At 298 K, the equilibrium constant is 2 × 10 for the reaction: Cu(s) + 2Ag (aq) ⇌ Cu (aq) + 2Ag(s) T — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Write the two reactions and identify their relationship** Reaction 1: Cu(s) + 2Ag⁺(aq) ⇌ Cu²⁺(aq) + 2Ag(s) K₁ = 2 × 10¹⁵ Reaction 2: Cu²⁺(aq) + Ag(s) ⇌ Cu(s) + Ag⁺(aq) K₂ = x × 10⁻⁸ **Step 2: Determine how Reaction 2 relates to Reaction 1** Reaction 2 is half of Reaction 1 (divide all coefficients by 2). **Step 3: Apply the relationship between K values** When a reaction is divided by n, the equilibrium constant becomes: **K_new = (K_original)^(1/n)** Since Reaction 2 is Reaction 1 divided by 2: K₂ = (K₁)^(1/2) **Step 4: Calculate K₂** K₂ = (2 × 10¹⁵)^(1/2) K₂ = √(2 × 10¹⁵) K₂ = √2 × √(10¹⁵) K₂ = √2 × 10^(15/2) K₂ = √2 × 10^7.5 K₂ = 1.414 × 10^7.5 **Step 5: Express in the form x × 10⁻⁸** 10^7.5 = 10⁸ × 10^(-0.5) = 10⁸/√10 K₂ = 1.414 × 10⁸/√10 = (1.414/√10) × 10⁸ ≈ 0.447 × 10⁸ = 4.47 × 10⁷ Alternatively: K