Cytochrome, a complicated molecule that we will represent as CyFe2+, reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At pH 7.0 the following reduction potentials pertain to this oxidation of CyFe2+: O21g2 + 4 H+1aq2 + 4 e- ¡ 2 H2O1l2 Ered ° = +0.82 V CyFe3+1aq2 + e- ¡ CyFe2+1aq2 E°red = +0.22 V (a) What is ∆G for the oxidation of CyFe2+ by air? (b) If the synthesis of 1.00 mol of ATP from adenosine diphosphate (ADP) requires a ∆G of 37.7 kJ, how many moles of ATP are synthesized per mole of O2?

Video transcript

Hello, everyone in this video, we're given these two reactions right here. And what's happening is that molecular oxygen? So oxygen gas is being reduced to a T. P. So first things first, I want to go ahead and apply Hess's law. And what we can do is find the overall reaction with that. So I can see you're here for my second equation. What I'm gonna do is actually flip my equation. So the products will now become the starting re agents and we're also going to multiply the whole reaction by two. Just so we can cancel some stuff out. So let's go ahead and do that. So again, I'm just multiplying this by two and doing a little flip here, if I do so I'll go ahead and write the resulting equations in green. So for the first equation, of course everything stays the same. So we have 02 gas were acting with four H plus cat ions. And for electrons to give us two moles of H +20. Which is of course in its liquid state. And then for the second equation now we have two moles of N A. D. H. And then we have two moles of N A D plus reacting now with two moles of H plus cat ion and for electrons. Alright, so you see her now that some stuff will definitely cancel out one of them, mainly being the four electrons and then we can see here that the two H pluses will cancel. All right. So now the overall reaction where we're going to get here on the starting material side is we have 02 guests with two moles of H plus and two moles of N A d H to give us two moles of H 20 liquid And two moles of NAD plus. Alright, so now we're gonna go ahead and use the given reduction standard reduction potentials from the problem and calculating for our overall Alright, so we're gonna do is we know that the equation just scroll down so we have a little bit more space. Okay, so new, you know, the overall reaction or the overall equation to use, we can just simply plug in the numbers and we'll get that value. So we know that E is equal to the state of reduction potential of our cathode minus that of our a node. So then plugging those numerical values in. We have 0.82 for the cathode And negative 0.32 for our an ode. So, if we put this into a calculator, we get that. The overall reactions Is going to be 1.1, And just a low Refresher here. That units of volts is also equals two jewels per cool. Um Alright. Now calculating for delta g. So delta g again and we just know the equation, we can simply plug in those values. So delta G is equal to negative and F and r E. All right. All right. So N is going to be of course our formals of our electrons And then our F is just a Um constant and that is 964, columns or C per moles of electrons. Okay. And then our energy. We just calculated which was 1.14 jules per. If you put everything into a calculator, we get that. The numerical value is going to be negative. 439971 units being jules. Let's go ahead and convert this into killer jewels with a simple and direct conversion for every 1000 jewels that we have. We have one killer jewels. So simply dividing this number by 1000. We do. So we can see that the final delta G value is going to be negative. 440 units being killed jules. All right. And this right here is going to be my final answer for this problem. Thank you all so much for watching.