Question

At a certain temperature, the Kp for the decomposition of H2S is 0.861.

H₂S (g) <----> H₂(g) + S(g)

Initially, only H2S is present at a pressure of 0.179 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer 2

Given:

• Reaction:

  $${\text{H}}_2\text{S (g)}\leftrightarrow {\text{H}}_2\text{(g)}+\text{S (g)}$$

• Equilibrium Constant ($K_p$): 0.861

• Initial Pressure of H₂S ($P_{{\text{H}}_2\text{S, initial}}$): 0.179 atm

• Initial Pressures of Products:
  $P_{{\text{H}}_2,\text{initial}}=0$ atm, $P_{\text{S},\text{initial}}=0$ atm

• 
  

Step 1: Set Up the ICE Table

Let $x$ be the amount of H₂S that decomposes at equilibrium.

Species	Initial (atm)	Change (atm)	Equilibrium (atm)
H₂S	0.179	$-x$	$0.179-x$
H₂	0	$+x$	$x$
S	0	$+x$	$x$

Step 2: Write the $K_p$ Expression

$$K_p=\frac{P_{{\text{H}}_2}\cdot P_{\text{S}}}{P_{{\text{H}}_2\text{S}}}$$

Substitute equilibrium pressures:

$$0.861=\frac{x\cdot x}{0.179-x}=\frac{x^2}{0.179-x}$$

Step 3: Solve for $x$

Rearrange the equation:

$$x^2+0.861x-0.154=0$$

Use the quadratic formula ($x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$):

$$x=\frac{-0.861\pm \sqrt{(0.861{)}^2+4\times 0.154}}{2}$$$$x=\frac{-0.861\pm \sqrt{0.741+0.616}}{2}=\frac{-0.861\pm 1.164}{2}$$

Discard the negative root ($x$ must be positive):

$$x=\frac{0.303}{2}=0.1515\text{atm}$$

Step 4: Calculate Equilibrium Pressures

• $P_{{\text{H}}_2\text{S}}=0.179-0.1515=0.0275\text{atm}$

• $P_{{\text{H}}_2}=0.1515\text{atm}$

• $P_{\text{S}}=0.1515\text{atm}$

Step 5: Total Pressure at Equilibrium

$$P_{\text{total}}=P_{{\text{H}}_2\text{S}}+P_{{\text{H}}_2}+P_{\text{S}}=0.0275+0.1515+0.1515=0.3305\text{atm}$$

Answer:

The total pressure in the container at equilibrium is 0.330 atm (rounded to 3 significant figures).