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A student investigated the enthalpy of combustion (deltaHc) of methanol under standard conditions using the apparatus shown in
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"If I wanted a dHrxn I would want to do it with the good data. It amazes me that so many instructures use this Bond Energy thing when it is only good for a good guess. And frankly, at least in my opinion, I think using the tabulated values are easier to
The equation posted wrong. How does the diagram come into the question? Is there anything I can say about it for part b (why the value is different to the one obtained in the practical)?
H I H- C - O - H I H I thought there is a C-O bond? Is that not used? If it is ignored because we used the full equation average bond enthalpy then why in the products side when the water equation average bond enthalpy is used do we also use the O-H
What are the reasons that the the theoretical and actual enthalpy changes are different (part b)
I forgot to post this diagram. I did not do calorimeter work. b) Suggest some reasons as to why this value is different to the one obtained in the practical.
_Clamp stand _ _ _I__Thermometer _ I I Beaker _ I-----I _ I___I Water _ I Wick _ ----- _ I I _ ----- Alcohol _ I __I ______________
(35 + 412 x 3 + 358 + 463 + 496 x 1.5) − (41 x 2 + 463 x 4 + 743 x 2) = -584 kJ/mol I looked online and found that someone said this for this question for the enthalpy of combustion.
Is the second post right?
3 x C-H = 3 x 412 = 1236 kJ 1.5 x O=O = 1.5 x 496 = 744 kJ 1 x O-H = 1 x 463 = 463 kJ 35 kJ 2478 kJ 2 x C=O = 2 x 743 = 1486 kJ 4 x O-H = 4 x 463 = 1852 kJ 2 x -41 = -82 kJ 3256 kJ 2478 - 3256 = -778
3 x C-H = 3 x 412 = 1236 kJ 1.5 x O=O = 1.5 x 496 = 744 kJ 1 x O-H = 1 x 463 = 463 kJ 2443 kJ 2 x C=O = 2 x 743 = 1486 kJ 4 x O-H = 4 x 463 = 1852 kJ 2 x -41 = -82 kJ 3256 kJ 2443 - 3256 = -813
a) Can you help with what I did b) Suggest some reasons as to why this value is different to the one obtained in the practical. 31024 J/mol or 31.024 kJ/mol I do not know reasons for this. Can you help me with all the reasons.
I did the enthalpy of combustion of 1 mole of methanol calculation again doing it your way and got the same answer. Why did you do this part: 1 mol we have 4560 x (1 mol/0.147) ? Why did you do this to get 1 mol?
How I got the enthalpy of combustion of 1 mole of methanol: 80.6 – 75.9 = 4.7g CH3OH: 12 + (3 x 1) + 16 + 1 = 32 75.9/32 = 2.371875 = 2.4 mole (1 d.p.) Q = m c deltaT Q = 100g x 4.18 J/g °C x (32.4°C – 21.5°C ) = 100g x 4.18 J/g °C x 10.9°C =
a) I'm told to calculate theoretical enthalpy of combustion. It says to include the enthalpy of vaporisation for the liquid components. Why does it say enthalpy of combustion and then to include enthalpy of vaporisation?
a) I think this is the way to find the average bond enthalpy: 3 x 412 = 1236 kJ 1.5 x 496 = 744 kJ 1 x 463 = 463 kJ 358 kJ 1486 kJ Formed 2 x 743 = 1486 kJ 4 x 463 = 1852 kJ I did not want to do anything else before checking this was right. The note says
4556.2 J is from my working out using the information and answering another question. I used Q=mcdeltaT. Part b is asking why this value I got using this equation is different to to the one obtained from the practical in part a. Part a has to be worked out
