Question

Calculate the standard enthalpy change for the phase transition reaction where 100.0 grams of solid water (ice) at -10.0 °C is heated to form 100.0 grams of liquid water at 10.0 °C, given the specific heat capacities of ice and water are 2.09 J/g·°C and 4.18 J/g·°C, respectively, and the enthalpy of fusion of water is 6.01 kJ/mol.

Answer 1

To calculate the standard enthalpy change for this phase transition, we need to consider three steps:1. Heating the ice from -10.0 C to 0 C2. Melting the ice at 0 C3. Heating the liquid water from 0 C to 10.0 CStep 1: Heating the ice from -10.0 C to 0 Cq1 = mass x specific heat capacity of ice x temperature changeq1 = 100.0 g x 2.09 J/gC x  0 -  -10.0 C  q1 = 100.0 g x 2.09 J/gC x 10.0 Cq1 = 2090 JStep 2: Melting the ice at 0 CFirst, we need to find the number of moles of water:Molar mass of water  H2O  = 18.02 g/molmoles = mass / molar massmoles = 100.0 g / 18.02 g/molmoles = 5.55 molNow, we can calculate the enthalpy change for melting the ice:q2 = moles x enthalpy of fusionq2 = 5.55 mol x 6.01 kJ/molq2 = 33.36 kJConvert to J: q2 = 33.36 kJ x 1000 J/kJ = 33360 JStep 3: Heating the liquid water from 0 C to 10.0 Cq3 = mass x specific heat capacity of water x temperature changeq3 = 100.0 g x 4.18 J/gC x  10.0 C - 0 C q3 = 100.0 g x 4.18 J/gC x 10.0 Cq3 = 4180 JNow, we can find the total enthalpy change for the entire process by adding the enthalpy changes for each step:Total enthalpy change = q1 + q2 + q3Total enthalpy change = 2090 J + 33360 J + 4180 JTotal enthalpy change = 39630 JThe standard enthalpy change for the phase transition reaction is 39,630 J or 39.63 kJ.