For more information, please see full course syllabus of Biochemistry

For more information, please see full course syllabus of Biochemistry

### Example Problems For Bioenergetics

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Example 1: Calculate the ∆G°' For The Following Reaction 1:04
- Example 1: Question
- Example 1: Solution
- Example 2: Calculate the Keq For the Following 4:20
- Example 2: Question
- Example 2: Solution
- Example 3: Calculate the ∆G°' For The Hydrolysis of ATP At 25°C 8:52
- Example 3: Question
- Example 3: Solution
- Example 3: Alternate Procedure
- Example 4: Problems For Bioenergetics 16:46
- Example 4: Questions
- Example 4: Part A Solution
- Example 4: Part B Solution
- Example 4: Part C Solution
- Example 5: Problems For Bioenergetics 29:27
- Example 5: Questions
- Example 5: Solution - Part 1
- Example 5: Solution - Part 2

### Biochemistry Online Course

### Transcription: Example Problems For Bioenergetics

*Hello and welcome back to Educator.com, and welcome back to Biochemistry.*0000

*We just finished our unit on bioenergetics, and today, we are going to start off with some example problems of bioenergetics.*0004

*I am actually going to be doing a couple of lessons on this; this is really, really, really important.*0011

*Bioenergetics is one of those topics that unfortunately, a lot of kids do not get a lot of practice in, and there is always going to be some sort of a hazy idea of what is it that is going on.*0016

*This idea of thermodynamics, this idea of ΔG, of spontaneity, of the coupling of an exergonic reaction with an endergonic reaction, the idea of oxidation-reduction, these are profoundly, profoundly important ideas.*0027

*And if you know what is happening both globally and in detail, so much of biochemistry and so much of everything in science is actually made so much easier.*0044

*I want to do a fair number of problems; some of them are going to be basic.*0055

*Some of them are going to be quite detailed, so let’s just jump right on in.*0059

*OK, example no. 1, let’s see.*0064

*Let’s start with black; how about that?*0069

*OK, so example 1, we would like to calculate the standard free energy change, the biochem’s standard free energy change for the following reaction.*0074

*We have aspartate + alpha-Ketoglutarate going to - actually let’s go ahead and leave it as an equilibrium here - an equilibrium with glutamate + oxaloacetate.*0095

*And we say that the K _{eq} for this reaction is 0.147 at 25°C.*0125

*We would like you to calculate the free energy change for this given its K _{eq}.*0137

*OK, well, we have a relation for this; it is really nice and straightforward.*0142

*The standard free energy change is equal to -R x T times the natural logarithm of the K _{eq}.*0146

*We just put the numbers in- really, really great.*0154

*We have minus, so this is going to be 8.315, and I am going to write this out with its units so you see everything.*0158

*This is J/mol-K, let me make that a little bit more clear, and then we have the 298K; and then we have the natural logarithm of the K _{eq}, which is 0.147, and then when we do this, we end up with, so notice the Kelvin cancels the Kelvin, we end up with -4,750J/mol, or if you prefer the kJ version, -4.75kJ/mol- there you go, that is it, really, really straight forward, definitely an exergonic reaction.*0167

*No problems there; OK, wait a minute.*0217

*No, I am sorry; this is not going to be negative because the logarithm of a number that is going to be less than 1 is going to be negative.*0225

*So, negative-negative makes it positive, so this is positive, which, of course, is confirmed by the fact that this is a really, really tiny K _{eq}.*0237

*That means, it is not very favorable; that means, it favors not the products, but it actually favors the reactants.*0247

*This is +4.75kJ/mol, my apologies.*0254

*OK, let’s go to example no. 2.*0260

*We would like you to, this time, calculate the equilibrium constant for the following reaction.*0266

*It is just some basic stuff to get us going.*0279

*This time, we have L-malate, and let’s go ahead and write the reaction this way, in biochemical format.*0284

