Answer:
Total pressure at equilibrium is 0.2798atm.
Explanation:
For the reaction:
H₂S(g) ⇄ H₂(g) + S(g)
Kp is defined as:
[tex]Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834[/tex]
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:
[tex]\frac{X*X}{0.150atm-X} = 0.834[/tex]
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:
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
