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# Gibbs Free Energy

Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not used as a criterion. To obviate this difficulty, we use a new thermodynamic function called the free energy whose symbol is G.

### Introduction

In 18th century, Josiah Willard Gibbs, one of the founders of thermodynamics, combined the first law with the second law of thermodynamics and created the free energy function of a system, which has the following equation:

$\Delta{G} = \Delta H - T \Delta S \tag{1}$

• $$\Delta G$$: The change in free energy undergoes a transformation at constant temperature and constant pressure.

• $$\Delta H$$: The change in enthalpy

• $$\Delta S$$: The change in entropy

If $$\left | \Delta H \right | >> \left | T\Delta S \right |$$: the reaction is enthalpy-driven

If $$\Delta H$$ << $$T\Delta S$$: the reaction is entropy-driven

Note: constant pressure q = $$\Delta H$$ and constant temperature are required to derive equation (1).

The enthalpy change is $$\Delta H = \Delta E -P \Delta V$$.

For most biochemical reactions, $$\Delta V$$ (the volume change), is considered very small so that we can ignore it. Thus, we have $$\Delta H$$ is equal to $$\Delta U$$, which leads to:

$\Delta G \cong \Delta U - T\Delta S$

Hence, there are two factors affect the change in free energy $$\Delta G$$:

• $$\Delta U$$ = the change in internal energy
• $$\Delta H$$ = the change in entropy of the system

These factors are contrast to each other, which is a important criterion to determine a reaction is spontaneous or not.

• $$\Delta G < 0$$: reaction is spontaneous in the direction written. (Exergonic)
• $$\Delta G =0$$: the system is at equilibrium and there is no net change either in forward or reverse direction.
• $$\Delta G > 0$$: reaction is not spontaneous and the process proceeds spontaneously in the reserse direction. To drive such a reaction, we need to have input of free energy. (Endergonic)

The affactors affect $$\Delta G$$of a reaction (assume $$\Delta H$$ and $$\Delta S$$ are independent of temperature):

$$\Delta H$$ $$\Delta S$$ $$\Delta G$$ Example
+ + at low temperature: + , at high temperature: - 2HgO(s) -> 2Hg (l) + O2 (g)
+ - at all temperature: + 3O2 (g) ->2O3 (g)
- + at all temperature: - 2H2O2 (l) -> 2H2O (l) + O2 (g)
- - at low temperature: - , at high temperature: + NH3 (g) + HCl (g) -> NH4Cl (s)

Note:

1. $$\Delta G$$ depends only on the difference in free energy of products and reactants (or final state and initial state).  $$\Delta G$$ is independent of the path of the transformation and is unaffected by the mechanism of a reaction.
2. $$\Delta G$$ cannot tell us anything about the rate of a reaction.

The standard Gibbs energy change $$\Delta G^o$$ (at which reactants are converted to products at 1 bar) for:

$aA + bB \rightarrow cC + dD$

$\Delta r G^o = c \Delta _fG^o (C) + d \Delta _fG^o (D) - a \Delta _fG^o (A) - b \Delta _fG^o (B)$

$\Delta _fG^0 = \sum v \Delta _f G^0 (\text {products}) - \sum v \Delta _f G^0 (\text {reactants})$

### Standard-State Free Energy Change

Let's consider the following reaction:

$A + B \leftrightharpoons C + D$

We will obtain $$\Delta{G}$$:

$\Delta{G} = \Delta{G}^o + RT\ ln \dfrac{[C][D]}{[A][B]} \tag{2}$

with

• $$\Delta{G}^o$$ = standard free energy change
• $$R$$ = gas constant = 1.98 * 10-3 kcal mol-1 deg-10
• $$T$$ = is usually room temperature = 298 K

The Gibb's free energy $$\Delta{G}$$ depends primarily on the reactants' nature and concentrations (expressed in the $$\Delta{G}^o$$ term and the logarithmic term of equation (2), respectively).

At the standard state, pH = 7, the standard free change is denoted as $$\Delta{G}^{o'}$$.

At equilibrium, $$\Delta{G} = 0$$: no driving force remains

$0 = \Delta{G}^{o'} + RTln\dfrac{[C][D]}{[A][B]}$

$\Delta{G}^{o'} = -RTln\dfrac{[C][D]}{[A][B]}$  (3)

The equilibrium constant, $$K'eq = \dfrac{[C][D]}{[A][B]}$$

When $$K'eq$$ is large, almost all reactants are converted to products.

Substituting $K'eq$ into equation (3), we have:

$\Delta{G}^{o'} = -RTlnK'eq$

or

$\Delta{G}^{o'} = -2.303RT log_{10} K'eq$

Rearrange,

$K'eq = 10^{-\Delta{G}^{o'}/(2.303RT)}$

Let's calculate $\Delta{G}^{o'}$. For example, what is $$\Delta{G}^{o'}$$ for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate? At equilibrium, we have $K'eq$ = 0.0475 at 298 K and pH 7. We can calculate:

$\Delta{G}^{o'} = -2.303RT log_{10} K'eq$ = (-2.303) * (1.98 * 10-3) * 298 * (log100.0475) = 1.8 kcal/mol

Given:

• The initial concentration of dihydroxyacetone phosphate = 2 * 10-4 M
• The initial concentration of glyceraldehyde 3-phosphate = 3 * 10-6 M

From equation (2),

$$\Delta{G}$$ = 1.8 kcal/mol + 2.303 RT log10(3*10-6 M/2*10-4 M) = -0.7 kcal/mol

Note: $$\Delta{G} < 0$$: the reaction occurs spontaneously when $$\Delta{G}^{o'} > 0$$. A reaction is determined spontaneous or not depends on $$\Delta{G}$$, not $$\Delta{G}^{o'}$$.

### The relationship between free energy and transition state

$S (Substrate) \leftrightharpoons S^{\ddagger} (Transition State) \rightarrow P (Product)$

The transition state has a higher free energy than either substrate or product.

The Gibbs free energy of activation is denoted as $$E(A$$ or $$\Delta{G}^{\ddagger}$$ is equal to transition state's free energy minus substrate's free energy. In catalyzed reaction, enzyme accelerate reaction by decreasing the Gibbs free energy of activation (the activation barrier) or by stabilizing transition states.

### References

1. Chang, Raymond. Physical Chemistry for the Biosciences. Sansalito, CA: University Sciences, 2005.
2. Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, NY: W. H. Freeman and Company, 2006. Page 153-163, 286.
3. Stryer, Lubert. Biochemistry (Third Edition). New York, NY: W.H. Freeman and Company, 1988. Page 181-184.

### Contributors

• Han Le

10:16, 16 Dec 2013

## Classifications

Vet1
Fundamental

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