Gibbs free energy change

In a nutshell

The Gibbs free energy change tells you if a reaction will occur. If a reaction is feasible, the value will be zero or negative. The temperature a reaction becomes feasible can be calculated. The enthalpy change can be calculated from the energy change and moles. This can be used to calculate Gibbs free energy change.

Equations

word equation	symbol equation
$Gibbs\> free\> energy\>change=enthalpy \>change\>-\>(temperature\>\times\>entropy\>change)$	$\triangle G=\triangle H-T\triangle S$
$temperature=\dfrac{enthalpy\>change}{entropy\>change}$	$T=\dfrac{\triangle H}{\triangle S}$
$Gibbs\>free\>energy\>change=-(gas\>constant\times temperature)\times(logarithm\>of\> {equilibrium \>constant})$	$\triangle G=-RT\>ln\>{K}$
$energy\>change=mass\>\times\>specific\>heat\>capacity\>\times\>temperature\>change$	$q=mc\triangle T$
$enthalpy\>change=\dfrac{energy\>change}{moles}$	$\triangle H =\dfrac{q}{n}$
$moles=\dfrac{mass}{Molar \ mass}$	$n=\dfrac{m}{M}$
$entropy\>change\>of\>a\>system\>=\>entropy\>of\>products\>-\>entropy\>of\>reactants$	$\triangle S_{system}=S_{products}-S_{reactants}$

Constants

constant	symbol	value
$gas\>constant$	$R$	$8.31\>J\>K^{-1}\>mol^{-1}$
$specific\>heat\>capacity\>of\>water$	$c$	$4.18\>J\>g^{-1}\>K^{-1}$

Variable units

quantity name	symbol	unit name	unit
$Gibbs \>free\>energy\newline change$	$\triangle G$	$joule\>per\>mole\>or\>kilo\>joule\>per\>mole$	$J\>mol^{-1}\>or \newline \>kJ\>mol^{-1}$
$entropy \>change$	$\triangle S$	$joule\>per\>kelvin\>per\>mole\>or\>kilo\> \newline joule\>per\>kelvin\>per\>mole$	$J\>K^{-1}\>mol^{-1}\>or \newline \>kJ\>K^{-1}\>mol^{-1}$
$enthalpy \>change$	$\triangle H$	$joule\>per\>mole\>or\>kilo\>joule \> \newline per\>mole$	$J\>mol^{-1}\>or \newline \>kJ\>mol^{-1}$
$temperature$	$T$	$kelvin$	$K$
$equilibrium\>constant$	$K$	$dependent\>on\>reaction$	$dependent\>on\>reaction$
$energy\>change$	$q$	$joule$	$J$
$moles$	$n$	$mole$	$mol$
$mass$	$m$	$gram$	$g$
$relative\>formula\>mass$	$M_r$	$gram\>per\>mole$	$g\>mol^{-1}$

Gibbs free energy

Gibbs free energy change

Gibbs free energy change tells you if a reaction will happen or not. If a reaction is feasible, the value will be zero or negative. It is dependent on enthalpy change, temperature and entropy change.

$Gibbs\> free\> energy\>change=enthalpy \>change\>-\>(temperature\>\times\>entropy\>change) \\ \ \\ \triangle G=\triangle H-T\triangle S$

The more feasible the reaction is, the Gibbs free energy change will be more negative. However, a negative Gibbs free energy change doesn't mean that the reaction will always be feasible in real life. It's possible that the activation energy is so high that the reaction occurs at a very slow pace.

Determine if a reaction is feasible.

procedure

1.	Calculate the Gibbs free energy change using the equation: $\triangle G=\triangle H-T\triangle S$
2.	Determine if the reaction is feasible: $feasible\>reaction=\triangle G\leq0$

Example

Determine if the oxidation of magnesium is a feasible reaction at $298\>K$ . The enthalpy change is $-601\>kJ\>mol^{-1}$ and the entropy change is $-533\>J\>K^{-1}\>mol^{-1}$ .

$Mg(s)+\dfrac{1}{2}O_2(g)→MgO(s)$

Write the equation for Gibbs free energy change:

$\triangle G=\triangle H-T\triangle S$

Convert the enthalpy change units to be the same as the entropy change units:

$\triangle H=-601\times10^3=-601 \, 000\>J\>mol^{-1}$

Note: You could have changed the entropy change units to $kJ\>K^{-1}\>mol^{-1}$ instead. Choose your units depending on what the question asks for, just make sure they're both in $J$ or in $kJ$ .

