Hello,
For this question, is it really okay to use Table 1 in order to calculate this? I was taught in biochemistry that ΔG °' is a different quantity than ΔG ° and that certain conversions were required in order to properly use these values, so using Hess's law directly wasn't a method I had in mind. Thanks!
NS FL9: Physical Sciences #39

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 Joined: Sat Mar 30, 2019 8:39 pm
Re: NS FL9: Physical Sciences #39
So you like asking the hard questions, huh?
You actually can, since ΔG′∘ refers to free energy at a constant pH of 7. Disclaimer: This isn't exactly MCAT material to know about in detail, so I'll be just buzzing through this real quick for reassurance sake.
Simply put, for a regular biological system and a reaction that changes the pH, ΔG′∘ is simply the deltaG assuming constant pH. It turns out that you can adjust your regular ΔG∘ expression the following way: ΔG′∘=−RTln(K/[H+])
We simply corrected our equilibrium expression so that [H+] would fall away. That is, initially this was K=[Species 1][H+]/[Protonated species 1]. We divide that by H+ and get [Species 1]/[Protonated species 1]. Again: Do not overthink this, it would be great if you could grok this for the sake of practice with equilibrium expressions, but this isn't directly testable.
But how does this justify using Hess' law? Well, just rewrite your ΔG′∘ expression using the law of logarithms: ΔG′∘=−RTln(K)  (RTln([H+]))
And since RTln([H+) is just a single constant (remember, we said that pH is being held constant, so this [H+] term is constant), we can just say that ΔG′∘=−RTln(K)+m  where m is that constant (RTln([H+]). Now it becomes a little obvious: The additive properties of Hess' law MUST hold if all we did was shift every single term by a constant factor.
That said, where you are right is that this should definitely say ΔG′∘ not ΔG∘, but for a simpler reason: Making this adjustment to the final value of ΔG∘ gives something closer to 73 kJ/mol. That in itself is a bit of an oddity, but is indeed what pops out with the givens. You would not be expected to make this adjustment on the MCAT unless given the actual formula I provided above.
You actually can, since ΔG′∘ refers to free energy at a constant pH of 7. Disclaimer: This isn't exactly MCAT material to know about in detail, so I'll be just buzzing through this real quick for reassurance sake.
Simply put, for a regular biological system and a reaction that changes the pH, ΔG′∘ is simply the deltaG assuming constant pH. It turns out that you can adjust your regular ΔG∘ expression the following way: ΔG′∘=−RTln(K/[H+])
We simply corrected our equilibrium expression so that [H+] would fall away. That is, initially this was K=[Species 1][H+]/[Protonated species 1]. We divide that by H+ and get [Species 1]/[Protonated species 1]. Again: Do not overthink this, it would be great if you could grok this for the sake of practice with equilibrium expressions, but this isn't directly testable.
But how does this justify using Hess' law? Well, just rewrite your ΔG′∘ expression using the law of logarithms: ΔG′∘=−RTln(K)  (RTln([H+]))
And since RTln([H+) is just a single constant (remember, we said that pH is being held constant, so this [H+] term is constant), we can just say that ΔG′∘=−RTln(K)+m  where m is that constant (RTln([H+]). Now it becomes a little obvious: The additive properties of Hess' law MUST hold if all we did was shift every single term by a constant factor.
That said, where you are right is that this should definitely say ΔG′∘ not ΔG∘, but for a simpler reason: Making this adjustment to the final value of ΔG∘ gives something closer to 73 kJ/mol. That in itself is a bit of an oddity, but is indeed what pops out with the givens. You would not be expected to make this adjustment on the MCAT unless given the actual formula I provided above.
Re: NS FL9: Physical Sciences #39
I just like making sure my content side of test prep is ready! But, I'm a bit more relieved on that part now. Thank you!