Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ?fusH = 6.03 kJ mol–1 at 0°C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 36.8 J mol–1 K– ?

To calculate the enthalpy change when freezing 1.0 mol of water at 10.0°C to ice at –10.0°C, we need to consider two main processes: the freezing of water at 0°C and the cooling of the ice from 0°C to –10°C. Let's break this down step by step.

Step 1: Freezing Water at 0°C

The first step involves the phase change from liquid water to solid ice at 0°C. The enthalpy change for this process is given by the heat of fusion, which is the energy required to change a substance from solid to liquid without changing its temperature. For water, this value is:

• ΔH_fus = 6.03 kJ/mol

Since we are freezing water, we will take this value as negative because energy is released during freezing:

• ΔH_freezing = -6.03 kJ/mol

Step 2: Cooling Ice from 0°C to –10°C

Next, we need to calculate the enthalpy change associated with cooling the ice from 0°C to –10°C. This involves using the specific heat capacity of ice:

• Cp [H2O(s)] = 36.8 J/mol·K

To find the enthalpy change for this cooling process, we can use the formula:

• ΔH = m × Cp × ΔT

Where:

• m = 1.0 mol
• ΔT = T_final - T_initial = -10°C - 0°C = -10 K

Now, substituting the values into the formula:

• ΔH_cooling = 1.0 mol × 36.8 J/mol·K × (-10 K)

This gives:

• ΔH_cooling = -368 J

To convert this to kJ, we divide by 1000:

• ΔH_cooling = -0.368 kJ

Step 3: Total Enthalpy Change

Now, we can find the total enthalpy change for the entire process by adding the enthalpy changes from both steps:

• Total ΔH = ΔH_freezing + ΔH_cooling

Substituting the values we calculated:

• Total ΔH = -6.03 kJ + (-0.368 kJ)

This results in:

• Total ΔH = -6.398 kJ

Final Result

The enthalpy change when freezing 1.0 mol of water at 10.0°C to ice at –10.0°C is approximately:

• ΔH = -6.40 kJ

This negative value indicates that the process is exothermic, meaning heat is released as the water freezes and cools down to the specified temperature.