Chapter 18, Problem 4SEPP

Physical and Biological Sciences

In chemistry, the standard state for a solution is 1M. This means that each solute concentration expressed in molarity is divided by 1M. In biological systems, however, we define the standard state for the ${\text{H}}^{\text{+}}\text{ ions to be 1 }\times {\text{ 10}}^{\text{-7}}M$. Consequently, the change in the standard Gibbs free energy according to these two conventions will be different for reactions involving the uptake or release of ${\text{H}}^{\text{+}}$ ions, depending on which convention is used. We trill therefore replace $\Delta Gº\text{ with }\Delta G\text{º,}$ where the prime denotes that it is the standard Gibbs free-energy change for a biological process.

${\text{NAD}}^{\text{+}}$ and NADH are the oxidized and reduced forms of nicotinamide adenine dinucleotide, two key compounds in the metabolic pathways. For the oxidation of NADH:

${\text{NADH+H}}^{\text{+}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}$

$\Delta G\text{º is -21}\text{.8 kJ/mol at 298 K}\text{. Calculate }\Delta G\text{º}$ for this process, and calculate $\Delta G$ using either the chemical or the biological convention when $\left[\text{NADH}\right]\text{ = 1}{\text{.5 x 10}}^{\text{-2}}M\text{,}\left[{\text{H}}^{\text{+}}\right]\text{ = 3}{\text{.0 x 10}}^{\text{-5}}M\text{,}\left[\text{NAD}\right]\text{ = 4}{\text{.6 x 10}}^{\text{-3}}\text{M, and}{P}_{{\text{H}}_{\text{2}}}\text{= 0}\text{.010 atm}\text{.}$

a)

$18.1\text{kJ/mol,50}\text{.3kJ/mol}$

b)

$21.8\text{kJ/mol,29}\text{.6kJ/mol}$

c)

$18.1\text{kJ/mol,10}\text{.3kJ/mol}$

d)

$21.8\text{kJ/mol,10}\text{.3kJ/mol}$