# Effect of temperature on free energy change of the oxidation of ethanol to ethanoic acid

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## Rationale

The idea of this investigation originated from my visit to a wine manufacturing unit during my summer break in Grade 11. The most surprising fact that intrigued me was storing the wine bottle in a dark place. While trying to find out an appropriate reason for this I got to know that ethanol, the principle organic compound in wine is susceptible to undergo areal oxidation and produce ethanoic acid. To prevent this, alcoholic beverages must be kept in a cool and dark place. Thus, this mean that the oxidation will be favored more if the temperature is increased. Applying the concepts studied in Topic 5 Higher Level portions, I made an attempt to figure out how can we calculate the free energy change of this reaction and predict its spontaneity. The main motive was to understand whether this reaction becomes more or less favorable if the temperature is increased or decreased. This research would allow us to understand an optimum value of the temperature that should be maintained with the most effective storage conditions. This inquiry led me to the research question stated below.

## Research question

How does the spontaneity of the of the oxidation of Ethanol (C2H2OH)to Ethanoic acid (CH3COOH)depends on the temperature in $$K$$at which it is carried out, determined using values of standard free energy change $$(ΔG°)$$in $$kJ.mol^{-1}$$ calculated based on secondary data?

## Background information

### Oxidation reaction

Oxidation Reaction may be defined as a reaction in which a reactant gains one or more oxygen atom or loses one or more hydrogen atom.

Moreover, if the oxidation number of a reactant increases during the process of conversion of reactant to product, it signifies the reaction is undergoes is an oxidation and the reactant has been oxidized.

### Chemical equation of oxidation of ethanol to ethanoic acid

2C2H5OH(l)+O2(g)→2CH3COOH(l)+2H2(g)

The reagents which are used to oxidize ethanol to ethanoic acid are mentioned below:

•  Acidified Potassium dichromate$$[K_2Cr_2O_7]$$
•  Acidified Potassium permanganate$$[KMnO_4]$$
•  Pyridinium chlorochromate (PCC) $$[C_5H_5NH]+[CrO_3Cl]^-$$

The number of hydrogen atom in ethanol is 6 and that of ethanoic acid is 4. So, there is an addition of 1 atom of oxygen and removal of 2 atoms of hydrogen during the conversion from ethanol (reactant) to ethanoic acid (product). So, it can be said to be an oxidation reaction.

Furthermore, the oxidation number of carbons with which the hydroxy group is attached in ethanol is – 1 and that of the carboxylic group in ethanoic acid is +3. As there is an increase in the oxidation number from ethanol and ethanoic acid by 4 units, the reaction can be recognized as an oxidation reaction involving the loss of 4 electrons.

### Enthalpy change of a reaction (∆H)

Enthalpy change of a reaction may be defined as the amount of heat change of a reaction at constant pressure. It is measured in the unit of $$kJ.mol^-1$$ Enthalpy change can have both positive and negative values. A positive value of ∆$$H$$ signifies that energy is absorbed in the reaction and the reaction is endothermic. On the other hand, a negative value of ∆$$H$$signifies that energy is released in the reaction and the reaction is exothermic.

### Entropy change of a reaction (∆S)

Entropy is an intensive thermodynamic property which indicates the randomness of any system. It is represented by the letter S. It is measured in the unit of $$J.mol^{-1}.K^{-1}$$ .$$ΔS$$ is the change in entropy which is determined by the formula -  $$ΔS\ =S_{final}\ -S_{initial}$$

Change in entropy of a system can also be determined by the formula mentioned below -

$$ΔS=\frac{q_{rev}}{T}$$

where,

$$q_{rev}=Amount\ of\ heat \ change\ in\ a\ reversibel\ reaction$$

### Gibb’s free energy (G)

Gibb’s Free Energy is an intensive thermodynamic property which indicates the amount of useful non- mechanical work available in a system. It is measured in the unit of kJ.mol-1. At constant pressure, the expression of Gibb’s Free Energy can be represented as:

G = H - TS

Differentiating both sides with respect to temperature -

∆G = ∆H - ∆(TS)

=>∆G = ∆H - T∆S - S∆T

At constant temperature, $$∆T= 0$$ -

∆G = ∆H - T∆S... ... ... (equation 1)

This above equation is known as Gibb’s Helmholtz Equation.

