Standard entropies of X₂, Y₂ and XY₅ are 70, 50 and 110 J K⁻¹ mol⁻¹ respectively. The temperature in Kelvin at which the reaction
1/2 X₂+5/2 Y₂ ⇌ XY₅ Δ H⊖=-35 kJ mol⁻¹
will be at equilibrium is __________ (Nearest integer)

Solution

Chemistry ma’am · 3 steps

At equilibrium Δ G = 0, so T = Δ H/Δ S. Just weight each entropy by its coefficient first.

1. Equilibrium condition. At equilibrium Δ G = Δ H - TΔ S = 0. So the temperature you need is T = Δ H/Δ S.
2. Entropy change. Multiply each standard entropy by its coefficient. Then take products minus reactants: Δ S = 110 - (1/2× 70 + 5/2× 50) = 110 - (35 + 125) = -50 J K⁻¹ mol⁻¹.

   T = (Δ H)/(Δ S) = (-35000 J mol⁻¹)/(-50 J K⁻¹ mol⁻¹) = 700 K

3. So. First convert Δ H to joules so that its units match Δ S. The reaction is at equilibrium at 700 K.

The common slip: 3500. Ignoring the coefficients 1/2 and 5/2 gives Δ S = 110 - 120 = -10 J K⁻¹ mol⁻¹. That leads to 3500 K, not 700 K.

The idea to keep. Equilibrium means Δ G = 0, so T = Δ H/Δ S. Always multiply entropies by their stoichiometric coefficients, and put Δ H and Δ S in the same energy unit.

Answer from the NTA final answer key. Solution by PrepWiser.