At a certain temperature, the Kp for the decomposition of H2S is 0.813.

H2S (Two wa arrow) H2(g) + S(g)
Initially, only H2S is present at a pressure of 0.280 atm in a closed container. What is the total pressure in the container at equilibrium?

Initially:

H2S = 0.28
H2(g) = 0
S(g) = 0
At equilibrium:
H2S = 0.28 - x
H2(g) = x
S(g) = x

Kp = Products/Reactants = ([x][x])/(0.28-x)
x^2 +0.813x-0.22764 = 0
x =(-b plus or minus sqrt(b^2 -4ac))/2a
x = 0.220303199 or approximately 0.220

What is the total pressure in the container at equilibrium?
Pressure of the reactants + the products
so
x + x + (0.28-x)
0.220 + 0.220 + (0.280 - 0.220)
= 0.5
This is the total pressure at equilibrium.
I hope this helps Janavi.

Thank you sooo much Karan!

To determine the total pressure at equilibrium, we need to use the equilibrium constant (Kp) and the initial pressure of the reactant.

The balanced equation for the decomposition of H2S is:
H2S (g) ⟶ H2 (g) + S (g)

Given:
Kp = 0.813
Initial Pressure of H2S (P(H2S)) = 0.280 atm

At equilibrium, we assume that x moles of H2S decompose, resulting in the formation of x moles of H2 and x moles of S.

Therefore,
[H2S] = (0.280 - x) atm
[H2] = x atm
[S] = x atm

Using the Kp expression:
Kp = ([H2] * [S]) / [H2S]

Plugging in the values, we get:
0.813 = (x * x) / (0.280 - x)

Now, we can solve this equation to find the value of x, which represents the number of moles of H2S that decomposed.

1. Multiply both sides by (0.280 - x):
0.813 * (0.280 - x) = x^2

2. Expand and rearrange the equation:
0.22944 - 0.813x = x^2

3. Move all terms to one side to form a quadratic equation:
x^2 + 0.813x - 0.22944 = 0

4. Solve the quadratic equation using the quadratic formula:
x = [-0.813 ± √(0.813^2 - 4 * 1 * -0.22944)] / (2 * 1)

By solving the quadratic equation, we find:
x ≈ 0.2009

Now, we can calculate the equilibrium pressures:
P(H2) = x ≈ 0.2009 atm
P(S) = x ≈ 0.2009 atm

To find the total pressure at equilibrium, we sum the partial pressures:
Total Pressure = P(H2) + P(S) + P(H2S)

Total Pressure = 0.2009 atm + 0.2009 atm + 0.280 atm

Total Pressure ≈ 0.6818 atm

Therefore, the total pressure in the container at equilibrium is approximately 0.6818 atm.

To find the total pressure in the container at equilibrium, we need to use the equilibrium constant (Kp) and the initial pressure of H2S. The equilibrium constant can be used to calculate the partial pressures of the products at equilibrium.

The equilibrium constant expression for the given reaction is:

Kp = (P(H2) * P(S)) / P(H2S)

We are given the Kp value as 0.813, and the initial pressure of H2S as 0.280 atm. We need to calculate the partial pressures of H2 and S at equilibrium.

Let's assume the equilibrium partial pressures of H2 and S as P(H2) and P(S), respectively.

Substitute these values into the equilibrium constant expression:

0.813 = (P(H2) * P(S)) / 0.280

Now, we need to make another assumption. Since we only have H2S initially, we can assume that the reaction proceeds to the extent that the initial pressure of H2S is completely consumed. Therefore, we can set P(H2S) = 0 - meaning it completely decomposes into H2 and S.

0.813 = (P(H2) * P(S)) / 0

Since any number divided by 0 is undefined, we can conclude that P(H2) and P(S) must also be 0 to satisfy the equilibrium constant expression. In other words, at equilibrium, the partial pressures of both H2 and S will be 0.

Therefore, the total pressure at equilibrium in the container is 0.