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,Δ H=-30 kJ, to be at equilibrium, the temperature will be:

Question

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,Δ H=-30 kJ, to be at equilibrium, the temperature will be: (A) 1250 K (B) 500 K (C) 750 K (D) 1000 K

Tardigrade Admin · Accepted Answer

(1/2)X2+(3/2)Y2→ XY3 Δ S textreaction=50-((3/2)×40+(1/2)×60)=-40J mol-1 Δ G=Δ H-TΔ S At equilibrium Δ G = 0 Δ H=TΔ S 30× 103=T× 40 ∴ T=750

Q. Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are $60, 40$ and $50 J K^{- 1} m o l^{- 1},$ respectively, for the reaction
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3,} Δ H = - 30 k J,$ to be at equilibrium, the temperature will be:

$1250 K$

$500 K$

$750 K$

$1000 K$

$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}$
$Δ S_{reaction} = 50 - (\frac{3}{2} \times 40 + \frac{1}{2} \times 60) = - 40 J m o l^{- 1}$
$Δ G = Δ H - T Δ S < b r / >$ At equilibrium $Δ G = 0$
$Δ H = T Δ S$
$30 \times 1 0^{3} = T \times 40$
$∴ T = 750$

Q. Standard entropy of X2​,Y2​ and XY3​ are 60,40 and 50JK−1mol−1, respectively, for the reaction 21​X2​+23​Y2​→XY3,​ΔH=−30kJ, to be at equilibrium, the temperature will be:

1250K

500K

750K

1000K