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 havent studied enthalpy change before my last lecture. I am expected to know how to work out enthalpy change however we were not taught how to do this during the lesson. I havent studied A level chemistry and am currently at university. I would appreciate if you could dumb it down for me. =] thanks
Sure, I can help explain how to calculate the enthalpy change for the given conversion. Enthalpy change, also known as heat of reaction, represents the amount of heat energy released or absorbed during a chemical reaction.
To calculate the enthalpy change for the conversion, you need to consider the enthalpy changes for the individual reactions involved.
First, let's consider the combustion reaction of glucose:
C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
ΔH1 = -2821 kJmol-1
This reaction shows that when one mole of glucose is burned with 6 moles of oxygen, it produces 6 moles of carbon dioxide and 6 moles of water, releasing 2821 kJ of heat energy.
Now, let's consider the reaction for the conversion of glucose to pyruvic acid:
C6H12O6 (s) + O2 (g) → 2C3H4O3 (l) + 2H2O (l)
From this equation, we can see that one mole of glucose reacts with one mole of oxygen to produce 2 moles of pyruvic acid and 2 moles of water.
To calculate the enthalpy change for this reaction, we can use a concept called Hess's Law. According to Hess's Law, the enthalpy change for a reaction is independent of the pathway and depends only on the initial and final states.
We can break down the reaction into two steps:
1. Combustion of glucose:
C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
ΔH1 = -2821 kJmol-1
2. Formation of pyruvic acid:
6CO2 (g) + 6H2O (l) → 2C3H4O3 (l) + 2H2O (l)
ΔH2 = -1170 kJmol-1
Since 6 moles of carbon dioxide and 6 moles of water are produced in the first step and then consumed in the second step, we can cancel them out:
C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)
6CO2 (g) + 6H2O (l) → 2C3H4O3 (l) + 2H2O (l)
By adding these two equations, we get the overall equation:
C6H12O6 (s) + 6O2 (g) → 2C3H4O3 (l) + 2H2O (l)
ΔH = ΔH1 + ΔH2 = -2821 kJmol-1 + (-1170 kJmol-1)
To find the enthalpy change (ΔH) for the overall reaction, you need to subtract the enthalpy of the reactants from the enthalpy of the products. In this case, you would subtract the enthalpy change for the combustion of glucose (-2821 kJmol-1) and the enthalpy change for the formation of pyruvic acid (-1170 kJmol-1).
So, the enthalpy change for the conversion of glucose to pyruvic acid would be:
ΔH = -2821 kJmol-1 + (-1170 kJmol-1) = -3991 kJmol-1
Therefore, the enthalpy change for this conversion is -3991 kJmol-1.