Chemistry HL

Sample Internal Assessment

Table of content

Rationale

Background information

Variables

Conclusion

Evaluation

References

To begin with, I was little disappointed when I got to know that performing an experiment in the laboratory and collecting primary data would not be feasible because of the pandemic situation. Situation brought us to an investigation based on secondary data. While exploring to find a suitable topic for an investigation based on secondary data, I got intrigued about how use of data analysis and mathematical modelling may be useful to make scientific predictions. Chemistry being an experimental science mostly relies on empirical evidences with rare cases where a set of pre-obtained data is used as prior knowledge in conjunction with statistical tools to forecast possibilities and make predictions or produce new scientific knowledge. In Topic-5, HL part when I studied how the feasibility of a process may depend on temperature, I got more interested about it understanding that how changing the temperature may make the reaction give us more products and make it more useful. This particular concept is widely used in metallurgy during extraction of metals and reduction of their ores. But who and how controls that to what extent will the spontaneity of a reaction depends on temperature? Is it only the macroscopic variables like enthalpy change and entropy change of the reaction? Thermodynamics mostly deals with the macroscopic properties of a system. I was more interested to connect this to the microscopic properties of a reactant system to make it more relevant for chemistry. So, the goal was to use mathematical modelling and a set of data to elucidate how the microscopic property or precisely the structural feature of a system would have an impact on the macroscopic property of a reactant system like free energy change. To make it more streamlined, I wondered does in any way the size of a molecule matter when it comes to justify the spontaneity of a chemical reaction which it participates in. This discussion brought me finally to the research question stated below:

I**s there a correlation between the temperature dependence of free energy change \((\frac{∆G° }{T})\) of combustion of straight chain alkanes (methane – CH _{4}, ethane – C_{2}H**

Alkanes are hydrocarbon with single bonds between C atoms. They are saturated and does not contain any multiple bonds (double or triple bonds). Alkanes are highly reactive and are extremely unreactive in nature as the C-H single bonds as well as the C-C single bonds are highly stable and requires huge amount of energy to be broken. Two main reactions of alkanes are substitutions and combustion. This investigation deals with combustion of alkanes.

The general reaction of combustion of alkanes are as shown below

C_{X}H_{Y} + \((x+\frac{y}{4})\) O_{2} -------🡪 x CO_{2} + \(\frac{y}{4}\) H_{2}O

This is the reaction of oxidation in presence of excess oxygen. In presence of limited amount of oxygen, the product is CO instead of CO_{2}.

This reaction is exothermic in nature and thus have negative values of enthalpy change (∆H<0 ) is negative. During these reactions, the number of gaseous products are more than the number of gaseous reactants and thus the disorder-ness increases as we move to the right. So, the entropy of the system increases as we move towards the product and consequently, the values of entropy change (∆S) is positive.

This investigation deals with the first five member of the homologous series of alkanes- methane – CH_{4}, ethane – C_{2}H_{6}, propane – C_{3}H_{6}, butane – C_{4}H_{10} and pentane – C_{5}H_{12}. The balanced equations of their oxidation are shown below:

CH_{4} (g) + 2 O_{2} (g) -----🡪 CO_{2} (g) + 2 H_{2}O (g)

C_{2}H_{6} (g) + \(\frac{7}{2}\)O_{2} (g) -----🡪 4 CO_{2} (g) + 6 H_{2}O (g)

C_{3}H_{8} (g) + 5 O_{2} (g) -----🡪 3 CO_{2} (g) + 4 H_{2}O (g)

C_{4}H_{10} (g) + \(\frac{13}{2}\) O_{2 }(g) -----🡪 4 CO_{2} (g) + 5 H_{2}O (g)

C_{5}H_{12} (g) + 8 O_{2} (g) -----🡪 5 CO_{2} (g) + 6 H_{2}O (g)

To keep the number of moles of hydrocarbon used as 1, fractions has been used as a coefficient for O_{2} in the balanced equations.

