Wize University Chemistry Textbook > Thermodynamics Part 2: (2nd Law-entropy, 3rd Law, Gibbs Free Energy, Spontaneity)

Connecting Gibbs Free Energy and Equilibrium

Chemical Equilibrium and ΔG

In general the Gibbs free energy can be related to the reaction quotient by the following expression

$\Delta G=\Delta G^o +RT\ln Q$

Recall:

∆G < 0 for a spontaneous process

∆G = 0 at equilibrium

If the reaction is at equilibrium then:

$\Delta G=0 \hspace{20pt} Q=K$

$\therefore$

$\Delta G^o =-RT\ln K$

We can rearrange this equation to solve for K:

$\ln K=\frac{\Delta G^o}{-RT}$

$e^{\ln K}=e^{\left(\frac{\Delta G^o}{-RT}\right)}$

$\ln e\left(K\right)=e^{\left(\frac{\Delta G^o}{-RT}\right)}$

$K=e^{\left(\frac{\Delta G^o}{-RT}\right)}$

Gibbs Free Energy and Chemical Reactions

Gibbs free energy is a sate function, so just like enthalpy we can use Gibbs free energies of formation to determine the change in Gibbs free energy for a reaction. This equation can only be used when the temperature is 298K!

$\Delta_rG^o=\sum_iv_i\Delta_fG^o_{m,i}(products) -\sum_jv_j\Delta_fG^o_{m,j}(reactants)$

van’t Hoff Equation

$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_{^{1_{ }}}}\right)$

**memorize this equation as well!

ΔGo and K

If ΔGo < 0 then K is...

K > 1

If ΔGo > 0 then K is...

0 < K < 1

If ΔGo = 0 then K is...

K = 1

Note: that ΔGo does not determine spontaneity (remember ΔG does!) ΔGo only tells us about the value of K

Only when Q = 1 or everything is in standard state, or everything is in standard state, will ΔGo tell us about spontaneity.

Consider the following reaction at 25°C:

𝐹𝑒(��𝐻)2(𝑠) ⇌ 𝐹𝑒2+(𝑎��) + 2 𝑂𝐻−(𝑎𝑞)

Calculate ΔG° for the reaction. Ksp for Fe(OH)2 is 1.6 × 10−14.

The reaction we are asked to calculate ΔG0 for is the solubility of Fe(OH)2 in water ie, Ksp reaction. Therefore, we can use the relationship between K and ΔG0 to find the answer.

\Delta G^o=-RT \ 1n\ K

\Delta G^o=-\Big(8.314\frac{J}{mol\ K}\Big)(298\ K)1n(1.6\times10^{-14})

\Delta G^{o}=78.7\ \frac{kJ}{mol}\ \left(79\ \frac{kJ}{mol}\ considering\ sig\ figs\right)

ΔrS° and ΔrH° are known to be –225.3 J K–1mol–1 and –32.8 kJ mol–1 respectively for the following reaction:

$CO_2+H_2\to HCOOH$

a. Is this reaction spontaneous under standard conditions at T = 298K and P = 1 atm?

b. Are reactants or products favoured in the above conditions?

c. At what temperature does the reaction favour the opposite?

ΔG = ΔH - T(ΔS)

ΔG = -32,800 J/mol - (298K)(-225.3 J/Kmol)

ΔG = +34k J/mol

Since the value is positive, the reaction is not spontaneous.

b.

Since ΔGo is positive it means that reactants are favoured over the product under STANDARD conditions.

A reaction transitions from reactants to products or vice versa when ΔGo = 0. In our case, ΔGo > 0 first so reactants are favoured. ΔGo = 0 is the middle point and when ΔGo < 0 products are favored.

Assuming ΔHo and ΔSo are temperature independent:

$\Delta G^o=0=\Delta H^o -T\Delta S^o$

$T=\cfrac{\Delta H^o}{\Delta S^o} =\Bigg(\cfrac{-32800\frac{J}{mol}}{-225.3\frac{J}{Kmol}}\Bigg)=145.6K$

So any temperature below 145.6K would result in a negative value for ΔGo , where products are favored over reactants!

