18 g of ice is converted into water at 0° C and 1 atm. The entropies of H2O (s) and H2O (l) are 38.2 and 60 J text/mol text K respectively. The enthalpy change for this conversion is

Question

18 g of ice is converted into water at 0° C and 1 atm. The entropies of H2O (s) and H2O (l) are 38.2 and 60 J text/mol text K respectively. The enthalpy change for this conversion is (A) 595.14 J/mol (B) - 5951.4 J/mol (C) 5951.4 J/mol (D) - 595.14 J/mol

Tardigrade Admin · Accepted Answer

At equilibrium Δ H=T⋅ Δ S H2O(s) xrightarrowH2O(l) Δ S=SH2O(l)-SH2O(s) =(60-38.2) J text/mol text-K=21.8 J text/mol text-K Δ H=273× (21.8)=5951.4 J text/mol

Q. 18 g of ice is converted into water at $0^{\circ} C$ and 1 atm. The entropies of $H_{2} O (s)$ and $H_{2} O (l)$ are 38.2 and $60 J / m o l K$ respectively. The enthalpy change for this conversion is

595.14 J/mol

- 5951.4 J/mol

5951.4 J/mol

- 595.14 J/mol

At equilibrium $Δ H = T \cdot Δ S$ $H_{2} O (s) H_{2} O (l)$ $Δ S = S_{H_{2} O (l)} - S_{H_{2} O (s)}$ $= (60 - 38.2) J / m o l - K = 21.8 J / m o l - K$ $Δ H = 273 \times (21.8) = 5951.4 J / m o l$

Q. 18 g of ice is converted into water at 0∘C and 1 atm. The entropies of H2​O(s) and H2​O(l) are 38.2 and 60J/molK respectively. The enthalpy change for this conversion is