Chapter 17, Problem 17.107QP

In chemistry, the standard state for as solution is 1 M (see Table17.2). This means that each solute concentration expressed in molarity is divided by 1 M. In biological systems, however, we define the standard state for the H⁺ ions to be 1×10⁻⁷ M because the physiological pH is about 7. Consequently, the change in the standard Gibbs free energy according to these two conventions will be different involving uptake or release of H⁺ ions, depending on which convention is used. We will therefore replaced ΔG° with ΔG°′, where the prime denotes that it is the standard Gibbs free-energy change for a biological process. (a) Consider the reaction

$\text{A}+\text{B}\to \text{C}+x{\text{H}}^{+}$

where x is a stoichiometric coefficient. Use Equation (17.13) to derive a relation between ΔG° and ΔG°′, keeping in mind that ΔG is the same for the process regardless of which convention is used. Repeat the derivation for the reverse process:

$\text{C}+x{\text{H}}^{+}\to \text{A}+\text{B}$

(b) NAD⁺ and NADH are the oxidized and reduced forms of nicotinamide adenine dinucleotide, two key compounds in the metabolic pathways. For the oxidation of NADH:

$\text{NADH}+{\text{H}}^{+}\to {\text{NAD}}^{+}+{\text{H}}_2$

ΔG° is −21.8 kJ/mol at 298 K. Calculate ΔG°′. Also calculate ΔG using both the chemical and biological conventions when [NADH] = 1.5 × 10⁻² M, [H⁺] = 3.0 × 10⁻⁵ M, [NAD] = 4.6 × 10⁻³ M, and $P_{H_2}=0.010\text{ atm}$.

Interpretation Introduction

Interpretation:

The standard free energy $\Delta G_{}^{°}and\Delta G_{}^{°\text{'}}$ value has to be derived following the reverse process reaction.

Concept Introduction:

Free energy$(\text{ΔG})$: In thermodynamics free energy or Gibbs free energy is the energy that is used to express the total energy content of a system.  According to second law of thermodynamics, in all spontaneous process is associated with the decrease in free energy of the system.  That is the change in free energy will be negative.

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}\Delta \text{G=}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.314\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{Q=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

The given reverse process reaction is,

$A+B\underset{}{\to }C+x{H}^{+}$

Let us consider the following free energy equation,

$\Delta G=\Delta G°+RTInQ$

Where $R$ is the gas constant $8.314J/K\cdot mol$, (T) is the absolute temperature of the reaction and (Q) is the reaction quotient, the free energy $\Delta G$ depends on two quantities $\Delta G°$ and $RTInQ$

Here the chemical standard of (1M), we can write has,

$\Delta G=\Delta G°+RTIn\frac{\left(\frac{[C]}{1M}\right){\left(\frac{[H^{+}]}{1M}\right)}^x}{\left(\frac{[A]}{1M}\right)\left(\frac{[B]}{1M}\right)}------[1]$

For the biological standard state, so we can write as,

$\Delta G=\Delta G°\text{'}+RTIn\frac{\left(\frac{[C]}{1M}\right){\left(\frac{[H^{+}]}{1\times {10}^{-7}M}\right)}^x}{\left(\frac{[A]}{1M}\right)\left(\frac{[B]}{1M}\right)}-----[2]$

We set the two equations (1) and (2) equal to each other,

$\Delta G°+RTIn\frac{\left(\frac{[C]}{1M}\right){\left(\frac{[H^{+}]}{1M}\right)}^x}{\left(\frac{[A]}{1M}\right)\left(\frac{[B]}{1M}\right)}=\Delta G°\text{'}+RTIn\frac{\left(\frac{[C]}{1M}\right){\left(\frac{[H^{+}]}{1\times {10}^{-7}M}\right)}^x}{\left(\frac{[A]}{1M}\right)\left(\frac{[B]}{1M}\right)}$

$\begin{array}{l}\Delta G°+RTIn{\left(\frac{[H^{+}]}{1M}\right)}^x=\Delta G°\text{'}+RTIn{\left(\frac{[H^{+}]}{1\times {10}^{-7}M}\right)}^x\\ \Delta G°=\Delta G°\text{'}+RTIn{\left(\frac{[H^{+}]}{1\times {10}^{-7}M}\right)}^x-RTIn{\left(\frac{[H^{+}]}{1M}\right)}^x\\ \Delta G°=\Delta G°\text{'}+RTIn{\left(\frac{\frac{\cancel{[H^{+}]}}{1\times {10}^{-7}M}}{\frac{\cancel{[H^{+}]}}{1M}}\right)}^x\end{array}$