If you get confused with deciding if the temperature should be below or above 145.6 to get non-spontaneous, think of the signs:

We want ΔG to be negative, ΔH is negative and ΔS is negative:

ΔG = ΔH - T(ΔS)

(-G) = (-H) - T(-S)

Now the minus sign multiplied by the negative sign in front of ΔS will multiply to create a positive sign:

(-G) = (-H) + T(S)

Now we know that when T = 145.6K, ΔG = 0 because the negative H is cancelled out perfectly by +T(S)

If I increased the temperature, then -H + TS would make ΔG positive

If i decreased the temperature, then -H + TS would make ΔG negative

Another way to figure out if ΔG is negative at a temperature lower or higher than 145.6K is by plugging in other temperature values:

a) If we plug in a higher temperature than 145.6K (can pick any number let's do 160K)

ΔG = ΔH - T(ΔS)

ΔG = -32,800J/mol - (160)(-225.3)

ΔG = +3,248 J/mol

Now we know that we get a +ΔG when we use a HIGHER temperature.

b) So we must get a -ΔG when we use a LOWER temperature, but let's check:

We can pick any temperature lower than 145.6K to check this. Let's check 135K

ΔG = ΔH - TΔS

ΔG = -32,800 - (135)(-225)

ΔG = -2,425 J/mol

Now we are sure that we get a -ΔG when we use a LOWER temperature (in this example)

The equilibrium constant for the decomposition of N2O4(g) into NO2(g) has an equilibrium constant K = 660021 at 298K.

Write the balanced chemical reaction for the decomposition of N2O4(g).
What is ΔGo for the above reaction?
It is known that ΔSo = +165 J mol-1 K -1 and ΔHo = +15.9 kJ mol-1 . Assuming ΔHo and ΔSo are temperature independent, calculate the equilibrium constant, K.

$N_2O_{4(g)}\to2NO_{2(g)}$

deltaG=deltaG0 + RTlnQ

At equilibrium, deltaG=0, deltaGo=-RTlnK

This equation can only be used in standard conditions:

$\Delta G^0=-RT \ln K=-\bigg(8.314\cfrac{J}{K mol}\bigg)(298K)\ln (660021)=-33.2kJ/mol$

c. Solve for K:

deltaGo = deltaHo - TdeltaSo

deltaGo = 15,900 J/mol - 298K(165 J/molK)

deltaGo = -33,270 J/mol

deltaG0=-RTlnK

lnK = deltaG0/-RT

elnK = edeltaG0/-RT

lne(K) = edeltaG0/-RT lne = 1

$K=e^{-\frac{\Delta G^0}{RT}}=e^{-\frac{-33270}{8.314\times298}}$

K = e13.428...

K = 679063

In British Columbia there has been fierce debate over whether we should increase or halt production of LNG liquefied natural gas (CH4). One alternative to liquefying CH4 (bp = -161.5 oC) is to convert it directly into methanol (CH3OH) which is a liquid at room temperature and pressure in the reaction shown below.

$CH_{4(g)}+1/2O_{2(g)}\to CH_3OH_{(1)}$

Image illustrating Chemical Equilibrium and ΔG Practice #3

Calculate ΔHo , ΔSo and ΔGo at 25oC for the conversion of natural gas to methanol. Is this process spontaneous?
Calculate the equilibrium constant for this reaction at 25oC.
What are three things you could do to favour the production of methanol?
Under the following conditions is the reaction spontaneous? (see below)

$\Delta H=\Delta_f H^o(products)-\Delta_f H^o(reactants)$

$=\bigg[-238.7\cfrac{kJ}{mol}\bigg]-\bigg[-74.8\cfrac{kJ}{mol}\bigg]=-163.9\cfrac{kJ}{mol}$

$\Delta S=S^o (products)-S^o(reactants)$

$=\bigg[126.8\cfrac{J}{mol K}\bigg]-\bigg[186\cfrac{J}{molK}+(0.5)205\cfrac{J}{molK}\bigg]=-161.7\cfrac{J}{Kmol}$

$\Delta G^o =\Delta H^o-T\Delta S^o =\bigg(-163900\cfrac{J}{mol}\bigg)-(298K)\bigg(-161.7\cfrac{J}{Kmol}\bigg)=-115713.4\cfrac{J}{mol}$

Since deltaGo is negative, it is spontaneous.

