# The Enthalpy Change of a Chemical Reaction

Jul 22nd, 2015
Anonymous
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Science
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Question description

# The Enthalpy Change of a Chemical Reaction

## Experiment 1

1.Record the following for each of the three trials:

·Trial 1

 a Mass of the empty calorimeter (g)18.6g b Initial temperature of the calorimeter (°C)21.5°C c Maximum temperature in the calorimeter from the reaction (°C)35.0°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum - Tinitial35.0°C-21.5°C=13.5°C e Mass of the calorimeter and its contents after the reaction (g)68.74g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.74g-18.6g=50.14g g Calculate the moles of Mg reacted (MW = 24.305 g/mole)?50mL of 1M HCI is 50/1000x1=0.05.15/24.305=.006g/mole

·Trial 2

 a Mass of the empty calorimeter (g)18.6g b Initial temperature of the calorimeter (°C)21.5°C c Maximum temperature in the calorimeter from the reaction (°C)44.0°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum - Tinitial44.0°C-21.5°C=22.5°C e Mass of the calorimeter and its contents after the reaction (g)68.83g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.83-18.6=50.23 g Calculate the moles of Mg reacted (MW = 24.305 g/mole).25/24.305=.010g/mole

·Trial 3

 a Mass of the empty calorimeter (g)18.6g b Initial temperature of the calorimeter (°C)21.5°C c Maximum temperature in the calorimeter from the reaction (°C)52.9°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum - Tinitial52.9°C-21.5°C=31.4°C e Mass of the calorimeter and its contents after the reaction (g)68.92g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.92g-18.6g=50.32 g Calculate the moles of Mg reacted (MW = 24.305 g/mole).35/24.305=.014g/mole

2.Calculate the heat released into the solution for the 3 reactions, according to the formula:

 qreaction= (Ccal * ΔT) + (mcontents* Cpcontents* ΔT)

3.
Assume Cpcontents= Cpwater= 4.18 J/g °C

 a Trial 1 (J)(4.18J/g°C)(68.74g)(35.0°C-21.5°C)=3878.9982 J/g°C b Trial 2 (J)(4.18J/g°C)(68.83g)(44.0°C-21.5°C)=6473.4615 J/g°C c Trial 3 (J)(4.18J/g°C)(68.92g)(52.9°C-21.5°C)=9045.88784 J/g°C

4.Find the molar heat of reaction for each experiment in units of kilojoules / (mole of Mg) by dividing the heat of reaction (converted to kJ by dividing by 1000) by the moles of Mg used.

 a Trial 1 (kJ/mol) b Trial 2 (kJ/mol) c Trial 3 (kJ/mol)

5.Calculate and record the average molar heat of reaction from the three results:

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