Question

At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) H2S(g)↽−−⇀H2(g)+S(g) Initially, only H 2 S H2S is present at a pressure of 0.150 atm 0.150 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer:

Total pressure at equilibrium is 0.2798atm.

Explanation:

For the reaction:

H₂S(g) ⇄ H₂(g) + S(g)

Kp is defined as:

$Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834$

If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:

H₂S(g): 0.150atm - x

H₂(g): x

S(g): x

Replacing in Kp:

$\frac{X*X}{0.150atm-X} = 0.834$

X² = 0.1251 - 0.834X

X² +  0.834X - 0.1251 = 0

Solving for X:

X = -0.964 → False solution: There is no negative pressures

X = 0.1298

Thus, pressures are:

H₂S(g): 0.150atm - 0.1298atm = 0.0202atm

H₂(g): 0.1298atm

S(g): 0.1298atm

Thus, total pressure in the container at equilibrium is:

0.0202atm + 0.1298atm + 0.1298atm = 0.2798atm

Answer 2

Answer:

Total pressure = 0.2798 atm

Explanation:

Step 1: Data given

Kp = 0.834

Initially, only H2S is present at a pressure of 0.150 atm

Step 2: The balanced equation

H2S(g)⇄ H2(g) + S(g)

Step 3: The initial pressure

pH2S = 0.150 atm

pH2 = 0 atm

pS = 0 atm

Step 3: The pressure at the equilibrium

For 1 mol H2S we'll have 1 mo lH2 and 1 mol S

pH2S = 0.150 - X atm

pH2 = X atm

pS = X atm

Step 4: Calculate Kp

Kp = (pS*pH2)/(pH2S)

0.834 = X² / (0.150 - X)

X = 0.1298

pH2S = 0.150 - 0.1298 = 0.0202 atm

pH2 = 0.1298 atm

pS =0.1298 atm

Kp = (0.1298*0.1298) / 0.0202

Kp = 0.834

The total pressure = pH2S + pH2 + pS

Total pressure = 0.0202 atm+ 0.1298 atm+ 0.1298 atm

Total pressure = 0.2798 atm