Question

Calculate the standard enthalpy change for the phase transition reaction, where 1 mole of solid water (ice) is transformed into 1 mole of liquid water at a constant pressure of 1 atm and a temperature of 298 K. Given the specific heat capacity of ice is 2.108 J g^-1 K^-1 and the specific heat capacity of water is 4.184 J g^-1 K^-1. The molar enthalpy of fusion of water is 6.01 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the phase transition reaction, we need to consider the following steps:1. Heating the ice from its initial temperature  assumed to be 273 K  to its melting point  273 K .2. Melting the ice at 273 K  fusion .3. Heating the liquid water from 273 K to the final temperature  298 K .Step 1: Heating the ice to its melting pointq1 = n * C_ice * Twhere n is the number of moles  1 mole , C_ice is the specific heat capacity of ice  2.108 J g^-1 K^-1 , and T is the temperature change  0 K, since the initial temperature is assumed to be 273 K .q1 = 1 * 2.108 * 0 = 0 JStep 2: Melting the ice  fusion q2 = n * H_fusionwhere H_fusion is the molar enthalpy of fusion  6.01 kJ/mol .q2 = 1 * 6.01 kJ = 6.01 kJStep 3: Heating the liquid water to the final temperatureq3 = n * C_water * Twhere C_water is the specific heat capacity of water  4.184 J g^-1 K^-1 , and T is the temperature change  298 K - 273 K = 25 K .q3 = 1 * 4.184 * 25 = 104.6 JNow, we can calculate the total enthalpy change for the phase transition reaction:H_total = q1 + q2 + q3H_total = 0 J + 6.01 kJ + 104.6 JConvert the joules to kilojoules:H_total = 0 kJ + 6.01 kJ + 0.1046 kJH_total = 6.1146 kJThe standard enthalpy change for the phase transition reaction is approximately 6.11 kJ.