### Variation of ∆G with respect to temperature

The variation of Gibb’s Free Energy with respect to temperature can be explained using four different cases. According to the Gibb’s Helmholtz Equation, the four different cases can be obtained by taking values with different signs of enthalpy change (∆$$H$$) and entropy change (∆$$S$$). The four different cases are shown below:

 Case Sign of ∆H Sign of ∆S Case 1 Positive (+) Positive (+) Case 2 Positive (+) Negative (-) Case 3 Negative (-) Positive (+) Case 4 Negative (-) Negative (-)

Figure 1 - Table On

To determine the variation of Gibb’s Free energy with respect to temperature, the Gibb’s Helmoltz Equation can be compared with the generalized equation of a straight line as shown below:

The generalized equation of a straight line:

y = mx + ... ... ... (equation 2)

y = Co - ordinate of y - Axis,

x = Co - ordinate of x - Axis,

m = slope of the curve,

c = y - Intercept of the curve

Comparing equation (1) with equation (2):

∆G = - T∆S + ∆H ... ... ... (equation 3)

Here, Gibb’s free energy (∆G) resembles with y as it is the dependent variable of the exploration, Temperature (T) resembles with x as it is the independent variable of the exploration, ∆S is compared with m which is the slope of the curve andH is compared with c which is the value of Y – Intercept of the curve.

### Investigation of case 1

Here, the sign of enthalpy change and entropy change, both are positive. Thus, the equation (3) for this case can be written as:

∆G = - T∆S + ∆H ... ... ... (equation 4)

As the value of enthalpy change in positive, the reaction is endothermic in nature. With an increase in temperature from Point A to Point B, the value of Gibb’s Free Energy decreases and it becomes more negative. As a result, in this phase (AB), the spontaneity of the reaction increases. However, as the value of Gibb’s Free Energy is positive, the reaction is not spontaneous. At B, the value of Gibb’s Free Energy is zero, thus this state can be considered as the equilibrium state. With an increase in temperature from Point B to Point C, the value of Gibb’s Free Energy decreases from zero and it becomes more negative. As a result, in this phase (BC), the spontaneity of the reaction again increases. Thus, it can be said that, in this case, with an increase in temperature the value of Gibb’s Free Energy decreases and the spontaneity increases.

### Investigation of case 2

Here, the sign of enthalpy change is positive and entropy change is negative. Thus, the equation (3) for this case can be written as:

∆G = T∆S + ∆H ... ... ... (equation 5)

As the value of enthalpy change in positive, the reaction is endothermic in nature. With an increase in temperature from Point A to Point B, the value of Gibb’s Free Energy increases and it becomes less negative. As a result, in this phase (AB), the spontaneity of the reaction decreases. However, as the value of Gibb’s Free Energy is positive throughout. Hence, the reaction is always non- spontaneous. However, from the graph, it can be said that with an increase in temperature, the spontaneity decreases.

### Investigation of case 3

Here, the sign of enthalpy change is negative and entropy change is positive. Thus, the equation (3) for this case can be written as:

∆G = T∆S + ∆H ... ... ... (equation 6)

As the value of enthalpy change in negative, the reaction is exothermic in nature. With an increase in temperature from Point A to Point B, the value of Gibb’s Free Energy decreases and it becomes more negative. As a result, in this phase (AB), the spontaneity of the reaction increases. However, as the value of Gibb’s Free Energy is negative throughout. Hence, the reaction is always spontaneous. However, from the graph, it can be said that with an increase in temperature, the spontaneity increases.

### Investigation of case 4

Here, the sign of enthalpy change and entropy change, both are negative. Thus, the equation (3) for this case can be written as:

∆G = + T∆S - ∆H ... ... ... (equation 7)

As the value of enthalpy change in negative, the reaction is exothermic in nature. With an increase in temperature from Point A to Point B, the value of Gibb’s Free Energy increases and it becomes less negative. As a result, in this phase (AB), the spontaneity of the reaction decreases. However, as the value of Gibb’s Free Energy is negative throughout this phase (AB), the reaction is spontaneous. At B, the value of Gibb’s Free Energy is zero, thus this state can be considered as the equilibrium state.

