Sign in
Please select an account to continue using cracku.in
↓ →
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).}Â
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.
Create a FREE account and get:
Educational materials for JEE preparation