[Aetos]

During aerobic respiration in animals and plants, glucose is oxidised to carbon dioxide and water:

C6H12O6  +  6O2      →      6CO2   +   6H2O

Calculate the entropy change per mole of glucose at 37oC given that the enthalpy of the reaction, ΔH = -2807.8 kJ mol-1 and the free energy change, ΔG = -3089.0 kJ mol-1


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please dumb it down for me as I havent actually studied chemistry/entropy at A level.
I'd also appreciate if you could explain the theory behind the working out so i can have a better understanding

[fledarmus]

What is the equation that relates free energy, enthalpy, and entropy?

[Aetos]

well if i had to guess it would be that funny looking equations that says:


delta G= delta H - T delta S


and if that is right I would have to guess that I would need to rearrange the equation to get delta S.
and again im guessing that the equation would look something like this:

delta S= delta H - delta G / T

and then im guessing that i would put the values for delta H and delta G into the equation. And im guessing T is for temperature? but would that be in farenhiet or celcius.


OR perhaps i just have complete misunderstood everything >.<

[Jeremy]

Just plug in the values. Make sure your temperature is in kelvins though.

[fledarmus]

That's the one!

And as Jeremy said,

Make sure your temperature is in kelvins though

[Jeremy]

ΔG = ΔH - TΔS

re-arranges to :

ΔS = (ΔH - ΔG) / T

Plug in the numbers, but convert celcius to kelvins by adding on 273, (37*C = 310 K).

[Aetos]

thank you!!!!!!! (is happy now)

[Aetos]

so it would be

Convert 37*C to K = 37 + 273 = 310 K
Convert ΔH kJ/mol-1 to J/mol = -2807.8 x1000 = -2807800 J/mol

 (ΔH - ΔG) / T = ΔS

(-2807800 – (-3089)) / 310 = -9047


what's the units?

[Jeremy]

Don't convert enthalpy to J/mol. Sorry, my mistake there. 

[Aetos]

oh so i dont multiply by 1000?  

may I ask why, it made sense to do so. but now im a little confused

[Jeremy]

ΔG and ΔH are almost always given in kJ/mol. The units are the same so you don't need to change anything there.

[Aetos]

so the answer is

Convert 37*C to K = 37 + 273 = 310 K
ΔH kJ/mol-1 = -2807.8

 (ΔH - ΔG) / T = ΔS

(-2807.8 – (-3089)) / 310 = 0.907


what are the units?

[Aetos]

ΔG and ΔH are almost always given in kJ/mol. The units are the same so you don't need to change anything there.

oh I see, thank you ^_^

[Jeremy]

The answer is 0.907 kJ mol^-1 K^-1. However the size of ΔS is normally very small, so it's almost always quoted in J mol^-1 K^-1 which would be 907 J mol^-1 K^-1

[Aetos]

ahh okay =] thanks again!

I think this is the only question I now know how to answer, partially because it involves an equation. otherwise i'd be even more confused than usual.