To calculate the standard enthalpy change for the fusion reaction, we need to consider the following steps:1. Melting the ice at 0C2. Heating the melted ice now water from 0C to the final temperature3. Cooling the liquid water from 25C to the final temperatureFirst, let's find the moles of ice and water:Molar mass of water H2O = 18.015 g/molMoles of ice = 50 g / 18.015 g/mol = 2.776 molMoles of water = 450 g / 18.015 g/mol = 24.986 molStep 1: Melting the ice at 0CEnthalpy change for melting ice = moles of ice molar enthalpy of fusionH1 = 2.776 mol 6.01 kJ/mol = 16.68 kJStep 2: Heating the melted ice now water from 0C to the final temperatureLet Tf be the final temperature.q2 = moles of melted ice specific heat capacity of water Tf - 0C H2 = 2.776 mol 4.18 J/gC 18.015 g/mol Tf - 0C Step 3: Cooling the liquid water from 25C to the final temperatureq3 = moles of water specific heat capacity of water Tf - 25C H3 = 24.986 mol 4.18 J/gC 18.015 g/mol Tf - 25C Since there are no heat losses to the surroundings, the heat gained by the ice must equal the heat lost by the water:H1 + H2 = -H316.68 kJ + 2.776 mol 4.18 J/gC 18.015 g/mol Tf - 0C = - 24.986 mol 4.18 J/gC 18.015 g/mol Tf - 25C Now, we need to solve for Tf:16.68 kJ 1000 J/kJ + 2.776 4.18 18.015 Tf = 24.986 4.18 18.015 Tf - 25 16680 J + 208.58 Tf = 1874.16 Tf - 1874.16 2516680 J = 1665.58 Tf - 46854Tf = 16680 + 46854 / 1665.58Tf 32.1CNow that we have the final temperature, we can calculate the enthalpy changes for steps 2 and 3:H2 = 2.776 mol 4.18 J/gC 18.015 g/mol 32.1 - 0 H2 208.58 32.1H2 6695.5 J = 6.695 kJH3 = 24.986 mol 4.18 J/gC 18.015 g/mol 32.1 - 25 H3 1874.16 7.1H3 -13307.5 J = -13.307 kJThe standard enthalpy change for the fusion reaction is the sum of the enthalpy changes for all three steps:H = H1 + H2 + H3H = 16.68 kJ + 6.695 kJ - 13.307 kJH 10.068 kJ