[riboswitch]

This is supposed to be a very easy problem from my Physical Chemistry book. However, my friends and I failed to solve this properly. Somehow our results differ from the one hinted in the book.

Problem:

Suppose that when you exercise, you consume 100 grams of glucose and that all the energy released as heat remains in your body at 37°C. What is the change in entropy of your body?

How would I solve the problem?

Entropy is supposed to be the measure of disorder of a thermodynamic system. However I'm not interested in calculating its absolute value. The problem wants me to calculate the change in entropy (ΔS) of our system (which is supposed to be the human body). According to my book, the change in entropy of a system is equal to "the energy transferred as heat to it reversibly divided by the temperature at which the transfer takes place".

[tex] \Delta S_{sys} \ = \  \frac{q_{rev}}{T} [/tex]

I already know the temperature of the system (37°C ≈ 310 K). I only have to determine the energy released from 100 grams of glucose. I do know the standard enthalpy of formation of our sugar.

[tex]\Delta_{f}H^{ \Theta }(C_ {6}H_{12}O_{6}) \ = \  - 1273 \  \frac{kJ}{mol}[/tex]

By dividing the mass of glucose (100 grams) to the molar mass of a typical glucose molecule (≈180 g/mol), I can calculate the moles of the substance of interest. Multiplying the standard enthalpy of formation of our sugar to the number of moles available, I can determine the total amount of energy released as heat by consuming 100 grams of glucose.

[tex]q_{rev} \ = \  - 1273 \  \frac{kJ}{mol} \  \times \ 0.556 \ mol[/tex]
[tex]q_{rev} \ = \  - 707.788 \ kJ[/tex]

Now that I know the temperature and the energy released as heat to the system, I can finally calculate the change in entropy by using the first equation that I wrote above.

[tex]\Delta S_{sys} \ = \  \frac{ - 707.788 \ kJ}{310 \ K} \ = \  - 2.28 \  \frac{kJ}{K}[/tex]

However my book suggests a different answer:

[tex]\Delta S_{sys} \ = \ 5.03 \  \frac{kJ}{K}[/tex]

Can someone help me with this one? Did I use the correct values?

[mjc123]

Heat of formation is the wrong thing to use. It refers to the formation of a compound from its elements:
6C + 6H₂ + 3O₂  C₆H₁₂O₆
This is certainly not what your body does with glucose - you are breaking it down, not forming it -and it's not the reverse reaction either. Now I'm not a biochemist, and I invite anyone to correct me if I'm wrong, but I suspect what happens overall is the oxidation of glucose to CO₂ and water:
C₆H₁₂O₆ + 6O₂  6CO₂ + 6H₂O
The heat of this reaction is called the heat of combustion of glucose and you should be able to find it readily. Try this and see if it gives you the right answer (careful with signs)

[Corribus]

The heat of this reaction is called the heat of combustion of glucose and you should be able to find it readily. Try this and see if it gives you the right answer (careful with signs)

Or, ironically, it can be calculated from heats of formation.
I agree, though, the question doesn't really make it clear what it means by "released from 100 g of glucose". I agree it's probably referring to the (net)  biochemical process of metabolism... and a quick check confirms it... but sometimes in problems like this you never know why wacky assumptions you're supposed to be making.

[riboswitch]

Heat of formation is the wrong thing to use. It refers to the formation of a compound from its elements.

The enthalpy of combustion of a mole of glucose to carbon dioxide and water is:

[tex] \Delta _{c}H^{ \Theta } \ = \  - 2808 \  \frac{kJ}{mol}[/tex]

I will do all the calculations necessary to determine the change of entropy of our reference system (the human body). To do that, I will assume that the standard enthalpy of combustion is equal to the heat transferred reversibly to our system.

[tex]q_{rev} \ = \  - 2808 \  \frac{kJ}{mol} \  \times  \ 0.556 \ mol \ = \  - 1561.25 \ kJ[/tex]

Now that I know the value of q_rev, it's time to calculate the change of entropy.

