[Aetos]

The biochemical conversion of Glucose, C6H12O6, into pyruvic acid, C3H4O3, is an important step in the TCA (Kreb’s) cycle which forms a part of the metabolic breakdown of glucose (to produce energy).

C6H12O6 (s) + O2 (g) → 2C3H4O3(l) + 2H2O (l)
glucose oxygen pyruvic acid water

Calculate the enthalpy change for this conversion given the following enthalpies of combustion:

1. C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
ΔH1 = -2821 kJmol-1


2. C3H4O3 (l) + 5/2 O2 (g) → 3CO2 (g) + 2H2O (l)
ΔH2 = -1170 kJmol-1




I've only had one lecture at university on enthalpy and thermodynamics and as i'm aware that is the only lecture I will get on it. Since most of the students have studied A level chemistry before, we are expected to know how to work this out already. But since I did not study A level chemistry I am finding difficulty in even understanding the question let alone working it out. I would appreciate it greatly if someone could at least work out one of the 2 questions above so I can see how it is done.

And just so you know this is not a assignment or homework, it is an optional task set by my lecturer. I really want to do well in my exams so please help me out

[Jeremy]

The enthalpy change of the reaction can be found indirectly using Hess's law which states that the energy change accompanying a chemical reaction is independent of the pathway.

So you can imagine the reaction occuring in two stages:

1) First, 2821 kJ energy is released when 1 mol of glucose is burnt to 6 mol of CO₂ and 6 mol of H₂O. (ΔH < 0)
2) Then, 2340 kJ (1170 kJ/mol * 2 mol) is taken in to convert the CO₂ and H₂O to 2 mol of pyruvic acid. (ΔH > 0).

So, ΔH = -2821 - (-1170*2) = -481 kJ/mol.

Reading the first three links here http://www.chemguide.co.uk/physical/energeticsmenu.html#top will definitely help you a lot.

[Aetos]

thank you so much for your help.

I will read through the link you provided. I understand how the working out was done but i have another question. I assumed that they were 2 questions but now I understand that it was 1 question? and also did you combine/rearrange the two equations from 1. and 2. to get 2340 kj because otherwise i dont quite understand why you multiplied 1170 by 2

[Jeremy]

The reason I multiplied 1170 by 2 is because you've been given standard enthalpy of combustion data which is the energy released when 1 mol of a substance is burnt. But here we want to find the energy released when 2 mol of pyruvic acid is burnt.

[Aetos]

*scratches head*

okay thank you again