Question 75

Standard entropies of  $$X_2,\ Y_2$$  and  $$XY_5$$  are  $$70,\ 50$$  and  $$110\,J\,K^{-1}mol^{-1}$$ respectively. The temperature in Kelvin at which the reaction $$\frac{1}{2}X_2 + \frac{5}{2}Y_2 \rightleftharpoons XY_5 \Delta H^\ominus = -35\,kJ\,mol^{-1}$$ will be at equilibrium is $$\underline{\hspace{2cm}}$$  (Nearest integer).} 


Correct Answer: 700

Standard entropies: X₂=70, Y₂=50, XY₅=110 J/K/mol. Find T where ½X₂ + 5/2Y₂ → XY₅ is at equilibrium with ΔH = -35 kJ/mol.

ΔS = 110 - ½(70) - 5/2(50) = 110 - 35 - 125 = -50 J/K/mol

At equilibrium: ΔG = 0, so T = ΔH/ΔS = -35000/(-50) = 700 K

The answer is 700.

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