*NAD ^{+} - oops, let me write this a little bit better - NAD^{+} comes in.*0297

*NADH + H ^{+} goes out.*0307

*The enzyme that is catalyzing this is a dehydrogenase, right?*0313

*We just finished discussing how the NAD coenzymes work with dehydrogenases.*0318

*This is malate dehydrogenase, and it is going to convert this to oxaloacetate; and we are given a ΔG for this.*0327

*The free energy change is going to be 29.7kJ/mol, and it looks like this is positive, so we would like you to find the K _{eq} for this.*0338

*Well, it is exactly the same as before, except now, you are just reversing the reaction.*0347

*You are finding the K _{eq} instead of the ΔG.*0351

*Let’s go ahead and draw out some structures, just so we have a good sense of what is going on.*0355

*Again, this is biochemistry; we want to deal with structures as much as possible.*0360

*This much is just reasonable; we have C, C, C, C.*0364

*This is O ^{-}; we will put an H here.*0369

*This is L-malate; this is CH _{2}, COO^{-}.*0373

*There is that; there is this.*0379

*The NAD ^{+} is there; the NADH plus the hydrogen ion.*0383

*And again, I will just write out the malate dehydrogenase, and what we get is our oxaloacetate, C, C, C, C, right?*0390

*What we have done, we have oxidized this; we have pulled off this hydrogen and this hydrogen.*0408

*We are left with that; we are left with COO ^{-} there, H_{2}, COO^{-} there.*0412

*Well, we know that the ΔG, our basic equation is -RT ln K _{eq}.*0418

*We just rearrange this equation, and we get that the K _{eq} is equal to E^{-ΔG} / RT.*0429

*And then when we put the numbers in, we get E ^{-29,700J/mol}, right?*0439

*Now, notice, they gave the ΔG in kJ, but the R 8.315 is in J/mol-K, so we have to convert the ΔG to Joules.*0457

*That is why I have the 29,700 - again, that is going to be the biggest issue- is conversions - over 8.315.*0469

*It is going to be J/mol-K times...and it looks like this is going to be at 25°C, so we are there.*0480

*This cancels that- mole, Joule, Joule, mole; all the units cancel.*0490

*And what you end up is a K _{eq} equal to 6.23 x 10^{-6}; if I have done my arithmetic correct.*0494

*Again, arithmetic is ultimately...well, it is kind of irrelevant.*0502

*The idea is to understand what is going on, but 6.23 x 10 ^{-6}- that is highly unfavorable.*0506

*That is confirmed by the ΔG, which was a +29.7kJ- highly unfavorable as written.*0512

*In other words, in this particular case, these are the things that are favored, not that- that is all this means.*0520

*OK, let’s see example no. 3.*0528

*Well, let’s go ahead and change colors here; I think I am going to go to blue for example no. 3.*0535

*Now, given the following information, given the following info, we would like you to calculate the standard free energy change for the hydrolysis of adenosine triphosphate, for the hydrolysis of ATP at 25°C.*0543

*If what we are given is the 2 following reactions, we are given glucose 6-phosphate + water, so its hydrolysis goes to glucose plus an inorganic phosphate, they tell us that the K1 of this one, the equilibrium constant, is equal to 270.*0582

*And they give us another reaction; they give us ATP + glucose goes to glucose 6-phosphate + ADP.*0601

*And they tell us that the equilibrium constant for this one equals 890.*0614

*What they want us to find using this information, is to find the free energy change for the hydrolysis of ATP.*0619

*OK, basically, what we are going to do is the following.*0626

*We are going to add the reactions, and then we are going to fiddle around, either with the K _{eq}s or the ΔGs.*0633

*We are actually going to do this in 2 different ways; I am going to do it one way, then I am going to give you an alternate procedure.*0642

*I am going to add the reactions to get a net reaction.*0648

*Now, the equilibrium constant for the net reaction that I add is equal to K1 x K2.*0655