Insert known values:

$\triangle G=(-601 \, 000\>J\>mol^{-1})-(298\>K)(-533\>J\>K^{-1}\>mol^{-1})=-442 \, 166\>J\>mol^{-1}$

Round the answer to three significant figures:

$\triangle G=-442 \, 000\>J\>mol^{-1}$

The Gibbs energy change is $\underline{-442 \, 000\>J\>mol^{-1}}$ .

Determine if $\triangle G\leq0$ :

$-442 \, 000<0$

The reaction is feasible as the Gibbs free energy change is less than zero.

Temperature

The temperature that a reaction becomes feasible can be calculated. A reaction becomes feasible when the Gibbs free energy change equals to zero. The rearranged equation for Gibbs free energy change to find temperature is seen below.

Example

Show the equation for working out the temperature for when a reaction becomes feasible.

Write the Gibbs free energy change equation:

$\triangle G=\triangle H-T\triangle S$

Equate $\triangle G$ to zero:

$0=\triangle H-T\triangle S$

Make $T$ the subject:

$T=\dfrac{\triangle H}{\triangle S}$

The temperature for when a reaction becomes feasible can be calculated.

Procedure

1.	Equate the Gibbs free energy change equation to zero.
2.	Rearrange the equation to make temperature the subject.
3.	Insert known values to calculate temperature.

Example

Calculate the temperature for when the production of sulfuric acid becomes a feasible reaction. The enthalpy change is $291\>kJ\>mol^{-1}$ and the entropy change is $299\>J\>K^{-1}\>mol^{-1}$ . Give your answer to three significant figures.

$H_2S_2O_7(aq)+H_2O(l) → 2H_2SO_4(aq)$

Equate the Gibbs energy change equation to zero:

$0=\triangle H-T\triangle S$

Rearrange the equation to make temperature the subject:

$T=\dfrac{\triangle H}{\triangle S}$

Convert the units so that enthalpy change and entropy change units are the same:

$\triangle H=291\times10^3=291 \, 000\>J\>mol^{-1}$

Insert known values to calculate temperature:

$T=\dfrac{291 \, 000\>J\>mol^{-1}}{299\>J\>K^{-1}\>mol^{-1}}=973.24\>K$

Round to three significant figures:

$T=973\>K$

The temperature for when the reaction becomes feasible is $\underline{973\>K}$ .

Equilibrium constant

The equilibrium constant is the ratio of the concentrations of the products and reactants in an equilibrium at a certain temperature. Theoretical feasible reactions tend to have large equilibrium constant values and theoretical non-feasible reactions tend to have very small equilibrium constant values.

$Gibbs\>free\>energy\>change=-(gas\>constant\times temperature)\times(logarithm\>of\> {equilibrium \>constant}) \\ \ \\ \triangle G=-RT\>ln\>{K}$

The Gibbs free energy change can be calculated if the temperature and equilibrium constant are known.

Example

Calculate the Gibbs free energy change for the production of carbon dioxide at $200\>K$ when the equilibrium constant is $3$ . Round your answer to three significant values.

$C(s)+O_2(g)⇌CO_2(g)$

Write the equation linking Gibbs free energy change and the equilibrium constant:

$\triangle G=-RT\>ln\>{K}$

Insert known values to calculate the Gibbs free energy change:

$\triangle G=-(8.31\>J\>K^{-1}\>mol^{-1})\times(200\>K)\times(\>ln\>{3})=-1825.893624\>J\>mol^{-1}$

Round your answer to three significant figures:

$\triangle G=-1830\>J\>mol^{-1}$ 

The Gibbs free energy change is $\underline{-1830\>J\>mol^{-1}}$ .

The equilibrium constant can be calculated with a known Gibbs free energy change and temperature values.

Procedure

1.	Write the equation linking Gibbs free energy change and the equilibrium constant.
2.	Rearrange to make the logarithm of the equilibrium constant the subject: $ln\>K=\dfrac{\triangle G}{-RT}$
3.	Insert known values to calculate the logarithm of the equilibrium constant.
4	Calculate the equilibrium constant using the equation: $K=e^{(ln\>K)}$
5.	Work out the units for the equilibrium constant.

Example

Calculate the equilibrium constant for the reaction to form hydrogen chloride at $950\>K$ . The Gibbs free energy change is $-31 \, 900\>J\>mol^{-1}$ . Round your answer to three significant figures.

