Calculate the enthalpy change on freezing of 1.0 mole of water at 10.0 °C to ice at –10.0°C.
$\begin{array}{l}{\mathrm{\Delta }}_{\text{fus }}\mathrm{H}=6.03\mathrm{k}{\mathrm{Jmol}}^{-1}\text{ at }{0}^{\circ }\mathrm{C}\text{. }\\ {\mathrm{C}}_{\mathrm{p}}\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right]=75.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\\ {\mathrm{C}}_{\mathrm{p}}\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\right]=36.8\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\end{array}$

Freezing of water at 10⁰C to ice at  -10⁰C involves the following steps:

1 mole H₂O(10⁰C) $\underset{∆{\mathrm{H}}_1}{\overset{1}{\to }}$ 1 mole H₂O(0⁰C) $\underset{∆{\mathrm{H}}_2}{\overset{2}{\to }}$1mole ice(0⁰C) $\underset{∆{\mathrm{H}}_3}{\overset{3}{\to }}$

1 mole ice(10⁰C)

$∆{\mathrm{H}}_1$= nC_P$∆\mathrm{T}$ = 1$\times$75.3 $\times$(0-10)$\times$10^-3 = - 0.753 kJ

$∆{\mathrm{H}}_2 =$$-{∆}_{\mathrm{f}}\mathrm{H}$ = $-$6.03 kJ

$∆{\mathrm{H}}_2 =$ nC_P$∆\mathrm{T}$ = 1 $\times$36.8 $\times$(-10-0) $\times$10^-3 = - 0.368 kJ

$∆\mathrm{H}=∆{\mathrm{H}}_1+∆{\mathrm{H}}_2+∆{\mathrm{H}}_3$ = - 0.753 - 6.03 - 0.368 = - 7.151 kJ/mole

Explanation:

Given Data:

• ΔH_fus: 6.03 kJ/mol at 0°C
• C_p for water (liquid): 75.3 J/mol/K
• C_p for ice (solid): 36.8 J/mol/K

Step 1: Cool liquid water from 10.0°C to 0.0°C

The temperature change is:

ΔT = T_final - T_initial = 0.0 - 10.0 = -10.0 K

The heat released during this step is:

q₁ = n × C_p × ΔT

Substitute values:

q₁ = 1.0 × 75.3 × (-10.0) = -753.0 J

Convert to kilojoules:

q₁ = -0.753 kJ

Step 2: Freeze water at 0.0°C

The heat released during freezing is:

q₂ = -ΔH_fus

Substitute values:

q₂ = -6.03 kJ

Step 3: Cool ice from 0.0°C to –10.0°C

The temperature change is:

ΔT = T_final - T_initial = -10.0 - 0.0 = -10.0 K

The heat released during this step is:

q₃ = n × C_p × ΔT

Substitute values:

q₃ = 1.0 × 36.8 × (-10.0) = -368.0 J

Convert to kilojoules:

q₃ = -0.368 kJ

Step 4: Total Enthalpy Change

The total enthalpy change is:

ΔH_total = q₁ + q₂ + q₃

Substitute values:

ΔH_total = -0.753 kJ + (-6.03 kJ) + (-0.368 kJ)

ΔH_total = -7.151 kJ

Final Answer:

The enthalpy change for freezing 1 mole of water at 10.0°C to ice at –10.0°C is approximately –7.15 kJ.