$K=e^{-\frac{\Delta G}{RT}}=e^{-\frac{-115713.4}{8.314*298}}=1.92x10^{20}$

Increase partial pressure of CH4 or O2
Remove Methanol from the reaction as formed
Cool the reaction down .
Increase Total Pressure / Decrease Total Volume

$T=1000K$

$p(O_2)=100atm$

$p(CH_4)=200atm$

p(CH3OH)=1atm

$\Delta G=\Delta G^o+RT \ln Q=-115713.4\cfrac{J}{mol}+\bigg(8.314\cfrac{J}{Kmol}\bigg) (1000K)\ln \frac{1}{(100)^{\frac{1}{2}}(200)}=-178.9\cfrac{kJ}{mol}$

Since deltaG <0 the reaction is spontaneous in the forward direction!

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Consider the following reaction

\rm 2\ {SO_2} \quad+\quad {O_2}\quad \xrightleftharpoons{\hspace{1.5cm}}\quad 2\ {SO_3}

$\Delta H^0=-180\ kJ/mol$

$K$ is measured to be $9.9 \times 10^2$ at $800\ K$ , what temperature would the reaction need to be run at to get an equilibrium constant of $1\times10^6$ ?

I have answered this question

Extra Practice

Gibbs Free Energy: Formation

The value of ΔGo for the following reaction is -2745 kJ. Use this value and the information on the Data Sheet to calculate the standard free energy of formation, ΔGfo of C4H10 (g) in kJ/mol.

C4H10 (g) + 13/2 O2 (g) $\rightarrow$ 4 CO2 (g) + 5 H2O (l)

ΔGfo (CO2(g)) = -394.4 kJ/mol ΔGfo (H2O(l)) = -237.13 kJ/mol

Thermodynamics

The Solubility Constant for AgI(s) at 25oC is 8.3x10-17. What is ΔG(rxn) given that the current concentrations of ions are: [Ag+]=9.1x10-9 & [I-]=9.1x10-9

1)Write the variables we know/trying to solve for

2) Out of the different equations for deltaG, which one should we use?

Gibbs Free Energy

Determine the value of K for these reactions

Gibbs free Energy: Equilibrium

At 699 K, $\Delta$ Go = -23.25 kJ for the reaction H2 (g) + I2 (g) ⇌ 2 HI (g). Calculate $\Delta$ G for this reaction if the reagents are both supplied at 10.0 atm pressure and the product HI is at 1.00 atm

pressure.

For a reaction with a negative ΔGo value, which of the following statements is/are FALSE?

1. The equilibrium constant K is greater than one.

2. The reaction is spontaneous when all the reactants and products are in their standard states.

Use the following reactions to answer the questions below

$Cu(s)+2Ag^+(aq)⇌Cu^{2+}(aq)+2Ag(s)$ Spontaneous; E = 0.46 V
$Cu(s)+Pb^{2+}(aq)⇌Cu^{2+}(aq)+Pb(s)$ Non-Spontaneous; E = -0.47 V

Connecting Gibbs Free Energy and Equilibrium

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Chemical Equilibrium and ΔG

Gibbs Free Energy and Chemical Reactions

van’t Hoff Equation

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ΔGo and K

b.

c. Solve for K:

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Extra Practice

The Solubility Constant for AgI(s) at 25oC is 8.3x10-17. What is ΔG(rxn) given that the current concentrations of ions are: [Ag+]=9.1x10-9 & [I-]=9.1x10-9

1)Write the variables we know/trying to solve for

2) Out of the different equations for deltaG, which one should we use?