Post Prac-Exam Query (1 Viewer)

[imoO]

Hey guys,

I was wondering if I got this correct for the calculation of the enthalpy of combustion of ethanol.

Mass of ethanol used 0.6g
Temperature change 21.5

From my calculations, I got like 8xxx kJ/mol...it seemed quite large to me. Can anybody clarify this?

Cheers,

 

[minijumbuk]

Hmmmmmmmmmmmmmmmmm looks incorrect.

I think ethanol has a theoretical molar heat of combustion of around 1XXX, but I can't remember. 8XXX is definitely wrong, since it's a lot bigger than the value of the theoretical molar heat of combustion.

It's true there's experimental error, but that should bring the value down, and not up.

You probably added one extra digit somewhere by mistake.

 

[Rachaek]

I think you only found the KJ released through combustion of your 0.6g of ethanol, rather than the KJ per mole

provided you used 100mL of water
h= m x c x change in temp
= 100 x 4.18 x 21.5
= 8987 J (which is the amount of energy released by combustion of 0.6g of ethanol, not one mole)

to find the KJ per mole, you need to find how many moles were used:
n = 0.6/(6+16+12*2)
= 0.013

therefore, KJ/mol
= 8987/0.013
=691307.7 J/mol
= 691.3 kJ/mol

(correct me if the calculations are wrong...)

 

[imoO]

I think you only found the KJ released through combustion of your 0.6g of ethanol, rather than the KJ per mole

provided you used 100mL of water
h= m x c x change in temp
= 100 x 4.18 x 21.5
= 8987 J (which is the amount of energy released by combustion of 0.6g of ethanol, not one mole)

to find the KJ per mole, you need to find how many moles were used:
n = 0.6/(6+16+12*2)
= 0.013

therefore, KJ/mol
= 8987/0.013
=691307.7 J/mol
= 691.3 kJ/mol

(correct me if the calculations are wrong...)

I think you're right. I did

100x4.18x10^3x21.5

and I got like 8987000joules/0.6grams ethanol used (i.e 8978kJ)

i went rong here.

I divided 46 by 0.6, and got 76 moles, so i multiplied 76 by 8978kJ, and said is was kJ/mol...rofl...

I'm very disappointed about this test. It was out of 25, and scaling 25%, so 1% per mark.

I know I've already lost one mark for not putting a title on my graph, 1 mark for this incorrect calculation, and god knows how many for not putting safety risk assessments and no results table in my report. Have to play that gay game we call 'catch-up' now.....

 

[minijumbuk]

Seems like you haven't understood how enthalpy calculations are done properly.

It's a pain explaining it over the internet, so try ask your teacher (or tutor) step-by-step how to do calculations for enthalpy change. I think I saw at least 3 errors in your working, so it's best to get some practice on these questions. There is usually one question that is worth 2/3 marks in the HSC on this.