Chemical Equilibrium — A/L Chemistry

How do I practise Chemical Equilibrium?

1 question with a worked solution

Check before you answer: State Symbols & Expression: Strict inclusion of state symbols (g, aq) and explicit exclusion of pure solids (s) and pure liquids (l) from Kc and Kp expressions.

1. At 700 K, the equilibrium constant K_c for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is 64.0. If 1.00 mol of H₂ and 1.00 mol of I₂ are introduced into an evacuated 1.00 dm³ rigid reaction vessel at 700 K, what is the equilibrium concentration of HI?
   1. A) 0.80 mol dm⁻³
   2. B) 0.40 mol dm⁻³
   3. C) 1.20 mol dm⁻³
   4. D) 1.60 mol dm⁻³
   Check my answer

   Correct answer: D) 1.60 mol dm⁻³

   Let x mol dm⁻³ of H₂ react. At equilibrium: [H₂] = 1.00 - x, [I₂] = 1.00 - x, and [HI] = 2x. K_c = [HI]² / ([H₂][I₂]) = (2x)² / (1.00 - x)² = 64.0. Taking square root of both sides: 2x / (1.00 - x) = 8.00 ⇒ 2x = 8.00 - 8.00x ⇒ 10.00x = 8.00 ⇒ x = 0.80 mol dm⁻³. Therefore, [HI] = 2x = 2(0.80) = 1.60 mol dm⁻³.

   Show worked solution

   Step 1: Set up the ICE table (Initial, Change, Equilibrium) in mol dm⁻³ (V = 1.00 dm³): H₂(g) + I₂(g) ⇌ 2HI(g) Initial: 1.00 1.00 0 Change: -x -x +2x Equil: 1.00 - x 1.00 - x 2x Step 2: Write expression for equilibrium constant: K_c = [HI]² / ([H₂][I₂]) = (2x)² / (1.00 - x)² = 64.0 Step 3: Solve algebraically taking positive square root: 2x / (1.00 - x) = √64.0 = 8.00 2x = 8.00(1.00 - x) = 8.00 - 8.00x 10.00x = 8.00 ⇒ x = 0.80 mol dm⁻³ Step 4: Compute equilibrium concentration of HI: [HI] = 2x = 2 × 0.80 = 1.60 mol dm⁻³. Answer: 1.60 mol dm⁻³.
   Modelled on: the A/L Chemistry examination pattern

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Grade 13 · Term 1