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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 - Tinitial

35.0°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 - Tinitial

44.0°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 - Tinitial

52.9°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:


Tutor Answer

(Top Tutor) Daniel C.
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School: University of Maryland
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