Coffee Cup Calorimetry: Heat of Solution

Question

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee cup calorimeter.

In the laboratory a general chemistry student finds that when 6.84 g of K₂SO₄(s) are dissolved in 118.80 g of water, the temperature of the solution drops from 25.42 to 23.37 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.77 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of K₂SO₄(s) in kJ/mol.

Assume the specific heat of the solution is equal to the specific heat of water.

ΔH_dissolution =  _____ kJ/mol

Answer 1

q = mC∆T + Ccal∆T

q = heat = ?

m = mass of water = 118.80 g

C = specific heat of water =4.184 J/gº

∆T = change in temperature = 2.05º (this value is actually negative)

Ccal = calorimeter constant = 1.77 J/º

q = (118.80 g)(4184 J/gº)(2.05º) + (1.77 J/º)(2.05º) = 1019 J + 3.6 J

q = 1023 J This is the heat absorbed for 6.84 g of K₂SO₄

molar mass K₂SO₄ = 174 g/mol

moles K₂SO₄ present = 6.84 g x 1 mol/174 g = 0.0393 moles

∆H_dissolution = 1023 J / 0.0393 mol = 26,031 J/mol = 26.0 kJ/mol