With an increase in temperature from Point B to Point C, the value of Gibb’s Free Energy increases from zero and it becomes less negative. As a result, in this phase (BC), the spontaneity of the reaction again decreases. As the value of Gibb’s Free Energy is positive throughout this phase (AB), the reaction is non-spontaneous. Thus, it can be said that, in this case, with an increase in temperature the value of Gibb’s Free Energy increases and the spontaneity decreases.

## Variables

### Independent variable

The temperature at which the reaction is performed is considered as the independent variable. A wide range of temperature has been chosen. The initial value of temperature has been chosen as 298 K as it represents the room temperature, the final value has been chosen as 412 K to maintain a range of 100 units. An interval of 2K has been chosen to get sufficient number of data points.

### Dependent variable

The standard Gibb’s Free Energy change ($$ΔG^0$$) measured in kJ.mol-1 for the oxidation of ethanol to ethanoic acid as oxygen as a reagent has been chosen as the dependent variable in this investigation. It will be computed using the Gibb’s Helmoltz Equation $$((ΔG^0\ =ΔH^0\ =TΔS^0)$$.

### Controlled variable

• It has been assumed that the values of the enthalpy change and entropy change of the reaction does not depend on the temperature. The magnitude of enthalpy change and entropy change of the reaction has been calculated and the value obtained will be used for all value of temperature in the independent variable.
• The Gibb’s Helmoltz Equation $$(ΔG^0\ =ΔH^0\ =TΔS^0)$$ is applicable only for isothermal and isobaric process. Hence it has been considered that the oxidation reaction has been occurring at a constant pressure (1 atm). Moreover, the temperature has also been assumed to remain the same as long as the reaction continues.

### Methodology

The investigation is based on secondary data. The value of enthalpy change and entropy change will be calculated using values from the IB Chemistry Data Booklet. The enthalpy change will be calculated in three different methods – using an energy cycle and applying Hess’s law, using values of bond energy and using enthalpy of formation. Following this, arithmetic mean of the values obtained from these three different sources will be used. The entropy change will be calculated using the values of entropy of the reactants and products at standard states. Using the values of enthalpy change and entropy change, the magnitude of standard Gibb’s free energy change at different values of temperature ranging from 298 K to 412 K will be determined. The Gibb’s Helmoltz equation - $$ΔG=ΔH-TΔS$$ will be used for this.

### Calculation of enthalpy change (ΔH) in kJ mol-1

• Using bond energy (use values from IB Data Booklet)

$$ΔH^0_{vap}\text{ of } CH_3CH_2OH = 38.56\text{ kJ mol}^{-1}$$

$$ΔH^0_{vap}\text{ of }CH_3COOH = 51.60\text{ kJ mol}^{-1}$$

ΔH(using bond energy) =[2(C - H) +(O = O)]-[(C = O)+2(O - H)]

= [2(414) + (498)] – [(804) + 2(463)] = -404 kJ mol-1

According to the energy cycle displayed in the image above,

$$ΔH^0_r=38.56 + (−404) − (51.60) = −417.04\text{ kJ mol}^{-1}$$

b. Using enthalpy of formation (use values from IB Data Booklet).

$$CH_3CH_2OH(l)+O_2(g) .......→ CH_3 COOH (l)+2H_2O(g)$$

Enthalpy change

$$(ΔH^0_{r})=[(ΔH^o_{f} \, \,of \, \,CH_3 COOH) +(\Delta H^o_f \, \,of \, \,H_2O)]-[(\Delta H^o_f \, \,of \, \,CH_3CH_2OH)]=[(-484)+(-241.8)]-[278]$$

As elements have standard enthalpy of formation as zero, H2 and O2 are

c. Using enthalpy of combustion (use values from IB Data Booklet).

$$CH_3CH_2OH(l)+3O_2(g)\longrightarrow 2 CO_2 (g) + 3 H_20 (g) ∆H^o_c=-1367\text{ kj mol}^{-1}$$ ............(equation-1)

$$CH_3COOH(l)+2O_2(g)\longrightarrow2 CO_2(g) +2H_2O(g)∆H^o_c=-874\text{ kJ mol}^{-1}$$ ......(equation-2)