*Do you remember from chemistry, if you have 2 reactions with the particular equilibrium constant, if you add them, the equilibrium constant of the net reaction is the product of the equilibrium constants?*0665

*Let’s go ahead and add them; let's see.*0679

*Let’s go ahead and take...well, let me rewrite it.*0683

*Glucose 6-phosphate + H _{2}O goes to glucose + inorganic phosphate, and then I have adenosine triphosphate + glucose goes to Glc, G6P, so glucose 6-phosphate + ADP.*0689

*Let me go ahead and add those; G6P cancels G6P.*0713

*Glucose cancels glucose, so what I am left with is my final reaction, which is the hydrolysis of ATP.*0718

*ATP + H _{2}O goes to ADP + inorganic phosphate.*0724

*Now, the K _{eq}...I will just write K_{net}.*0732

*I do not want to write _{eq}; we know we are taking about an equilibrium constant.*0739

*So, the K _{net} is equal to K1 x K2, well, which is equal to 270 times the 890.*0742

*I just multiply the...right, I am going to get a K _{net} here.*0751

*It is going to be the product of the equilibrium constants of the individual reactions.*0755

*This is going to equal 240,300.*0759

*Now, I can go ahead and use my equation for ΔG to calculate the ΔG.*0763

*Now, my ΔG standard is equal to -RT ln K _{net}, which is equal to -8.315 x 298 times the logarithm of this number right here, 240,300.*0769

*When we run this calculation, we end up with -30.6kJ/mol.*0794

*You are going to get -30,600; I just converted it to kJ- that is it.*0804

*You have given 2 reactions; if you need to know something about another reaction that can come from the 2 reactions that you are given, the 2 or 3 or 4, when you add them up, I worked directly with the equilibrium constants because that was what was given to me.*0810

*I just multiplied the equilibrium constants and dealt with the equation.*0825

*Now, I will go ahead and do the alternate procedure, where I will take the individual reactions; I will use the K _{eq}s to find the ΔGs, then I will just add the ΔGs.*0829

*Again, you are here; you want to go here.*0837

*You can go this way, or you can go this way; it does not really matter.*0840

*Let’s go ahead and do this; let’s do this one in red.*0845

*Again, this is just a question of personal taste; you know that is all it is- alternate procedure.*0849

*Some people’s minds work one way; some people’s minds work another.*0855

*As long as we end up in the same place and understand what is happening, that is what we are after.*0858

*We want you to understand what is happening, so alternate procedure.*0862

*Let me go ahead and calculate for reaction 1.*0866

*Our ΔG is equal to, well, -RT ln K.*0874

*So, it is -8.315 x 298 times the logarithm of the - let’s see, what have I got - logarithm of the 890.*0879

*OK, and I end up with -16.8kJ/mol.*0897

*That is one of the reactions; now, I will deal with the other reaction, reaction no. 2*0905

*ΔG = -8.315 x 298 times the logarithm of 270, I believe, correct?*0910

*Yes, OK, 270, and for this one, I get -13.8kJ/mol.*0918

*Well, when I add those together, my ΔG net is equal to this, plus this, -30.6kJ/mol- that is it, very simple.*0930

*You can either work with the K _{eq}s, and then use the ΔG equation, or you can use the K_{eq}s that were given for each individual equation.*0947

*Get the ΔGs, and then add the ΔGs- that is all it is.*0955

*The only thing you have to watch out for is, again, with the K _{eq}.*0959

*When you are adding one equation to another, you add their thermodynamic quantities, whether it is ΔG, ΔS, or ΔH.*0962

*You add the thermodynamic quantities to get the thermodynamic quantity of the net reaction.*0968

*When you are adding 2 equations together, the equilibrium constant for the net reaction is the product of the equilibrium constants for those 2 reactions- that is all you have to watch out for.*0974

*It is the thermodynamic quantities that are added, it is the K _{eq}s that are multiplied.*0987

