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?

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

i'm assuming that it's c.) red bank..