Question

Calculate the enthalpy change, ΔH, for the process in which 31.6 g of water is converted from liquid at 3.2 ∘C to vapor at 25.0°C . For water, ΔHvap = 44.0 kJ/mol at 25.0°C and Cs = 4.18 J/(g*°C) for H2O(l).

Answer 1

Enthalpy change = 74.36 kJ

Step-by-step explanation:

Enthalpy change is defines as the heat absorbed or evolved during a chemical reaction at constant pressure and volume

Reaction:

H2O(l) --> H2O(g)

DHr = mCsDT

Where m is the mass of water

Cs is the specific heat capacity

DT is the temperature difference

= 31.6 * 4.18 * (25 - 3.2)

= 2879.518 J

DHvap = 44kJ/mol

Moles of water = mass/molecular weight

Molecular weight = (1*2) + 16

= 18g/mol

Moles of water = 31.6/18

= 1.756 moles

DHvap = 44 * 1.756

= 77.244kJ

DH = DHproduct - DHreactant

DHproduct = DHvap = 77.24kJ

DHreactant = DHr = 2879.518J = 2.880kJ

= 77.24 - 2.880

DH = 74.36kJ

Enthalpy change = 74.36 kJ