*OK, let’s see what we have got here.*0991

*OK, a little bit of a long problem here but a good one, nevertheless.*0998

*Let’s go ahead and stay with red, so example no. 4.*1004

*Now, given the following half reactions, we have crotonyl-coenzyme A + 2 electrons + 2 hydrogen ions goes to butyryl-coenzyme A, and we have a standard reduction potential of minus...let me write this a little bit better here.*1014

*The standard reduction potential on this is equal to -0.015V for that half reaction.*1050

*Now, we also have our NAD ^{+} reaction + H^{+} + 2 electrons goes to NADH, and the standard reduction potential on this is -0.320.*1059

*Here are the questions that we would like you to deal with.*1084

*Now, let me make this M a little bit bigger, at 1M for each species and a pH equal to 7.0 - basically under standard conditions - we would like you to write the spontaneous reaction that will take place.*1089

*Write the spontaneous reaction that will occur.*1116

*That is going to be the first part, OK?*1122

*Now, part B, what we would like you to do is calculate what are the ΔG standard and the K _{eq} for this reaction that you have written.*1125

*And part C, here is the interesting one.*1146

*Describe and demonstrate quantitatively what happens when we begin with the following concentrations, with a crotonyl-CoA concentration of 0.5, a but-CoA concentration of 1.0, an NAD ^{+} concentration of 1.0 and an NADH concentration of 0.1.*1151

*We are going to switch the concentrations around; we would like you to describe what it is that is going to happen, and actually do it for us quantitatively, confirm what is going to happen quantitatively.*1215

*OK, well, let’s go ahead and take a look.*1224

*Part A, we notice that, again, we are looking at a reduction potential; so in a table of reduction potentials, they are all written as reductions.*1229

*You are going to take a look at the one that has the higher reduction potential.*1237

*The other reaction, that is the one that is going to be oxidized; you are going to switch that one.*1241

*In this case, the crotonyl-CoA reaction is -0.015, the NAD ^{+} reduction is -0.320.*1245

*This is more negative, so this second reaction is the one that is actually going to switch.*1257

*This first reaction will stay as is; this will stay a reduction.*1264

*This will become an oxidation; that is what we are going to write.*1269

*We are going to write those 2, and we are just going to add them; and then we are going to add the reduction potentials to get a net reduction potential for that reaction.*1272

*Let’s write this out, so part A.*1280

*We have crotonyl-CoA + 2 electrons + 2 hydrogen ions goes to but-CoA.*1284

*And again, our standard reduction potential is -0.015.*1298

*And, of course, we switch the other one, so now it is going to be NADH.*1305

*Now, it is going to be on the left, and it is going to go to NAD ^{+} + H^{+} + 2 electrons.*1310

*And, of course, when we switch the equation, we switch the sign, so it becomes now, +0.320V.*1318

*Now, we are going to add straight down; 2 electrons cancels 2 electrons.*1327

*One H cancels one of the Hs, and we are left with the following net reaction.*1333

*We have crotonyl-CoA plus the NADH; this is being reduced, right?*1337

*The crotonyl-CoA is being reduced to butyryl-CoA plus this thing, right there; and it is going to go but-CoA + NAD ^{+} and its net.*1347

*I will write that, and I will write net; you see, just add those 2, and you get a +0.305.*1365

*This is the spontaneous reaction that is going to take place, if you put these together under the appropriate circumstances and with the right enzyme.*1373

*The crotonyl-CoA is going to be reduced by the NADH to but-CoA.*1380

*The NADH is going to be oxidized to NAD ^{+}- that is it.*1385

*It is always going to be like this; just take the one that has the higher reduction potential.*1389

*Leave it alone; take the one that has the lower reduction potential.*1394

*Switch it around, and then cancel the electrons; I mean, you might have to multiply by something to cancel electrons, but in general, it is always going to be 2 electrons 2, 4, 6- something like that.*1397

*Now, part B, well, part B is actually really, really easy; we wanted the ΔG, and we wanted the K _{eq}.*1407