[tex]\Delta S \ = \  \frac{q_{rev}}{T} \ = \  \frac{1561.25 \ kJ}{310 \ K} \   \approx   \ 5.03  \frac{kJ}{K}[/tex]

The heat transferred, by convention, is positive. It's because heat enters the system.
Thanks for helping me out. 

Or, ironically, it can be calculated from heats of formation.

I already used the standard enthalpy of formation of a mole of glucose from its elemental constituents but the answer doesn't correspond to the one written in my book.

Or probably you're talking about applying Hess's Law? I'm still struggling about applying Hess's Law. I tried using it to solve this problem, but I'm having difficulty applying it. I can't tell whether I'm doing it right or I'm making mistakes.

What I would do if I used Hess's Law:

C₆H₁₂O_6(s)  6C_(s) + 6H_2(g) + 3O_2(g)  Δ_fH = + 1273 kJ/mol

6C_(s) + 6O_2(g)  6CO_2(g)             Δ_fH = - 6(393.51) kJ/mol

6H_2(g) + 3O_2(g)  6H₂O_(l)            Δ_fH = - 3(241.818) kJ/mol 

Then I would stop here, because the sum of the enthalpies isn't equal to the standard enthalpy of combustion of glucose. Plus, I still have to review Hess's Law...

[mjc123]

Two points:
i. The enthalpy of your third reaction should be 6Δ_fH(H₂O)
ii. You have used the heat of formation of H₂O(g), despite writing H₂O(l). The heat of formation of H₂O(l) is -285.83 kJ/mol. Try using that.

[riboswitch]

Two points:
i. The enthalpy of your third reaction should be 6Δ_fH(H₂O)
ii. You have used the heat of formation of H₂O(g), despite writing H₂O(l). The heat of formation of H₂O(l) is -285.83 kJ/mol. Try using that.

I'm sorry for "bumping" this thread. It's been a long time since I last posted here but I decided to just finish this topic for formality's sake. I hope I'm not breaking rules by doing this.

I wanted to calculate the value of heat (or the enthalpy of reaction) added to the system in order to determine the value of entropy at 310 K. In order to do that, I'm using Hess's Law.

C₆H₁₂O_6(s)     6C_(s) + 3O_2(g) + 6H_2(g)  Δ_fH = + 1273 kJ/mol

6C_(s) + 6O_2(g)  6CO_2(g)  Δ_fH = - 6(393.51) kJ/mol

6H_2(g) + 3O_2(g)     6H₂O_(l)  Δ_fH = - 6(285.83) kJ/mol       

So the total reaction would be:

C₆H₁₂O_6(s)  +  6C_(s) + 6O_2(g) + 6H_2(g) + 3O_2(g)     6C_(s) + 3O_2(g) + 6H_2(g) + 6CO_2(g) + 6H₂O_(l)

or....

C₆H₁₂O_6(s)  +  6O_2(g)   6CO_2(g) + 6H₂O_(l)

with Δ_fH = -2802.74 kJ/mol 

The sign is negative, which means that the reaction is exothermic. The heat released from the reaction is added to our system (human body)  that has a temperature of 37 °C. Since we are adding heat to the system, the value of q_rev is positive.

With all the data I have gathered so far, I can now determine the increase of entropy in our system:

[tex]\Delta S \ = \   \frac{q_{rev} \  \times \ n}{T} \ = \  \frac{2802.74 \  \frac{KJ}{mol} \  \times \ 0.556 \ mol }{310 \ K} \ = \ 5.03 \  \frac{KJ}{mol} [/tex]

Thanks to everyone for the collaboration!

[mjc123]

Careful with units! What are the units of entropy?

[riboswitch]

Careful with units! What are the units of entropy?

Oh! My mistake!

[tex] \Delta S \ = \ 5.03 \  \frac{KJ}{K} [/tex]
I hope I've done everything correctly.

[mjc123]

Careful with the notation (you got it right earlier on). Small k for kilo, capital K for kelvin. So 5.03 kJ/K.