The value of Delta S for freezing of 10 g of | vedclass

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(D) The process is the freezing of water at its melting point,which is a reversible process at equilibrium.$\Delta H_{	ext{freezing}} = -\Delta H_{

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$-12.25$

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(D) The process is the freezing of water at its melting point,which is a reversible process at equilibrium.$\Delta H_{	ext{freezing}} = -\Delta H_{	ext{fusion}} = -80 \ cal/g 	imes 10 \ g = -800 \ cal$.The temperature $T = 0 \ ^\circ C = 273 \ K$.$\Delta S_{	ext{freezing}} = \frac{\Delta H_{	ext{freezing}}}{T} = \frac{-800 \ cal}{273 \ K} \approx -2.93 \ cal/K$.To convert to $J/K$,use the conversion factor $1 \ cal = 4.184 \ J$.$\Delta S = -2.93 \ cal/K 	imes 4.184 \ J/cal \approx -12.25 \ J/K$.

The value of $\Delta S$ for freezing of $10 \ g$ of $H_2O_{(l)}$ (enthalpy of fusion is $80 \ cal/g$) at $0 \ ^\circ C$ and $1 \ atm$ pressure is.......$J/K$

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