*Well, we have this, and we have a relation for that.*1414

*The standard free energy change is equal to -N x F times the E standard for the reaction.*1419

*We just put the numbers in; there are 2 electrons that are transferred in this reaction, so it is 2 there.*1428

*The Faraday constant is 96,485 C/mol of electrons transferred, and the E of the reaction is the 0.305.*1433

*When we do that, we end up with -58,856J or if you prefer -58.9kJ- that is it.*1446

*Here is your answer; that is our ΔG.*1465

*Well, we want the K _{eq}; we want the K for the net reaction.*1469

*Well, that is easy; that is just the equation for ΔG rearranged.*1473

*It is going to be E to the -ΔG / RT.*1478

*When we put these values in E ^{-58,856} divided by 8.315 - and it looks like the temperature is still 298, that is 25°C - we end up with a K_{eq}.*1485

*The K _{net} is equal to a huge number: 2.06 x 10^{10}- highly favorable reaction.*1507

*We know this already from the ΔG; the ΔG and the K _{eq}, they are just alternate ways of describing where a reaction is, how far from equilibrium- that is it.*1517

*You have a coin; you have heads or tails.*1529

*This is just 2 sides of a coin; the relationship between the two is, it is just a constant RT- that is it.*1531

*K _{eq} and ΔG, they tell us the same thing; they tell us how far from equilibrium something is.*1539

*In this particular case, it is really far from equilibrium; it wants to be over here, on the right-hand side- that is all it says.*1544

*OK, now, let’s see what we can do with part C.*1552

*Part C says, instead of starting with these 1M concentrations, the standard concentrations, we start with different concentrations.*1558

*So, we are going to have to use the Nernst equation and a reaction quotient, different numbers in the reaction quotient.*1565

*Let’s go ahead and do that; Part C, we are going to be using this.*1572

*E of the reaction is going to equal E standard - RT / nF x the ln of Q.*1579

*Now, Q here is going to be the following; It is going to be the concentration of the but-CoA, times the concentration of the NAD ^{+}, over the concentration of the crot-CoA, times the concentration of the NADH.*1589

*That is our reaction quotient; let’s go ahead and put the values in.*1612

*The E of the reaction is equal to...well, we found the E standard; sorry, not the chemical standard- the biochemical standard.*1617

*It is 0.305 - 8.315 x 298 divided by their 2 electrons that are transferred.*1625

*96,485 is the Faraday constant; the logarithm of...well, the but-CoA concentration was 1.*1639

*The NAD ^{+} concentration was 1; the crot-CoA was going to be 0.5, and the NADH is going to be 0.5*1651

*When we put that in, we get 0.305 - 0.0385 = 0.267- there we go.*1663

*Now, we had an initial, under standard conditions, the reduction potential for the net reaction was 0.0305.*1682

*It is going down a little bit; it is still spontaneous.*1693

*This is still positive; it is still a spontaneous reaction, but now, it is less spontaneous.*1696

*Under these conditions, there is less free energy available to do useful work.*1702

*There is still plenty of free energy available; I mean, 0.267 is a huge number relatively speaking, but there is less than there was under standard conditions.*1707

*Now, our qualitative response is the reaction is still spontaneous.*1717

*It will still go in the direction that we have written it but less so- that is it; that is all that is going on here.*1724

*OK, let’s go ahead and finish off with a very interesting kind of example.*1737

*Let’s see if we can, sort of, make sense of this.*1744

*Yes, Alright; I think I am going to do this one in red, and I think I am going to actually start on another page.*1748

*Let me go ahead and move to the next page here; OK, so example 5.*1763

*Let me see if I have enough, alright, I do, so example 5.*1768

*Now, consider the following reaction.*1775

*It is going to be ATP + H _{2}O, the hydrolysis of the adenosine triphosphate going to ADP, + inorganic phosphate.*1787

