18 g of ice is converted into water at 0 ^c | vedclass

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(A) The process is the fusion of ice: $H_2O_{(s)} \rightarrow H_2O_{(l)}$ at $T = 0 \ ^\circ C = 273 \ K$.The molar entropy change is $\Delta S_{fus}

Vedclass · Accepted Answer

$5951.4$

Vedclass · Answer

(A) The process is the fusion of ice: $H_2O_{(s)} ightarrow H_2O_{(l)}$ at $T = 0 \ ^\circ C = 273 \ K$.The molar entropy change is $\Delta S_{fus} = S_{product} - S_{reactant} = 60 - 38.2 = 21.8 \ J/mol \ K$.For a reversible phase transition at equilibrium,$\Delta S = \frac{\Delta H}{T}$.Therefore,$\Delta H = \Delta S 	imes T = 21.8 \ J/mol \ K 	imes 273 \ K = 5951.4 \ J/mol$.

$18 \ g$ of ice is converted into water at $0 \ ^\circ C$ and $1 \ atm$. The entropies of $H_2O_{(s)}$ and $H_2O_{(l)}$ are $38.2$ and $60 \ J/mol \ K$ respectively. The enthalpy change for this conversion is ..... $J/mol$.

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