Answer :

Given: Δ_{fus}H = 6.03 kJ mol^{-1} at 0^{°} C

C_{p} [H_{2}O(l)] = 75.3 J mol^{–1} K^{–1}

C_{p} [H_{2}O(s)] = 36.8 J mol^{–1} K^{–1}

To calculate the enthalpy change, we use Hess’s Law

Total ΔH = ΔH_{1} + ΔH_{2} + ΔH_{3}

Where ΔH = C_{p}ΔT

ΔC_{p} = Heat capacity at constant pressure

ΔT = T_{2}-T_{1}

∴ Total ΔH = (1 mol water at 10^{°} C →1 mol water at 0^{°} C) +

(1 mol water at 0^{°} C → 1 mol water at 0^{°} C) +

(1 mol water at 0^{°} C → 1 mol water at -10^{°} C)

∴ Total ΔH = C_{p}[H_{2}O (l)] × ΔT + ΔH_{freezing} + C_{p} [H_{2}O (s)] × ΔT

By substituting the given values, we get

Total ΔH = 75.3 J mol^{–1} K^{–1} × (0-10) K +(-6.03 kJ mol^{-1}) +36.8 J mol^{–1} K^{–1}× (-10) K

⇒Total ΔH = -753 Jmol^{–1} -6.03 kJ mol^{-1} -368 Jmol^{-1}

⇒Total ΔH = -0.753 kJmol^{–1} - 6.03 kJ mol^{-1} - 0.368 kJmol^{-1} (1J=)

⇒Total ΔH = -7.151 kJmol^{-1}

Thus, the enthalpy change involved in the process is -7.151 kJ/mol

Note: Hess’s Law states that “ if a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e., the total enthalpy change is independent of intermediate steps involved in the change. The enthalpy depends on the initial and final stages only.

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