heat of Combustion and Enthalpy - in dire need of help (1 Viewer)

[DistantCube]

Long story short, any help is greatly appreciated, I'm not sure exactly what to do for these questions;

Q1.
a group of students set up an apparatus to perform an experiment to find the enthalpy of combustion of ethanol.
They placed 200mL of water in the flask and ethanol in the spirit burner.
The results are as follows:
initial temp of water: 18
final temp of water: 23
initial mass of burner and ethanol: 131.42g
final mass of burner and ethanol: 130.28g

Use specific heat capacity of water as 4.18

a) calculate the heat gained by the water

b) i) the enthalpy of combustion of ethanol in kJ g^-1
ii) and in kJ mol^-1

c) The result they obtained (kJ mol^-1) was considerably less than the data book figure
i) suggest three reasons for this discrepancy
ii) suggest one improvement for the experiment to improve accuracy

-----------------------------I dont have answers for the above.

I have answers for these, but don't know how to achieve them.

Q2. The heat of combustion of ethanol is 1360kJ/mol. what is the mass of ethanol that needs to be burnt to raise the temp of 350 g of water through 77 degrees if 50% of the heat released by the ethanol is lost to the surroundings? The specific heat capacity of water is 4.2. (answer is 7.7g, but i dont know how to get it)

 

[DistantCube]

sorry to double post, but I tried working it out after some research, and I got Q1, but still nothing on Q2.

Q1.
a) Q=mct
Q=200 x 4.18 x 5
=4.18 kJ released

I got tripped out by this, it was 3 marks, but i think thats all you need, and yuo need to write an equation etc for all the marks.

b)i) 4.18 / (m of ethanol used)

ii) 4.18 / (m/M (i.e moles) of ethanol)

c) i) heat absorbed by thermometer, heat lost to air, accuracy of readings and of equipment

ii) use a bomb calorometer

any ideas on 2?

edit: and 2:

50% of temp is 77
so 100% temp is 154
heat = -154 x 4.2 x 350
/1000
= 226.38 KJ

molar heat = 1360 = 226.38/moles

moles = 226.38/1360
= 0.16646

n = m/M
m = ?
M = 46

So 0.16646 = m / 46
So m = 0.16646 x 46
= 7.7

 

[richz]

ok u got the enthalpy change u work out the KJ/mol by dividing 4.18KJ by (131.42 -130.28) this gives u KJ/g which gives 3.67KJ/g then u multiply by the molar mass of ethanol which i think is 44 of the top of my head. So if its 44 its 161.33KJ/mol

 

[m_isk]

Three reasons for the discrepancy: um well, obviously heat is lost to the environment (air, can, themometer, retort stand, clamp etc), maybe the distibution of heat to the water wasn't uniform because it wasn't stirred (ie different "parts" of the water had different temperatures and thus an error occured) and perhaps there was an error in the reading of the themometer, or the wieghing of the alkanol.
Some ways to improve the experiment: use draft excluders (just a block of something you put up around the apparatus to prevent gusts of wind - or a draught, as the name suggests). You should also stir the water for uniform distribution of heat, and you should also ensure that the flame from the spirit burner is as close as possible to the base of the can/calorimeter. Furthermore, you could factor in the specific heat capacity of the container into your calculations. Hope that helps

 

[LaCe]

So that's how u got the answer eh A?
yes good idea, its is very easy, i thought even u could have got that

 

[DistantCube]

then u multiply by the molar mass of ethanol which i think is 44 of the top of my head. So if its 44 its 161.33KJ/mol

Sure you're not mistaken there? I don't think you multiply by the molar mass, and I think you're supposed to use the 4.18kJ not the new value of kJ/g. I'm guessing it's divide since the units are kJ/mol ... help me out if I'm wrong here.

thanks for your help guys, particularly m_isk.

 

[m_isk]

Distant: just think about it like this.
I released 4.18kJ of energy from (131.42g - 130.28g)=1.14 grams
but 1.14 grams = (1.14/46)=0.025 mols
Therefore, 4.18kJ is released from 0.025 mols.
Now divide both sides of this "equation" by 0.025, and the molar heat of combustion is determined =4.18/0.025 = 167.2kJ/mol (and don't forget the minus )

 

[DistantCube]

oh touche.

 

[m_isk]

whats touche?

 

[james_chappo]

M_isk two things. 1. I wasn't being sarcastic in my reason for giving you rep points.

2. Touche is a word used in a situation where one person insults someone else only to be taken aback by a powerful comeback. It can also be used as a synonym of 'point taken'.