A) In the reaction: At temperature, and pH of 7.0. From Table 10.3, the Gibbs free energy, for hydrolysis of Glycerol 3-phosphate is found to be or. At standard temperature of 298 K and hydrogen ion concentration of pH 7.0, the value of ideal gas constant would be. When we aren't sure if our reaction is at equilibrium, we can calculate the reaction. The formula for calculating Kc or K or Keq doesn't seem to incorporate the.
![How To Calculate Keq How To Calculate Keq](/uploads/1/2/5/6/125619321/216560010.jpg)
![Keq Keq](/uploads/1/2/5/6/125619321/601671165.jpg)
- [Voiceover] Our goal is to calculate the equilibrium constantK for this reaction, so for this reaction right here. Now we're gonna use the standard reductionpotentials to do so. So in the previous video, wetalked about the relationship between the equilibrium constant K and the standard cell potential. So E zero. So if we can find Ezero for this reaction, we can calculate theequilibrium constant K. And we've seen how to find E zero, the standard cell potentialin earlier videos. So, let's start with thisfirst half reaction here, where we see solidiodine gaining electrons, so it's being reduced toturn into iodide anions. The standard reductionpotential for this half reaction is positive .54 volts. Now we can see that's what's happening in our redox reaction up here, right? We can see see that we're going from solid iodine on the left side and we have iodideanions on the right side. So we're gonna keep this, we're gonna keep thisreduction half reaction. If we look at what'shappening with aluminum, we're going from solid aluminumto aluminum three plus, so solid aluminum must be losing electrons to turn into aluminum three plus. So aluminum is being oxidized here. But this half reaction is written as a reduction half reaction. So we need to reverse it, right? We need to start with solid aluminum. So we reverse our half reaction to start with solid aluminum, so aluminum turns into aluminum three plus and to do that it losesthree electrons, right, so loss of electrons is oxidation. Here is our oxidation half reaction. So we reversed, we reversedour half reaction, all right. And remember what we do to thestandard reduction potential, we just change the sign, all right. So, for the reduction half reaction, the standard reductionpotential is negative 1.66. We reversed the reaction, sowe need to change the signs. So the standard oxidationpotential is positive 1.66 volts. For our reduction half reaction, we left it how it was written, all right. So we're just gonna write ourstandard reduction potential as positive .54 volts. Next, we need to look at ourbalanced equation, all right. We need to make thenumber of electrons equal for our half reactions. So for this first half reaction, I'm just gonna draw a line through, I'm just gonna draw a linethrough this half reaction so we don't get ourselves confused. For our first half reactionhere, we have two electrons. Then over here, right, forour oxidation half reaction, we have three electrons. We need to have the samenumber of electrons. So we need to have six electronsfor both half reactions, because remember theelectrons that are lost are the same electrons that are gained. So we need to multiply ourfirst half reaction by three. All right, if you multiply ourfirst half reaction by three, we'll end up with six electrons. And our second half reaction,we would need to multiply the oxidation half reaction by two, in order to end up with six electrons. So let's rewrite our half reactions. So, first we'll do thereduction half reaction, so we have, let mechange colors again here. And let's do this color. So we have three times, sincewe have three I two now, We have three I two, and three times twogives us six electrons. So three I two plus six electrons, and then three timestwo, all right, gives us, three times two gives us six I minus. All right, so we multipliedour half reaction by three, but remember, we don'tmultiply the voltage by three, 'cause voltage is an intense of property. So the standard reduction potential is still positive .54 volts. So we have positive .54volts for this half reaction. Next, we need to multiply our oxidation half reaction by two. So we have two Al, so this isour oxidation half reaction, so two Al, so two aluminum, and then we have two Al three plus, so two Al three plus, and then two times threegives us six electrons. So now we have our sixelectrons and once again, we do not multiply our standardoxidation potential by two, so we leave that, so thestandard oxidation potential is still positive 1.66 volts. Next we add our twohalf reactions together. And if we did everything right, we should get back our overall equation. So our overall equation here. We have six electronson the reactant side, six electrons on the product side, so the electrons cancel out. And so we have for ourreactants three I two, so we have three I two, plus two Al, and for our products right here, we have six I minus, so six I minus, plus two Al three plus,plus two Al three plus. So this should be ouroverall reaction, all right, this should be the overall reaction that we were given in our problem. Let's double check that real fast. So three I two plus two Al, all right, so right up here, sothree I two plus two Al, should give us six I minusplus two Al three plus. So six I minus plus two Al three plus. So we got back our original reaction. Remember our goal was,our goal was to find the standard cell potential E zero, because from E zero we can calculate the equilibrium constant K. So we know how to do that,again, from an earlier video. To find the standard cellpotential, all right, so to find the standard cell potential, all we have to do is add ourstandard reduction potential and our standard oxidation potential. So if we add our standardreduction potential and our standard oxidation potential, we'll get the standard cell potential. So that would be positive .54 volts, so positive .54 plus 1.66, plus positive 1.66 volts. So the standard potential for the cell, so E zero cell is equal to .54 plus 1.66 which is equal to 2.20 volts. All right, now that we've foundthe standard cell potential, we can calculate the equilibrium constant. So we can use one of theequations we talked about in the last video that relatesthe standard cell potential to the equilibrium constant. So I'm gonna choose, I'm gonnachoose one of those forms, so E zero is equal to 0.0592 volts over n times log of theequilibrium constant. So again, this is from the previous video. So, the standard cellpotential is 2.20 volts, so we're gonna plug thatin, we're gonna plug that in over here, so now we have 2.20 volts is equal to 0.0592 volts over n. Remember what n is, n is the number of moles transferredin our redox reaction. So we go back up here and welook at our half reactions and how many moles ofelectrons were transferred? Well, six electrons were lost, right, and then six electrons were gained. So n is equal to six. So we plug in n is equalto six into our equation. So n is equal to six. And now we have the log ofthe equilibrium constant K, log of the equilibrium constant K. So we just need to solve for K now. So let's get out the calculator and solve for K. So we would have 2.20times six, all right, divided by .0592 and that gives us, let's say 223. So this gives us 223. So 223 is equal to, so this is equal to log ofour equilibrium constant K. So we need to get rid of our log. And we can do that by taking10 to both sides, all right. So if we take 10 to the223 and 10 to the log of K, then that gives us K,the equilibrium constant. So K, the equilibrium constant, is equal to 10 to the 223rd power, which is obviously a huge number. So, huge number, we get a huge value for the equilibrium constant, which is a little bitsurprising, because we only had 2.20 volts, which doesn'tsound like that much. So from only 2.20 volts,we get a huge number for the equilibrium constant. So the reaction goes tocompletion, all right, with a huge value for theequilibrium constant like that, you pretty much don't have anything for your reverse reaction, which is why, which is why we're not drawing any kind of arrow going backwards here. We only have this arrowhere going forwards. So because of our huge numberfor the equilibrium constant.