Question

Calculate delta G at pH 7 for the LDH catalyzed reduction of pyruvate under the following conditions. (See attachment for entire question).

Answer 1

I do not even know where to start.

Answer 2

I think I did it, but I do not know if it is right. First I balanced the equation and I ended up with
\[pyruvate +NADH + H^{+} \rightarrow Lactate + NAD ^{}\].

And Then I used this formula to find E.
\[E= DeltaE -\frac{ RT }{ Fn } \ln \frac{ [A _{red}] [B _{oc}^{n+}]}{ [A _{ox}^{n+}] [B _{red}]}\]

Then plugged what I got for E into \[\Delta G = -nF \Delta E\]

Answer 3

for part a= \[E=0.13V -(\frac{ 8.314*298 }{ 2*96485 }) \ln(1*1) = 0.13V\]
Then, \[\Delta G = \frac{ -2*96,485*0.13 }{ 1000 } =-25.08kJ/mol\]


For part B:
\[E=0.13V -(\frac{ 8.314*298 }{ 2*96,485 }) *\ln(160*160)= -3.21*10^{-4}V \]
So, plug that into the formula for G
\[Delta G= \frac{ -2 \times 96,48 5\times (-3.21*10^{-4}) }{ 1000 }=61.9kJ/m\]

Answer 4

it's mostly right, the ratio should look like this though

a) \(E=0.13V−(\dfrac{8.314∗298}{2*96485})ln(1)=0.13V\)

b)\(E=0.13V−(\dfrac{8.314∗298}{2*96485})ln(160)\)

Answer 5

oohhh, so I do not multiply them. ok. Can you explain why I do not multiply them? And thanks by the way!!

Answer 6

crap, i'm sorry i read the question wrong, you were right the first ime