Subtracting equation (2) from equation (1),

$$CH_3CH_2OH(l)+O_2(g)\longrightarrow CH_3COOH(l)+H_2O(g)$$

$$ΔH^o_{r} =- 1367-(-874)=-493\text{ kJ mol}^{-1}$$

Enthalpy change

$$(ΔH^o_r)=\frac {ΔH^o_r\ using\ bond\ energy\ +\ ΔH^o_r\ using\ heat\ of\ formation\ +\ ΔH^o_r\ using\ heat \ of\ combustion }{3}$$

$$=\frac{(-417.80)+(-447.80)+(-493.00)}{3}= −452.60\text{ kJ mol}^{-1}$$

Standard deviation (SD)

$$=\frac{(417.80-452.60)^2+(447.80-452.60)^2+(493.00-452.60)^2}{3}=37.85$$

Fractional uncertainty in value of enthalpy change

$$(ΔH^o_r)=\frac{absolute\ error}{valu\ calculated}=\frac{ \pm 37.85}{452.60}$$

### Calculation of standard entropy change

$$CH_3CH_2OH (l) + O_2(g)\longrightarrow CH_3COOH (l) + H_2O (g)$$

∆So =[(so  of CH3COOH) + (So of H2O)]-[(So of CH3 CH2OH) +(So of O2 )]

= [ (+160.00) + (+188.80)] – [ (+161.00) + (205.00)]

= (+348.80) – (+366.00)

= - 18.00 J K-1 mol-1

The values of molar entropy used here have been taken from the Table-12 of the IB Chemistry Data Booklet. These values are originally taken from the textbook – “Aylward and Findlay's SI Chemical Data, 7th Edition” by Allan Blackman and Lawrie Grahan. As mentioned in the Section-“Properties of Organic Compounds” in Page 95, the data has been presented with an uncertainty level of ± 0.01. Thus, the absolute error of the entropy is considered as ± 0.01.

Thus, fractional error of entropy change

$$=±\frac{0.01}{18.00}$$

## Data processing

Temperature (T in K)

Standard free energy change (∆ $$G^o_r$$)  in kJ mol-1

298.00-447.24
300.00-447.20
302.00-447.16
304.00 -447.13
306.00-447.09
308.00-447.06
310.00-447.02
312.00 -446.98
314.00-446.95
316.00-446.91
318.00 -446.88
320.00-446.84
322.00-446.80
324.00 -446.77
326.00-446.73
328.00-446.70
330.00-446.66
332.00-446.62
334.00-446.59
336.00-446.55
338.00 -446.52
340.00 -446.48
342.00-446.44
344.00 -446.41
346.00-446.37
348.00-446.34
350.00-446.30
352.00-446.26
354.00-446.23
356.00-446.19
358.00-446.16
360.00-446.12
362.00-446.08
364.00-446.05
366.00-446.01
368.00-445.98
370.00-445.94
372.00-445.90
374.00-445.87
376.00-445.83
378.00-445.80
380.00-445.76
382.00-445.72
384.00-445.69
386.00-445.65
388.00-445.62
390.00-445.58
392.00-445.54
394.00-445.51
396.00-445.47
398.00-445.44
400.00-445.40
402.00-445.36
404.00-445.33
406.00-445.29
408.00-445.26
410.00-445.22
412.00-445.18

Figure 7 - Table On Calculated Values Of Standard Free Energy Change Of The Reaction Against Temperature

Sample Calculation

For 298.00 K:

Go = ∆ HoTS=-452.60 -298 × 0.018 =−447.24 kj. mol-1

### Error analysis

The values for the magnitude of free energy change have been calculated using: G= ∆ H-T                                                                                                               SThe values of temperature have been taken over a range of 298.00 K to 412.00 K at an interval of 2.00 K. These values have not been computed or extracted from a secondary source. Thus, any error is not associated with this.

Therefore, Fractional error in values of ∆Go

= Fractional error in ∆ H+ + fractional error in∆ So

$$=\frac{±37.85}{452.60}+ \frac{±0.01}{18.00}= ± 0.08$$

Refer to Page -6 for fractional uncertainty in value of enthalpy change and Page- 7 for fractional uncertainty in value of entropy change.

