Standard entropies of x2,y2 and xy3 are 70, 50 and 60 J K- 1mol- 1 respectively. For the reaction (1/2)x2+(3/2)y2leftharpoons xy3.Δ H=-30kJ to be at equilibrium, the temperature should be

Question

Standard entropies of x2,y2 and xy3 are 70, 50 and 60 J K- 1mol- 1 respectively. For the reaction (1/2)x2+(3/2)y2leftharpoons xy3.Δ H=-30kJ to be at equilibrium, the temperature should be (A) 450 K (B) 600 K (C) 1200 K (D) 300 K

Tardigrade Admin · Accepted Answer

Δ S=Sx y3-(1/2)Sx2-(3/2)Sy2 Δ S=60-(1/2)× 70-(3/2)× 50 =-50 JK- 1 mol- 1 At equilibrium Δ G=0 Δ G=Δ H-TΔ S 0=-30-(.-50× 10- 3× T.) T=600 textK

Q. Standard entropies of $x_{2}, y_{2}$ and $x y_{3}$ are 70, 50 and 60 J $K^{- 1} m o l^{- 1}$ respectively. For the reaction $\frac{1}{2} x_{2} + \frac{3}{2} y_{2} ⇌ x y_{3} .Δ H = - 30 k J$ to be at equilibrium, the temperature should be

450 K

600 K

1200 K

300 K

$Δ S = S_{x y_{3}} - \frac{1}{2} S_{x_{2}} - \frac{3}{2} S_{y_{2}}$

$Δ S = 60 - \frac{1}{2} \times 70 - \frac{3}{2} \times 50$

$= - 50$ $J K^{- 1} m o l^{- 1}$

At equilibrium $Δ G = 0$

$Δ G = Δ H - T Δ S$

$0 = - 30 - (- 50 \times 1 0^{- 3} \times T)$

$T = 600 K$

Q. Standard entropies of x2​,y2​ and xy3​ are 70, 50 and 60 J K−1mol−1 respectively. For the reaction 21​x2​+23​y2​⇌xy3​.ΔH=−30kJ to be at equilibrium, the temperature should be