$\Delta G°=\Delta G°\text{'}+xRTIn\left(\frac{1}{1\times {10}^{-7}M}\right)-----[3]$

Given the reverse reaction $C+x{H}^{+}\underset{}{\to }A+B$, so we can write as

$\Delta G°=\Delta G°\text{'}-xRTIn\left(\frac{1}{1\times {10}^{-7}M}\right)-----[4]$

Interpretation Introduction

Interpretation:

The standard free energy value $\Delta G_{}^{°}$ has to be calculated fallowing the bio chemical reaction at $298K$ .

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG = }\Delta {\rm H}\text{- T}\Delta \text{S}$

Where,

$\text{ΔG }$ is the change in free energy of the system

$\Delta {\rm H}$ is the change in enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta \text{S}$ is the change in entropy in the system

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}\Delta \text{G=}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.314\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{Q=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

Explanation of Solution

Next we calculate the standard free energy $\Delta G°$, we use equation (4) from part (a), the equation (4) becomes,

$\begin{array}{l}\Delta G°=\Delta G°\text{'}-xRTIn\left(\frac{1}{1\times {10}^{-7}M}\right)\\ \Delta G°=\Delta G°\text{'}-(1)(8.314J/mol\cdot K)(298)In\left(\frac{1}{1\times {10}^{-7}}\right)\\ \Delta G°=\Delta G°\text{'}-39.93\text{kJ/mol}\\ \text{(or),}\\ \Delta G°\text{'}=-21.8\text{kJ/mol}\text{+}39.93\text{kJ/mol}\\ \Delta G°\text{'}=18.13\text{kJ/mol}\end{array}$

We can now calculate $\Delta G$ using both conversions,

The chemical standard state is,

$\Delta G=\Delta G°\text{'}+RTIn\frac{\left(\frac{[NA{D}^{+}]}{1M}\right){\left(\frac{[P_{H_2}]}{atm}\right)}^x}{\left(\frac{[NADH]}{1M}\right)\left(\frac{[H^{+}]}{1M}\right)}$

$\begin{array}{l}\Delta G=-21.8\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)In\left(\frac{(4.6\times {10}^{-3})(0.010)}{(1.5\times {10}^{-2})(3.0\times {10}^{-5})}\right)\\ \Delta G=-21.8\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)In\left(\frac{4.6\times {10}^{-5}}{4.5\times {10}^{-7}}\right)\\ \Delta G=-21.8\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)In\left(1.022\times {10}^2\right)\\ \Delta G=-21.8\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)\left(4.62715\right)\\ \Delta G=-10336J/mol\\ ConvertJ/mol\text{into}kJ/mol\\ \Delta G=-10.3kJ/mol\end{array}$

The biological standard state is,

$\Delta G=\Delta G°\text{'}+RTIn\frac{\left(\frac{[NA{D}^{+}]}{1M}\right){\left(\frac{[P_{H_2}]}{atm}\right)}^x}{\left(\frac{[NADH]}{1M}\right)\left(\frac{[H^{+}]}{1\times {10}^{-7}M}\right)}$

$\begin{array}{l}\Delta G=18.1\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)In\left(\frac{(4.6\times {10}^{-3})(0.010)}{(1.5\times {10}^{-2})(3.0\times {10}^{-5}/1\times {10}^{-7})}\right)\\ \Delta G=18.1\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)ln\left(\frac{4.6\times {10}^{-5}}{(4.5)}\right)\\ \Delta G=18.1\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)ln\left(1.022\times {10}^{-5}\right)\\ \Delta G=18.1\times {10}^{3}J/mol+(8.314J/mol\cdot K)(298K)\left(-11.4909\right)\\ \Delta G=-10369J/mol\\ ConvertJ/mol\text{into}kJ/mol\\ \Delta G=-10.3kJ/mol\end{array}$

The expected free energy $\Delta G$ is the same regardless of which standard state we employ.