The enthalpy change on freezing of 1 mol of | vedclass

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(C) The process involves three steps:$1$. Cooling $1 \ mol$ of liquid water from $5 \ ^\circ C$ to $0 \ ^\circ C$: $\Delta H_1 = C_p(l) 	imes \Delta

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$6.56$

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(C) The process involves three steps:$1$. Cooling $1 \ mol$ of liquid water from $5 \ ^\circ C$ to $0 \ ^\circ C$: $\Delta H_1 = C_p(l) 	imes \Delta T = 75.3 \ J \ mol^{-1} \ K^{-1} 	imes (0 - 5) \ K = -376.5 \ J \ mol^{-1} = -0.3765 \ kJ \ mol^{-1}$.$2$. Freezing $1 \ mol$ of liquid water at $0 \ ^\circ C$ to ice at $0 \ ^\circ C$: $\Delta H_2 = -\Delta _{fus}H = -6 \ kJ \ mol^{-1}$.$3$. Cooling $1 \ mol$ of ice from $0 \ ^\circ C$ to $-5 \ ^\circ C$: $\Delta H_3 = C_p(s) 	imes \Delta T = 36.8 \ J \ mol^{-1} \ K^{-1} 	imes (-5 - 0) \ K = -184 \ J \ mol^{-1} = -0.184 \ kJ \ mol^{-1}$.Total enthalpy change $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -0.3765 - 6 - 0.184 = -6.5605 \ kJ \ mol^{-1}$.The magnitude of the enthalpy change is $6.56 \ kJ \ mol^{-1}$.

The enthalpy change on freezing of $1 \ mol$ of water at $5 \ ^\circ C$ to ice at $-5 \ ^\circ C$ is ..... $kJ \ mol^{-1}$.
(Given $\Delta _{fus}H = 6 \ kJ \ mol^{-1}$ at $0 \ ^\circ C$,
$C_p(H_2O, l) = 75.3 \ J \ mol^{-1} \ K^{-1}$,
$C_p(H_2O, s) = 36.8 \ J \ mol^{-1} \ K^{-1}$)

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The enthalpy change on freezing of $1 \ mol$ of water at $5 \ ^\circ C$ to ice at $-5 \ ^\circ C$ is ..... $kJ \ mol^{-1}$.(Given $\Delta _{fus}H = 6 \ kJ \ mol^{-1}$ at $0 \ ^\circ C$,$C_p(H_2O, l) = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p(H_2O, s) = 36.8 \ J \ mol^{-1} \ K^{-1}$)