Question

a. 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 16.11 g of BaCl₂(s) are dissolved in 118.70 g of water, the temperature of the solution increases from 22.98 to 25.41 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.71 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of BaCl₂(s) in kJ/mol.

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

ΔH_dissolution =  kJ/mol

b.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 3.22 g of CuSO₄(s) are dissolved in 110.60 g of water, the temperature of the solution increases from 24.18 to 27.36 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.64 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

Mass of BaCl2 = 16.77 g Number of moles of BaCl, = mass/mol.wt. = 16.77 g/(208.23 g/mol) = 0.080536 mol Mass of water = 118.70 g Mass of the solution = 118.70 + 16.77 = 135.47 g the specific heat of solution is same as that of water (4.184 J/(g.°C) Change in temperature = AT = 25.41 – 22.98 = 2.43 °C Heat associated with the calorimeter = C* AT = 1.71 J/°C * 2.43 °C = 4.1553 J Heat energy associated with this rise in temperature of solution = m*c* AT = 135.47 g x 4.184J.1.°C-1.x 2.43 °C = 1377.34 J At thermal equilibrium, qrxn + qwater + acal = 0 qrxn = -(water + qcal) = -(1377.34 + 4.1553 ) = - 1381.495 J Enthalpy of reaction = - 1381.495 J Since the temperature increases by the dissolution of BaCl2, this is an exothermic reaction. So AH solution=-q = -1381.495 J For 1 mole of BaCl2 AH solution = – 1381.495 / 0.080536 mol =-17153.77 J/mol = -17.15377 kJ/mol = - 17.15 kJ/mol AH solution = - 17.15 kJ/mol

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Mass of CuSO4 = 3.22 g Number of moles of CuSO4 = mass/mol.wt. = 3.22 g/(159.61 g/mol) = 0.020174 mol Mass of water = 110.60 g Mass of the solution = 110.60 + 3.22 = 113.82 g the specific heat of solution is same as that of water (4.184 J/(g.°C) Change in temperature = AT = 27.36 - 24.18 = 3.18 °C Heat associated with the calorimeter = C* AT = 1.64 J/°C * 3.18 °C = 5.2152 J Heat energy associated with this rise in temperature of solution = m*c* AT = 113.82 g x 4.184 J.g 1.°C-1, 3.18 °C = 1514.389 J At thermal equilibrium, qrxn + qwater + acal = 0 qrxn = -(water + qal) = -(1514.389 + 5.2152 ) = - 1519.604 J Enthalpy of reaction = - 1519.604 J Since the temperature increases by the dissolution of CuSO4, this is an exothermic reaction. So AH solution=-q = - 1519.604 J For 1 mole of CuSO, AH solution = – 1519.604 J/ 0.020174 mol =-75324.22 J/mol = - 75.32422 kJ/mol = - 75.32 kJ/mol AH solution = - 75.32 kJ/mol