*We have the ΔG standard for this, is equal to -30.5kJ/mol of ATP.*1801

*Now, ΔG is under biochem standards - right - of pH equal to 7.0 or a hydrogen ion concentration equal to 10 ^{-7}.*1810

*You know, I have never liked the notion of pH; it just really, really bothers me.*1837

*It has always bothered me; we are dealing with concentrations.*1840

*Let’s just deal with concentrations, but pH is everywhere, so that is fine.*1844

*So, pH7, that just means that the hydrogen concentration is 10 ^{-7}.*1849

*OK, now, here is our question.*1853

*If we drop the pH to 4.0 from 7.0, in other words, if we make this environment more acidic, pH to 4.0, our question is: will the reaction above become more exergonic or less exergonic.*1859

*In other words, if we drop the pH and make it more acidic, is there going to be more free energy available to do work, or is there going to be less free energy able to do work.*1896

*Is this going to get negative, more free energy; or is it going to get positive, less free energy- that is our question.*1905

*And, of course, the quantitative version: what will be - there is the qualitative and the quantitative - the new ΔG standard under conditions of pH4 instead of pH7- that is the real question.*1913

*OK, well, qualitatively, we can answer the first part as follows.*1935

*Let me go ahead and draw a little bit of a line here.*1940

*Qualitatively, we can answer the first part as follows.*1944

*Here is where we are getting into the details of what it is that is going on; often in biochemistry or in chemistry, there are going to be different things within a given reaction.*1963

*There are going to be different aspects that we are interested in; we do not want all of the details all the time.*1970

*It is just going to clutter things up; in this particular case, we want the details, and what I mean by details is we want to talk about what is going on where- where is each particle going, and what is the charge on each particle.*1975

*Well, the hydrolysis of ATP actually takes place like this.*1985

*ATP is carrying a 4- charge.*1991

*When it is hydrolyzed by a water molecule, it is going to form an ADP molecule, which is carrying a 3- charge plus an inorganic phosphate, which in general, is carrying a 2- charge; and it is going to have this.*1996

*It is going to release some hydrogen ion into the aqueous medium- this is what is really going on.*2013

*We did not write that up here; notice, we did not have this up there.*2018

*Now, dropping the pH means raising the hydrogen ion concentration*2022

*Well, if you raise the hydrogen ion concentration, you know by le Chatelier's principle that you are going to end up pushing the reaction that way.*2028

*That is what is going to happen here; qualitatively, you can answer this question.*2035

*It is actually going to become less exergonic; there is going to be less free energy available.*2039

*This -30.5 is going to get less negative; it is going to go up to maybe a -20, -15, -10, who knows- that is what happens.*2045

*It is good to know exactly what is going on in a particular reaction.*2052

*In general, we do not really concern ourselves with stuff like that, but it really is a great idea to understand the details.*2056

*So, yes, we want you to have a big picture, but there are certain things, certain reactions, that are so ubiquitous, we need you to know exactly what is happening- this is what is going on.*2063

*Dropping the pH will push the reaction that way.*2075

*OK, now, let’s do the quantitative; now, let’s do the math.*2080

*OK, let’s see what we can do here; I think I am going to start that on the next page.*2085

*Now, the biochem - so, let’s write this all out - standard for ΔG, it already accounts for a hydrogen ion concentration equal to 10...let me just write it here.*2091

*It already accounts for the fact that the hydrogen ion concentration is equal to 10 ^{-7}M.*2120

*Now, what we have to do is...well, let me write it out.*2130

*Well, I will say it; what we have to do is recover not the biochem standard but the normal chemical standard that does not account for this, where we say 1M concentration of every species.*2135

*Well, that 1M concentration also includes the hydrogen ion.*2151

*A 1M concentration, hydrogen ion, -log of 1, you are going to get a pH of 0; but we cannot do pH of 0.*2155

*This is why the biochemical standard uses a pH of 7; that is why we have this thing right here, the standard plus that little mark that lets us know that we are at pH7.*2165