$H_2(g)+Cl_2(g)⇌2HCl(g)$

Write the equation linking Gibbs free energy and the equilibrium constant:

$\triangle G=-RT\>ln\>{K}$

Rearrange for the logarithm of the equilibrium constant:

$ln\>K=\dfrac{\triangle G}{-RT}$

Insert known values to calculate the logarithm of the equilibrium constant:

$ln\>K=\dfrac{-31 \, 900\>J\>mol^{-1}}{-(8.31\>J\>K^{-1}\>mol^{-1})\times(950\>K)}=4.04078789$

Calculate the equilibrium constant:

$K=e^{(ln\>K)}=e^{(4.04078789)}=56.87113337$

Round to three significant figures:

 $K=56.9$ 

Work out the units for the equilibrium constant:

Write the expression for the equilibrium constant for the production of hydrogen chloride.

$K=\dfrac{[products]^{number\> of \>product \>moles}}{[reactants]^{number\> of reactants \>moles}}=\dfrac{[HCl]^2}{[H_2][O_2]}$

Insert the concentrations and calculate the units:

 $K=\dfrac{(mol\>dm^{-3})^2}{(mol\>dm^{-3})(mol\>dm^{-3})}=\dfrac{(mol\>dm^{-3})^2}{(mol\>dm^{-3})^2}=no\>units$

The equilibrium constant is $\underline{56.9}$ .

Enthalpy change

The enthalpy change, $\triangle H$ , can be calculated from the energy change and moles. The energy change equation is given below.

$energy\>change=mass\>\times\>specific\>heat\>capacity\>\times\>temperature\>change \newline q=mc\triangle T$

This can be used to calculate the enthalpy change by dividing the energy change by moles.

$enthalpy\>change=\dfrac{energy\>change}{moles} \\ \ \\ \triangle H =\dfrac{q}{n}$

The calculated enthalpy change is used to calculate the Gibbs free energy change.

Procedure

1.	Write the equation for energy change: $q=mc\triangle T$
2.	Calculate the energy change.
3.	Calculate the moles: $n=\dfrac{m}{M_r}$
4.	Calculate the enthalpy change by dividing the energy change by moles: $enthalpy\>change=\dfrac{energy\>change}{moles} \newline \triangle H =\dfrac{q}{n}$
5.	Write the Gibbs free energy change equation: $\triangle G=\triangle H-T\triangle S$
6.	Insert known values to calculate Gibbs free energy change.

Example

Calculate the Gibbs free energy change in $J\>mol^{-1}$ when $25.0\>g$ of ammonium bromide is dissolved in water at $290\>K$ . The total mass of solution is $950\>g$ . The temperature change is $2.50\>K$ . The entropy values for ammonium bromide is $91.0\>J\>K^{-1}\>mol^{-1}$ , ammonium is $115\>J\>K^{-1}\>mol^{-1}$ and bromide is $60.0\>J\>K^{-1}\>mol^{-1}$ . The specific heat capacity of water is $4.18\>J\>g^{-1}\>K^{-1}$ .

$NH_4Br(s)→NH_4^+(aq)+Br^-(aq)$ 

Write the equation for energy change:

$q=mc\triangle T$

Insert known values to calculate energy change:

$q=(950\>g)\times(4.18\>J\>g^{-1}\>K^{-1}) \times(2.50\>K)=9927.5\>J$

Calculate the moles:

$moles=\dfrac{m}{M_r}=\dfrac{25.0\>g}{97.9 \>g\>mol^{-1}}=0.2553626149\>mol$

Calculate the enthalpy change:

$\triangle H=\dfrac{q}{n}=\dfrac{9927.5\>J}{0.2553626149\>mol}=38 \, 876.09\>J\>mol^{-1}$

Write the equation linking Gibbs free energy change and enthalpy change:

 $\triangle G=\triangle H-T\triangle S$

Work out the entropy change by using the equation for entropy change of a system:

$\triangle S_{system}=S_{products}-S_{reactants}=((115)+(60))-(91)=84\>J\>K^{-1}\>mol^{-1}$ 

Insert known values to calculate Gibbs free energy change:

$\triangle G=(38 \, 876.09\>J\>mol^{-1})-((290\>K)\times(84\>J\>K^{-1}\>mol^{-1}))=14 \, 516.09\>J\>mol^{-1}$

Round to three significant figures::

 $\triangle G=14 \, 500\>J\>mol^{-1}$

The Gibbs free energy change is $\underline{14 \, 500\>J\>mol^{-1}}$ .

The Gibbs free energy change is positive. This suggests this reaction isn't feasible.