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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: shell52080 on October 14, 2007, 01:35:38 PM

Title: Biochemical Standard Conditions
Post by: shell52080 on October 14, 2007, 01:35:38 PM

This is a question for my p.chem homework, but it seems like more of a biochem question. I'm supposed to find the percentage efficiency of aerobic respiration under biochemical standard conditions. I've never learned what the standard conditions were. I'm guessing 1 atm for pressure and 1M concentration for all components. Is that correct? Is there anything I'm missing?

Thanks

Title: Re: Biochemical Standard Conditions
Post by: Yggdrasil on October 14, 2007, 01:46:29 PM

pH 7, 37^oC

Title: Re: Biochemical Standard Conditions
Post by: shell52080 on October 14, 2007, 02:05:23 PM

Thanks!

Title: Re: Biochemical Standard Conditions
Post by: shell52080 on October 14, 2007, 05:00:08 PM

Now I can't figure out how to even start the problem. Any hints?

Title: Re: Biochemical Standard Conditions
Post by: Yggdrasil on October 14, 2007, 09:10:10 PM

What is the overall chemical reaction for aerobic respiration?  What is the maximum ΔH of this reaction?  What energy-carrying molecules are created during aerobic respiration?  How much energy do these molecules carry?

Title: Re: Biochemical Standard Conditions
Post by: shell52080 on October 15, 2007, 03:51:51 PM

Thanks for the hints, but I think I may need to calculate it using a different approach.

The overall chemical reaction is
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O + 38ATP

The second part of the problem gives pCO2 = 0.053atm, pO2 = 0.132atm, [glucose] = 0.056M, [ATP]=[ADP]=[Pi]=1x10^-4M, pH=7.4, T=310K. Take activities to be the molar concentrations.

I don't understand where to start and where I'm supposed to get to. Do I have the right equation?

Edited here- I tried a little more work, but I still don't get how to do the part above.

Using the hints you provided, I determined that deltaH= -2880kJ for
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O + 38ATP

31.38kJ are consumed by each ATP. So 31.38 x 38 = 1192.44

1192.44/2880 = 41%

Is that correct?

By the way, are you a Valkyrie Profile fan?

Thanks

Title: Re: Biochemical Standard Conditions
Post by: Yggdrasil on October 15, 2007, 08:56:52 PM

First, your equation is not balanced as the ATP seemly pops up from nowhere (reactants should include ADP and P_i.

Second, I would say that the theoretical maximum of the reaction is the equation for the combustion of glucose:

C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O

So, how does this value compare to the energy stored in 38 molecules of ATP?

Also, I've never heard of Valkyrie Profile.

Title: Re: Biochemical Standard Conditions
Post by: shell52080 on October 16, 2007, 09:47:24 AM

I still don't understand what I'm doing.

Thanks for your *delete me*

Title: Re: Biochemical Standard Conditions
Post by: crazycarl1 on October 16, 2007, 03:55:29 PM

Do you go to West Chester shell? I have exact same problem for my p.chem class...same numbers in 2nd part too...still no clue how to do it though.

If we take the info from the 2nd part and just use
pCO2 = 1.00 atm, pO2 = 1.00 atm, [glucose] = 1.00 M, [ATP]=[ADP]=[Pi]=1.00 M, pH=7.4, T=310K we should be able to find the max # of ATP we can form but I'm not sure how to do it.

Also if by chance you are in the same class as me somehow how do you do #5 on that assignment with the triple point

Title: Re: Biochemical Standard Conditions
Post by: shell52080 on October 16, 2007, 10:16:28 PM

Hey,

Yeah. I go to West Chester so I'm guessing we're in the same class. I'm going to go talk to our professor about it tomorrow. I did figure out number 5. Set the 2 equations equal to each other and solve for T. Then solve for lnp using the T you solved for. I think that is right.

I'm so tired of looking at number 2. It's completely beyond me.