Answer to Question #107100 in General Chemistry for mya 13

Question

At a certain temperature, the Kp for the decomposition of H2S is 0.719.

H2S(g)−⇀H2(g)+S(g)

Initially, only H2S is present at a pressure of 0.193 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

$H_2S(g)\longrightarrow H_2(g) + S(g)$

here it is given as K_p=0.719 atm and p(H₂S)=.193 atm

so we will take x part of H₂S is change,

so we can write the equilibrium reaction as

$H_2S(g)\longrightarrow H_2(g) + S(g)$

0.193 0 0

0.193-x x x

we can write for equilibrium constant Kp as

$K_p= \frac{P_{H_2}\times P_{S}}{P_{H_2S}}$

$0.719= \frac{x\times x}{0.193-x}$

0.139-0.719$x=x^2$

$x^2+0.719x-0.139=0$

using quadratic formula we can find

$x=\frac{-0.719 \pm\sqrt{(.719^2-(4\times1\times(-.139)))}}{2}$

x= $\frac{-0.719+1.04}{2}=0.1605,x=\frac{-0.719-1.04}{2}=-0.8795$

so here , $P_{H_2}=P_S=0.1605 atm, and P_{H_2S}=0.719-0.1605=0.5585 atm$

Here total pressure will be sum of all

P_T= 0.1605+0.1605+0.5585=0.8795 atm

$P_T= 0.8795 atm$