{(g)}$ is placed in a flask at a pressure of $0.2$ atm. At equilibrium, partial pressure of $H I_{(g)}$ is $0.04$ atm. What is $K_{p}$ for the given equilibrium ?
$2 H I_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}$

$0.04$

$0.4$

$40$

$4$

$0.2 - 2 x = 0.04$
$\Rightarrow 2 x = 0.2 - 0.04$
$x = 0.08 a t m$
$K_{p} = \frac{( 0.08 ) ( 0.08 )}{( 0.04 ) ^{2}} = 4$

Q. A sample of HI(g)​ is placed in a flask at a pressure of 0.2 atm. At equilibrium, partial pressure of HI(g)​ is 0.04 atm. What is Kp​ for the given equilibrium ? 2HI(g)​⇌H2(g)​+I2(g)​

0.04

0.4

40

4

0.2−2x=0.04 ⇒2x=0.2−0.04 x=0.08atm Kp​=(0.04)2(0.08)(0.08)​=4

Q. A sample of $H I_{(g)}$ is placed in a flask at a pressure of $0.2$ atm. At equilibrium, partial pressure of $H I_{(g)}$ is $0.04$ atm. What is $K_{p}$ for the given equilibrium ?
$2 H I_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}$

$0.04$

$0.4$

$40$

$4$

$0.2 - 2 x = 0.04$
$\Rightarrow 2 x = 0.2 - 0.04$
$x = 0.08 a t m$
$K_{p} = \frac{( 0.08 ) ( 0.08 )}{( 0.04 ) ^{2}} = 4$