Percentage uncertainty in value of

Go = fractional uncertainty × 100 = ± 0.08 × 100 = ± 8.00 %

Graph-1 is a scattered plot with values of temperature in K along the x axes and the magnitude of standard free energy change in kJ mol-1 along the y axes. The data values are not displayed in the graph. The blue circles in the graph represents the data points while the blue dotted line indicates the linear trend line that the data fits into.

It is clearly depicted in the graph that as the temperature increases from 298 K to 412 K, the magnitude of free energy change at standard state changes from - 447.24 kJ mol-1 to - 445.18 kJ mol-1 . This shows that with the increase of temperature, the magnitude of free energy change becomes less negative and thus the oxidation of ethanol can be claimed to become less spontaneous. This shows that ethanol has a greater tendency to get oxidized to ethanoic acid in presence of oxygen at room temperature in comparison to high temperatures. Thus, the oxidation is a favorable process at room temperature and does not require any heating.

The increase in the value of free energy change at standard state with the increase in temperature is gradual and linear in nature. The difference between the consecutive data points remains the same throughout the graph. If the data values are inspected, it can be seen that the values are increasing mostly in the decimal places.

Maximum value of

Go(at 298 K) = - 447.24 kJ mol-1

Minimum value of

Go(at 412 K) = - 445.18 kJ mol-1

Difference in value = mod [(-447.24) – (-445.18) ] = 2.06 kJ mol-1

It means that as temperature changes by (412 – 298 ) = 114 units, the magnitude of free energy change at standard state changes by 2.06 units.

Thus, for change of temperature by 1 unit,

Standard free energy change alters by

$$\frac{2.06}{114}= 0.0108 ≅ 0.01 \, \, \,unit,$$

Thus, it can be claimed that the change in value of free energy change with respect to temperature is really small and not that significant. Hence, it can be concluded that the temperature has an impact on the magnitude of free energy change but not that significantly.

The data points obey a linear trend line as indicated in the graph. The equation of trend line has been obtained using MS-Excel:

y = 0.018 x452.6

Here, y represents ∆ Go and x represents temperature (T)

At equilibriumGo,0.00

Putting y = 0.00 in equation (1),

0.0 = 0.018 x 452.6

0.018 x = 452.6

$$X=\frac{452.6}{0.018}= 25144.44 \, \, \,K= 24871.44 \, \, ^0 C$$

This shows that at a temperature of 25144.44 K, the oxidation of ethanol to ethanoic acid attains equilibrium. However, this data does not make sense as the process itself is not reversible in nature and thus there is no scope of attaining equilibrium.

According to Gibb’s Helmoltz equation,

G= ∆ H- T  ∆ So

G= - ∆ S(T) + ∆ Ho

Thus, if ∆ Gois plotted against T,

Intercept (c ) =Ho

According to the equation of trend line :

y = 0.18 x 452.60

Entropy change at standard state (So) = - Gradient (m) = - 0.018 kJ mol-1

Enthalpy change at standard state (Ho) = Intercept = - 452.60 kJ mol-1

The negative sign in the magnitude of entropy change (So) indicates that the reaction has decrease in the disorderness of the system. The reactant side is more disordered than the product side.

The negative value of enthalpy change at standard state (Ho) indicates that the reaction is exothermic in nature and involves the release of heat.

Reaction favors increase in disorder ness and release of heat. So, reactions with negative values of enthalpy change and positive values of entropy change are always favored as shown in Section- of background information. This allows us to claim that the oxidation of ethanol to ethanoic acid is favored thermally (as heat is released and the reaction is exothermic in nature) but not entropically as entropy is decreasing showing decrease in disorderness.

## Conclusion

How does the standard free energy change (Go) in kj. mol-1 of the oxidation of Ethanol (C2H5OH) to Ethanoic acid (CH3COOH) depends on the temperature in K at which it is carried out?

