Sarah182
Herpes Member
Ok its the molar heat of combustion prac and I'm stressing cause my results are VERY different from the standard enthalpies of combustion but we didnt use anything to prevent heat loss in the experiment sooo yeah.
Propan-1-ol: (standard enthalpy of heat of combustion is -2020 kJ/mol
I have to work out heats of combustion per gram and per mole.
1.97g of propan-1-ol raised the temperature of 25g of water by 81.5 degrees
(delta) H= -mC (delta) T
= -25 x 4.18 x 81.5
= -8.51675 kJ (per 1.97g of propan-1-ol)
Heat of combustion per gram:
-8.51675/1.97 = -4.32 (to 2 d.p)
Molar mass of propan-1-ol: 58.078
n (propan-1-ol): mass/molar mass
= 1.97/ 58.078
=0.0339 mol (to 4 d.p)
As 0.0339 moles of propan-1-ol produces -8.51675 kJ of heat . The molar heat of combustion of propan-1-ol is:
(delta) H = -8.51675/ 0.0339
= -251.08 kJ/mol
Percentage error:
2020-251/82= 1769.92
1768.82/2020 x 100
= 85.57
= 86% error
ok so is that right? just make sure i go through the right PROCESS thats where I'll get my marks.
Thanks guys.
Propan-1-ol: (standard enthalpy of heat of combustion is -2020 kJ/mol
I have to work out heats of combustion per gram and per mole.
1.97g of propan-1-ol raised the temperature of 25g of water by 81.5 degrees
(delta) H= -mC (delta) T
= -25 x 4.18 x 81.5
= -8.51675 kJ (per 1.97g of propan-1-ol)
Heat of combustion per gram:
-8.51675/1.97 = -4.32 (to 2 d.p)
Molar mass of propan-1-ol: 58.078
n (propan-1-ol): mass/molar mass
= 1.97/ 58.078
=0.0339 mol (to 4 d.p)
As 0.0339 moles of propan-1-ol produces -8.51675 kJ of heat . The molar heat of combustion of propan-1-ol is:
(delta) H = -8.51675/ 0.0339
= -251.08 kJ/mol
Percentage error:
2020-251/82= 1769.92
1768.82/2020 x 100
= 85.57
= 86% error
ok so is that right? just make sure i go through the right PROCESS thats where I'll get my marks.
Thanks guys.