Coffee Calorimetry

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 2.79 g of CuSO₄(s) are dissolved in 108.20 g of water, the temperature of the solution increases from 25.00 to 27.85 °C. 

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.65 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of CuSO₄(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

heat of reaction = C_cal x ∆T_cal + (m_H2O)(C_H2O)(∆T_H2O)

C_cal = calorimeter constant = 1.65 J/º

∆T_cal = change in temperature of calorimeter = 27.85º - 25.00º = 2.85º

m_H2O = mass H2O = 108.20 g

C_H2O = specific heat H₂O = 4.184 J/gº

∆T_H2O = change in temperature of H₂O = 27.85º - 2.85º = 2.85º

heat of reaction = ∆H_rxn = (1.65 J/º)(2.85º) + (108.20 g)(4.184 J/gº)(2.85º)

∆H_rxn = 4.70 J + 1290 J

∆H_rxn = 1295 J

Since the question asks for the value of ∆H_rxn in terms of kJ/mol, we must find the mols of reaction and also convert J to kJ

Reaction: CuSO₄(s) ==> Cu²⁺(aq) + SO₄^2-(aq) ... dissolution of copper sulfate

2.79 g CuSO₄ x 1 mol CuSO₄ / 159.6 g = 0.01748 mols

∆H_rxn = 1295 J / 0.01748 mols x 1 kJ/1000 J = 74.1 kJ/mol (3 sig. figs.)