• As the temperature increases from 298 K to 412 K, the magnitude of free energy change at standard state changes from - 447.24 kJ mol-1 to - 445.18 kJ mol-1 . This shows that with the increase of temperature, the magnitude of free energy change becomes less negative and thus the oxidation of ethanol can be claimed to become less spontaneous.
• The increase in the value of free energy change at standard state with the increase in temperature is gradual and linear in nature.
• It can be claimed that the change in value of free energy change with respect to temperature is really small and not that significant. Hence, it can be concluded that the temperature has an impact on the magnitude of free energy change but not that significantly.
• The oxidation of ethanol to ethanoic acid is favored thermally (as heat is released and the reaction is exothermic in nature) but not entropically as entropy is decreasing showing decrease in disorderness.

## Evaluation

### Strengths

• The values of entropy change and enthalpy change has been calculated using the values of gradient and intercept in the equation of the linear trend line obtained in Figure- 8 and the values are -0.018 kJ mol-1 and -452.60 kJ mol-1 respectively. The same values have been calculated in the controlled variable using literature values of bond enthalpy, Hess’s law and heat of formation for enthalpy change and values of entropy for entropy change. The values obtained from the graph and that from the calculations in the section- Controlled variable section are identical. This indicates that the data processing is coherent.
• A wide range of independent variable has been chosen for the investigation. The values of temperature have been chosen from 298 K to 412 K at an interval of 2.00 K. This has allowed the results obtained to be more reliable and accurate.
• In the controlled variable section, the values of enthalpy change have been calculated in three different methods – using bond energy, using Hess’s law and using the values of enthalpy of formation. Finally, a mean value has been considered and thus the value of enthalpy change obtained can be considered to be more reliable and accurate in comparison to the value that would have been obtained if only one of the methods amongst – using bond energy, heat of formation and Hess’s law was used.

### Limitations

• The equationG=∆ Ho - TSo is applicable only at constant pressure and temperature. To be clearer, the value of free energy change Go can be used to predict the feasibility of a reaction only if it occurs at constant pressure and temperature. The reaction under investigation is not assured to be conducted in an isothermal (constant temperature) or isobaric (constant volume) condition. This fact limits the application of this equation for the reaction of oxidation of ethanol to ethanoic acid and hence questions the reliability or accuracy of the data collected.
• The reaction under investigation uses oxygen as the reactant. This represents aerial oxidation of ethanol to ethanoic acid. In reality, this process demands the use of strong oxidizing agent like acidified potassium permanganate, acidified potassium dichromate, Pyridinium chlorochromate and so on. Thus, this is more like an idealistic case. It is also clearly understood that a reaction which requires strong oxidizing agents like acidified potassium dichromate or acidified potassium permanganate cannot be a spontaneous process in the presence of aerial oxygen as an oxidizing agent.
• The values of enthalpy change (∆ Ho) has been calculated in three different methods – using bond energy and Hess’s law, using enthalpy of formation and using values of enthalpy of combustion. The values obtained are: - 417.04 kJ mol-1 , -447.80 kJ mol-1 and -452.60 kJ mol-1 respectively.

Mean enthalpy change​​​​​​​ $$(∆H^o_r)$$

$$=\frac{(∆H^o_r using \ bond\ energy+∆H^o_r using \ heat \ of\ formation\ ∆H^o_r\ using\ heat\ of\ combustion)}{3 }$$

$$=\frac{(-417.80)+(-447.80)+(-493.00)}{3}=−452.60\text{ kJ mol}^{-1}$$

Standard deviation

$$(SD)=\frac{(417.80-452.60)^2+(447.80-452.60)^2+(493.00-452.60)^2}{3}= 37.85$$

The value of standard deviation is really high and this indicates the lack of precision in the dataset obtained.

• It has been assumed that the magnitude of enthalpy change and entropy change does not change with temperature. This assumption limits the accuracy of the data collected.

### Extension

To extend the investigation, I would like to study the effect of structural features like chain length, inductive effect on the magnitude of free energy at standard state. To do this, the combustion of alkanes can be chosen. The magnitude of enthalpy change and entropy change for this reaction can be calculated at a temperature of 298 K using the method shown here. Using the values obtained, the value of free energy change at standard state can be obtained. The same process can be done for all alkanes from methane to decane. Thus, we can plot the value of free energy change against the chain length of alkanes (1 for methane to 10 for decane). This graph will allow us to understand how the magnitude of free energy change at a particular temperature varies with the chain length of alkanes.

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