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In the laboratory a general chemistry student finds that when 21.19 g BaBr2(s) is dissolved in 101.00 g water, the temperature of the solution increases from 24.95 to 28.33 °c. the heat capacity of the calorimeter was determined in a separate experiment to be 1.74 J/°C. Based on the student's observation, calculate the dissolution enthalpy of BaBr2(s) in kj/mol.
Accepted Answer
The dissolution enthalpy of BaBr2(s) can be calculated using the following steps:
1. Calculate the heat absorbed by the solution:
ΔT = 28.33 °C - 24.95 °C = 3.38 °C
q_solution = m
c
ΔT
q_solution = (101.00 g + 21.19 g)
4.184 J/(g °C)
3.38 °C
q_solution = 1755.6 J
2. Calculate the heat absorbed by the calorimeter:
q_calorimeter = C
ΔT
q_calorimeter = 1.74 J/°C
3.38 °C
q_calorimeter = 5.88 J
3. Calculate the total heat absorbed (q_rxn):
q_rxn = q_solution + q_calorimeter
q_rxn = 1755.6 J + 5.88 J
q_rxn = 1761.48 J
4. Calculate the moles of BaBr2:
moles of BaBr2 = 21.19 g / 297.14 g/mol
moles of BaBr2 = 0.0713 mol
5. Calculate the enthalpy change per mole of BaBr2:
ΔH = q_rxn / moles of BaBr2
ΔH = 1761.48 J / 0.0713 mol
ΔH = 24717.5 J/mol
ΔH = 24.72 kJ/mol
Therefore, the dissolution enthalpy of BaBr2(s) is 24.72 kJ/mol. Note that since the temperature of the solution increased, the dissolution process is exothermic, and the enthalpy change is negative. Therefore, the actual enthalpy change is -24.72 kJ/mol.
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