I need help with question 4 and 5.
The Enthalpy Change of a Chemical ReactionExperiment 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 = T_{maximum} - T_{initial} _{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 = T_{maximum} - T_{initial} _{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 = T_{maximum} - T_{initial} _{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:
q_{reaction}= (Ccal * ΔT) + (m_{contents}* Cp_{contents}* ΔT) |
3.
Assume Cp_{contents}= Cp_{water}= 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:
Top Answer
QUESTION 4 SINCE THE AMOUNT OF HCL USED FOR ALL THE THREE TRIALS WERE THE SAME 50mL of 1M HCI is 50/1000x1= 0.05 moles SINCE... View the full answer