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.

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&nbsp;\(kJ.mol^{-1}\) calculated based on secondary data?

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.

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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.

2C2H5OH(l)+O2(g)&rarr;2CH3COOH(l)+2H2(g)

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

<ul>
	<li>&nbsp;Acidified Potassium dichromate\([K_2Cr_2O_7]\)</li>
	<li>&nbsp;Acidified Potassium permanganate\([KMnO_4]\)</li>
	<li>&nbsp;Pyridinium chlorochromate (PCC)&nbsp;\([C_5H_5NH]+[CrO_3Cl]^-\)</li>
</ul>

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.

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Furthermore, the oxidation number of carbons with which the hydroxy group is attached in ethanol is &ndash; 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 may be defined as the amount of heat change of a reaction at constant pressure. It is measured in the unit of&nbsp;\(kJ.mol^-1\) Enthalpy change can have both positive and negative values. A positive value of ∆\(H\)&nbsp;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 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&nbsp;\(J.mol^{-1}.K^{-1}\)&nbsp;.\(ΔS\)&nbsp;is the change in entropy which is determined by the formula -&nbsp;&nbsp;\(ΔS\ =S_{final}\ -S_{initial}\)

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

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\(ΔS=\frac{q_{rev}}{T}\)

where,

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\(q_{rev}=Amount\ of\ heat \ change\ in\ a\ reversibel\ reaction\)&nbsp;

Gibb&rsquo;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&rsquo;s Free Energy can be represented as:

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G = H - TS

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Differentiating both sides with respect to temperature -

&nbsp;

∆G = ∆H - ∆(TS)

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&nbsp;

=&gt;∆G = ∆H - T∆S - S∆T

At constant temperature, \(∆T= 0\)&nbsp;-

&nbsp;

∆G = ∆H - T∆S... ... ... (equation 1)

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This above equation is known as Gibb&rsquo;s Helmholtz Equation.

The variation of Gibb&rsquo;s Free Energy with respect to temperature can be explained using four different cases. According to the Gibb&rsquo;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:

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

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The generalized equation of a straight line:

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y = mx + c&nbsp;... ... ... (equation 2)

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y = Co - ordinate of&nbsp;y - Axis,

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x = Co - ordinate of x - Axis,

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m = slope of the curve,

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c = y - Intercept of the curve

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Comparing equation (1) with equation (2):

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

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Here, Gibb&rsquo;s free energy (∆G) resembles with y as it is the dependent variable of the exploration, Temperature (T) resembles with x&nbsp;as it is the independent variable of the exploration,&nbsp;∆S&nbsp;is compared with m which is the slope of the curve and ∆H&nbsp;is compared with c which is the value of Y &ndash; Intercept of the curve.

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

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

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

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

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

&nbsp;

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

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)

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.

The standard Gibb&rsquo;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&rsquo;s Helmoltz Equation \(((ΔG^0\ =ΔH^0\ =TΔS^0)\).

<ul>
	<li>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.</li>
	<li>The Gibb&rsquo;s Helmoltz Equation \((ΔG^0\ =ΔH^0\ =TΔS^0)\)&nbsp;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.</li>
</ul>

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 &ndash; using an energy cycle and applying Hess&rsquo;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&rsquo;s free energy change at different values of temperature ranging from 298 K to 412 K will be determined. The Gibb&rsquo;s Helmoltz equation - \(ΔG=ΔH-TΔS\)&nbsp;will be used for this.

<ul>
	<li>Using bond energy (use values from IB Data Booklet)</li>
</ul>

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

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

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&Delta;H(using bond energy) =[2(C - H) +(O = O)]-[(C = O)+2(O - H)]

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

 
According to the energy cycle displayed in the image above,

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

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

&nbsp;

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

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Enthalpy change

&nbsp;

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

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

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

&nbsp;

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

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

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Subtracting equation (2) from equation (1),

&nbsp;

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

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

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Enthalpy change

&nbsp;

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

&nbsp;

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

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Standard deviation (SD)

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

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Fractional uncertainty in value of enthalpy change

&nbsp;

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

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

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

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= [ (+160.00) + (+188.80)] &ndash; [ (+161.00) + (205.00)]

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= (+348.80) &ndash; (+366.00)

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= - 18.00 J K-1 mol-1

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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 &ndash; &ldquo;Aylward and Findlay&#39;s SI Chemical Data, 7th Edition&rdquo; by Allan Blackman and Lawrie Grahan. As mentioned in the Section-&ldquo;Properties of Organic Compounds&rdquo; in Page 95, the data has been presented with an uncertainty level of &plusmn; 0.01. Thus, the absolute error of the entropy is considered as &plusmn; 0.01.

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Thus, fractional error of entropy change

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\(=±\frac{0.01}{18.00}\)

Sample Calculation

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For 298.00 K:

&nbsp;

∆ Go =&nbsp;∆ Ho -&nbsp;T∆ So&nbsp;=-452.60 -298&nbsp;&times; 0.018 =&minus;447.24 kj. mol-1

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