At 0° C, ice and water are in equilibrium and Δ S and Δ G for the conversion of ice to liquid water are

Question

At 0° C, ice and water are in equilibrium and Δ S and Δ G for the conversion of ice to liquid water are (A) 0,21.98 JK -1 mol -1 (B) -ve, zero (C) +v e, 21.98 JK -1 mol -1 (D) Zero, zero

Tardigrade Admin · Accepted Answer

Since the given process is in equilibrium, i.e., Δ G=0 Δ G=Δ H-T Δ S or, 0=Δ H-T Δ S or Δ H=T Δ S or, Δ S=(Δ H /T)=(6/273) × 103 JK -1 mol -1=21.98 JK -1 mol -1

Q. At $0^{\circ} C$ , ice and water are in equilibrium and $Δ S$ and $Δ G$ for the conversion of ice to liquid water are

$0, 21.98 J K^{- 1} m o l^{- 1}$

-ve, zero

$+ v e, 21.98 J K^{- 1} m o l^{- 1}$

Zero, zero

Since the given process is in equilibrium, i.e., $Δ G = 0$

$Δ G = Δ H - T Δ S$

or, $0 = Δ H - T Δ S$ or $Δ H = T Δ S$ or, $Δ S = \frac{Δ H}{T} = \frac{6}{273} \times 1 0^{3} J K^{- 1}$

$m o l^{- 1} = 21.98 J K^{- 1} m o l^{- 1}$

Q. At 0∘C, ice and water are in equilibrium and ΔS and ΔG for the conversion of ice to liquid water are

0,21.98JK−1mol−1