For a reversible reaction X(g)+3Y(g)2Z(g); Δ H=-40 kJ, the standard entropies of X, Y and Z are 60,40 and 50 text JK-1mol-1 respectively. The temperature at which the above reaction attains equilibrium is about

Question

For a reversible reaction X(g)+3Y(g)2Z(g); Δ H=-40 kJ, the standard entropies of X, Y and Z are 60,40 and 50 text JK-1mol-1 respectively. The temperature at which the above reaction attains equilibrium is about (A) 400 K (B) 500 K (C) 273 K (D) 373 K

Tardigrade Admin · Accepted Answer

X(g)+3Y(g)2Z(g) Δ so=2Δ so(Z)- so(X)+3so(Y) =2× 50- 60+3× 40 =100-180 =-80 JK-1mol-1 Given Δ Ho=-40 kJ =40,000 J Δ Go=Δ Ho-TΔ So At equilibrium, Δ Go=0 ∴ Δ Ho=TΔ So or T=(Δ Ho/Δ so) =(40,000/80) =500K

Q. For a reversible reaction X(g)+3Y(g)2Z(g); ΔH=−40kJ, the standard entropies of X, Y and Z are 60,40 and 50JK−1mol−1 respectively. The temperature at which the above reaction attains equilibrium is about

400 K

500 K

273 K

373 K

X(g)​+3Y(g)2Z(g) Δso=2Δso(Z)−{so(X)+3so(Y)} =2×50−{60+3×40} =100−180 =−80JK−1mol−1 Given ΔHo=−40kJ =40,000J ΔGo=ΔHo−TΔSo At equilibrium, ΔGo=0 ∴ ΔHo=TΔSo or T=ΔsoΔHo​ =8040,000​ =500K

Q. For a reversible reaction $X (g) + 3 Y (g) 2 Z (g);$ $Δ H = - 40 k J,$ the standard entropies of X, Y and Z are 60,40 and $50 J K^{- 1} m o l^{- 1}$ respectively. The temperature at which the above reaction attains equilibrium is about