Question

Calculate the standard enthalpy change for the reaction that occurs when 50 grams of ice at -10°C is heated to water at 25°C. The molar enthalpy of fusion of ice is 6.01 kJ/mol, and the specific heat capacity of water and ice are 4.184 J/g°C and 2.09  J/g°C, respectively. (Assume that the entire sample of ice is converted to water and that no heat is lost to the surroundings.)

Answer 1

To calculate the standard enthalpy change for this reaction, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the water from 0C to 25CFirst, let's calculate the energy required for each step:Step 1: Heating the ice from -10C to 0Cq1 = mass x specific heat capacity of ice x temperature changeq1 = 50 g x 2.09 J/gC x  0 -  -10  Cq1 = 50 g x 2.09 J/gC x 10Cq1 = 1045 JStep 2: Melting the ice at 0CFirst, we need to find the moles of ice:Molar mass of water  H2O  = 18.015 g/molmoles of ice = mass / molar massmoles of ice = 50 g / 18.015 g/molmoles of ice  2.775 molNow, we can calculate the energy required to melt the ice:q2 = moles x molar enthalpy of fusionq2 = 2.775 mol x 6.01 kJ/molq2 = 16.67 kJq2 = 16670 J  converted to Joules Step 3: Heating the water from 0C to 25Cq3 = mass x specific heat capacity of water x temperature changeq3 = 50 g x 4.184 J/gC x  25 - 0 Cq3 = 50 g x 4.184 J/gC x 25Cq3 = 5220 JNow, we can find the total enthalpy change by adding the energy required for each step:Total enthalpy change = q1 + q2 + q3Total enthalpy change = 1045 J + 16670 J + 5220 JTotal enthalpy change = 21935 JThe standard enthalpy change for the reaction is 21.935 kJ  rounded to three decimal places .