Chemical Thermodynamics

Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 J K–1 mol–1, respectively. For the reaction, 1/2 X2 + 3/2 Y2 → XY3, ΔH = –30 kJ, to be at equilibrium, the temperature will be

  1. 500 K
  2. 750 K
  3. 1000 K
  4. 1250 K

Answer

1/2X2 + 3/2Y2 → XY3

ΔS°reaction  = ΔS°products - ΔS°reactants

= 50 – 60/2 - 3*40/2

= -40 J/mol

ΔG = ΔH – TΔS

ΔH = -30 kJ, ΔS = -40J/(K mol)

At equilibrium, ΔG = 0

T = ΔH/ΔS = 750 K

The correct option is B.