Question

Unit 5 Activity 2: Calculations Involving Keq and Factors [7 marks] 2. 1.2 mol of gaseous hydrogen fluoride (HF(g)) is injected into a 1.9 L reaction vessel and allowed to reach equilibrium. If the equilibrium constant Keq = 7.5, calculate the equilibrium concentrations of all species in the reaction. [7] HF(g) ⇌ H2(g) + F2(g)

Answer 1

To find the equilibrium concentrations of the species in the reaction \( HF(g) \rightleftharpoons H_2(g) + F_2(g) \), we'll follow these steps:

1. Calculate the initial concentration of HF. Given:

   • Amount of \( HF \) = 2.1 mol
   • Volume of the reaction vessel = 1.9 L

   The initial concentration of \( HF \): \[ [HF]_{initial} = \frac{n}{V} = \frac{2.1 , \text{mol}}{1.9 , \text{L}} \approx 1.105 , \text{M} \]

2. Set up the expression for the equilibrium. Let \( x \) be the amount of \( HF \) that dissociates at equilibrium. The stoichiometry of the reaction tells us that for every 1 mole of \( HF \) that reacts, 1 mole of \( H_2 \) and 1 mole of \( F_2 \) are produced.

   At equilibrium, the concentrations will be:

   • \([HF] = [HF]_{initial} - x = 1.105 - x\)
   • \([H_2] = x\)
   • \([F_2] = x\)

3. Write the equilibrium expression using \( K_{eq} \). The equilibrium constant for the reaction is given by: \[ K_{eq} = \frac{[H_2][F_2]}{[HF]} \] Substituting the concentrations at equilibrium into the expression gives: \[ 7.5 = \frac{x \cdot x}{1.105 - x} \] This simplifies to: \[ 7.5 = \frac{x^2}{1.105 - x} \]

4. Clear the fraction by multiplying both sides by \( (1.105 - x) \): \[ 7.5(1.105 - x) = x^2 \] Expanding this, we have: \[ 8.2875 - 7.5x = x^2 \] Rearranging gives us a quadratic equation: \[ x^2 + 7.5x - 8.2875 = 0 \]

5. Use the quadratic formula to solve for \( x \): The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1 \), \( b = 7.5 \), and \( c = -8.2875 \). Plugging in the values: \[ x = \frac{-7.5 \pm \sqrt{(7.5)^2 - 4 \cdot 1 \cdot (-8.2875)}}{2 \cdot 1} \] \[ x = \frac{-7.5 \pm \sqrt{56.25 + 33.15}}{2} \] \[ x = \frac{-7.5 \pm \sqrt{89.4}}{2} \] \[ x = \frac{-7.5 \pm 9.43}{2} \]

   Calculating the two possible values for \( x \): \[ x_1 = \frac{-7.5 + 9.43}{2} \approx 0.965 \quad (positive ,solution) \] \[ x_2 = \frac{-7.5 - 9.43}{2} \quad (negative ,solution, ,not ,valid) \] Thus, we take \( x \approx 0.965 , \text{M} \).

6. Calculate the equilibrium concentrations:

   • \([HF] = 1.105 - x \approx 1.105 - 0.965 \approx 0.140 , \text{M}\)
   • \([H_2] = x \approx 0.965 , \text{M}\)
   • \([F_2] = x \approx 0.965 , \text{M}\)

7. Final Results for equilibrium concentrations:

   • \([HF]_{eq} \approx 0.140 , \text{M}\)
   • \([H_2]_{eq} \approx 0.965 , \text{M}\)
   • \([F_2]_{eq} \approx 0.965 , \text{M}\)

These are the equilibrium concentrations for the reaction.