A certain reaction is at equilibrium at 82° C and the enthalpy change for this reaction is 21.3 kJ. The value of Δ S(JK-1mol-1) for the reaction is

Question

A certain reaction is at equilibrium at 82° C and the enthalpy change for this reaction is 21.3 kJ. The value of Δ S(JK-1mol-1) for the reaction is (A) 55.0 (B) 60.0 (C) 68.5 (D) 120.0

Tardigrade Admin · Accepted Answer

Gibbs free energy. Δ G=Δ H-TΔ S But, at equilibrium, Δ G=0 ∴ Δ H=TΔ S or Δ S=(Δ H/T)=(21.3× 103/355)=60 JK-1mol-1

Q. A certain reaction is at equilibrium at $82^{\circ} C$ and the enthalpy change for this reaction is 21.3 kJ. The value of $Δ S (J K^{- 1} m o l^{- 1})$ for the reaction is

55.0

60.0

68.5

120.0

Gibbs free energy. $Δ G = Δ H - T Δ S$ But, at equilibrium, $Δ G = 0$ $∴$ $Δ H = T Δ S$ or $Δ S = \frac{Δ H}{T} = \frac{21.3 \times 10 ^{3}}{355} = 60 J K^{- 1} m o l^{- 1}$

Q. A certain reaction is at equilibrium at 82∘C and the enthalpy change for this reaction is 21.3 kJ. The value of ΔS(JK−1mol−1) for the reaction is