Question

In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2(g) + I2(g)

At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places.

HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values into the Kc expression.

Answer 1

The equilibrium constant (Kc) for the reaction where 2HI decomposes into H2 and I2 at a given temperature can be calculated using the ICE method. After setting up the Kc expression and solving for x, the concentration changes of HI, H2, and I2 can be used to find Kc, which is approximately 0.1176.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂), we must first write the Kc expression for the reaction:

2HI (g) ⇌ H₂ (g) + I₂ (g)

The Kc expression is Kc = [H₂][I₂] / [HI]^2.

Next, we use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction:

• Initial: [HI] = 0.822 mol / 1.11 L = 0.7409 M (since the flask is evacuated, [H₂] and [I₂] initially are 0)
• Change: At equilibrium, [HI] has decreased by x to (0.7409 - 2x) M, since HI decomposes into one mole of H₂ and one mole of I₂ for every two moles of HI that decompose.
• Equilibrium: [HI] = 0.055 M, so 0.7409 - 2x = 0.055 M. [H₂] and [I₂] both increase by x to x M given the stoichiometry of the reaction.

To calculate x, solve the equation 0.7409 - 2x = 0.055, yielding x = 0.34295 M. Now we know at equilibrium:

• [HI] = 0.055 M
• [H₂] = x = 0.34295 M
• [I₂] = x = 0.34295 M

Inserting these values into the Kc expression we get:Kc = (0.34295)(0.34295) / (0.055)^2

Simplifying gives us:Kc = 0.1176 to 4 decimal places.

Answer 2

Kc = 168.0749

Step-by-step explanation:

• 2HI(g) ↔ H2(g) + I2(g)

initial mol: 0.822 0 0

equil. mol: 2(0.822 - x) x x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749