*However, in order to actually find what the new ΔG standard is at a different pH - pH4 in this case - we have to recover the original chemical standard and then work from there forward, and find a new biochemical standard.*2174

*That is what we are going to do; we are going to run 2 calculations.*2190

*What we have to do, and this is the kind of analysis that you want to think about.*2193

*You want to understand what it is that we mean when we say biochemical standard.*2198

*When you see some number like a reduction potential that says that we have accounted for a pH of 7, what does that mean?*2203

*It means that we have taken the chemical standard; we have switched the number around.*2211

*We have recalculated, and we have entered new numbers in this table.*2215

*Well, in order to find a new standard, we have to go back to the original that we came from and then work forward from there.*2219

*What we have to do is recover that ΔG standard, then calculate a new ΔG standard biochem at a pH equal to 4.*2226

*That is what we are doing; this pH7, this pH4, it is a new biochemical standard.*2252

*Let’s go ahead and actually recover our ΔG; well, here is the equation.*2260

*The ΔG biochemical standard is equal to the ΔG chemical standard + RT ln.*2267

*Everything is the same, ADP, PI; but at this time, the H is put in there.*2277

*Now, we are using the entire equation; see, this is where it comes from.*2288

*We take the chemical standard; we recalculate using this equation for finding the new ΔG by including the H.*2293

*When we get this number, that is the number we put into the tables put into biochemistry texts.*2299

*It is the ΔG chemical standard + RT ln with the actual hydrogen ion in the reaction quotient- there, over ATP.*2304

*OK, now, let’s find...so, this is our variable.*2318

*We want to find this so that we can go backward from that; well, we want to know the ΔG standard.*2322

*It is the -30.5; our equation is -30,500 - remember, we are working in Joules - equals the chemical standard + 8.315 x 298 times the logarithm of...well, the concentration of ADP is 1.*2329

*The concentration of PI is 1; the hydrogen ion concentration at pH7 is 10 ^{-7}.*2351

*The concentration of ATP is 1; when we do this and solve for ΔG, what we get is the following: ΔG chemical standard = 9,438J/mol.*2360

*Notice, this is positive; this is our chemical standard.*2380

*We have recovered it from the biochem standard; we have just worked this equation backwards.*2384

*Instead of looking for something on the left side of the equation, we are looking for this thing.*2388

*Now, we take this number, and we readjust using a pH of 4 or a hydrogen ion concentration of 10 ^{-4}.*2393

*We go, ΔG, biochem standard for pH equal to 4, is going to equal 9,438 plus same thing, 8.315, the RT, + 8.315, times our 298, times the logarithm of, again, this time it is 1 for ADP.*2407

*It is 1 for PI, except now, it is going to be 10 ^{-4} / 1 for ATP.*2444

*OK, when we run this calculation, we end up with the following number.*2454

*ΔG standard for a pH equals 4 is equal to -13,384J/mol or -13.4kJ/mol.*2461

*Notice, this has confirmed the fact that it is less exergonic than the -30.5.*2483

*We knew what is going to happen qualitatively; now, we took care of it quantitatively.*2490

*This confirms it, and that is it.*2496

*Again, the biochem standard accounts for the pH equalling 7, the hydrogen ion concentration being 10 ^{-7}.*2499

*I have to use that equation to actually recover the chemical standard, and from there, calculate a new biochemical standard by including the pH now, which is 4 or the hydrogen ion concentration being 10 ^{-4}.*2509

*I hope that makes sense; thank you for joining us here at Educator.com.*2522

*We will see you next time for more problems on bioenergetics.*2526

*Take care, bye-bye.*2531

2 answers

Last reply by: Professor Hovasapian

Mon Feb 25, 2013 4:00 PM

Post by Nigel Hessing on February 25, 2013

Hmm, you didn't convert the gas constant into kilojoules first it should be 0.08315 as 8.315 is in joules.