The chain length of the alkane is an independent variable in this investigation. To understand a trend down the homologous series, the first five member has been chosen. The chain lengths are – T

Methane (CH_{4}) - 1

Ethane (C_{2}H_{6}) - 2

Propane (C_{3}H_{8}) - 3

Butane (C_{4}H_{10}) - 4

Pentane (C_{5}H_{12}) - 5

Temperature dependence of free energy change \(\frac{∆G° }{T}\) in kJ mol^{-1} K^{-1}

The enthalpy change of all of the combustion reactions of alkanes will be calculated in three different methods – using data for bond energy, using values of enthalpy of formation and directly using the values of enthalpy of combustion. An average of these three values will be considered as the enthalpy change of the reaction. The magnitude of entropy change will be calculated in J mol ^{-1} K^{-1} using the values of standard entropy values of the reactants and products. Using the values of enthalpy change (∆H°) and entropy change ∆S°, the values of ∆ G° will be computed at different values of temperature. The Gibb’s Helmholtz equation, ∆ G°= ∆ H°-T ∆ S° will be used for this. A temperature range of 298 K to 398 K will be taken. Following this, a scatter plot of ∆ G° against T will be made and the temperature dependence of the free energy change \((\frac{∆G° }{T})\) will be calculated from the gradient of the curve.

According to Gibb’s Helmholtz equation

∆ G^{0} = ∆ H^{0} - T ∆ S^{0}

Rearranging the equation, we get

∆ G^{0} = (-∆ S^{0})T + ∆ H^{0}

Comparing this to the standard form of the straight line equation, y = mx + c, we can get that if y axes is ∆ G^{0} and temperature (T) is the x axes, the gradient of the straight line is – \(\frac{y}{x}\) , which is \(\frac{∆G° }{T}\).

Figure - 2 Calculation of enthalpy changes

The general equation for combustion of alkanes:

C_{n}H_{2n+2} + \((n\ +\ \frac{2n+2}{4})\) O_{2} -------🡪 n CO_{2} + \(\frac{2n+2}{2}\) H_{2}O

The general formula for alkanes is C_{n}H_{2n+2} . The table above shows that the number of C-C bonds in an alkane may be represented by the number (n-1) and the number of C-H bonds may be represented as 2n+2. For example, if we consider, propane (C_{3}H_{8}), the value of n (number of C) is 3. Thus, the number of C-C bonds should be (n-1) = (3-1) = 2 and the number of C-H bonds should be (2n+2) = 2(3) + 2 = 8. The structural formula of propane shown below confirms that the above generalization is correct.

Alkanes

Molecular formula

Number of C atoms (n)

Number of C-C bonds (n-1)

Number of C-H bonds (2n+2)

Methane

CH_{4}

1

0

4

Ethane

C_{2}H_{6}

2

1

6

Propane

C_{3}H_{8}

3

2

8

Butane

C_{4}H_{10}

4

3

10

Pentane

C_{5}H_{12}

5

4

12

Using bond energy,

Enthalpy change (∆H^{0})

Energy supplied to break bonds in reactants

= [ {(n-1)(BE C-C) + (2n + 2) (C-H) } + { \((n\ +\ \frac{2n+2}{4})\) (BE O=O)]

= [ { (n-1) (346) + (2n + 2 ) (414) } + { \((n\ +\ \frac{2n+2}{4})\) (498) } ]

= [ { 346 n – 346 + 828 n + 828 } + \((\frac{(6n+2)}{4})\) (498)} ]

= [ { 1174 n + 482 } + { 747 n + 249} ]

= [ (1174 n + 747 n ) + (249 + 482) ] = 1921 n + 731

Energy released when the bonds are formed in the products

= [2 n (BE C = O) + \((\frac{2n+2}{2}\ × \ 2)\) (BE O-H)]

= [ 2n (804) + (2n + 2 ) (463) ]

= [ 1608 n + 926 n + 926 ]

= 2534 n + 926

Enthalpy change (∆ H^{o})= Energy supplied – Energy released

= (1921 n + 731) – (2534 n + 926)

= (1921 – 2534) n + (731 - 926)

= - 613 n – 195

Enthalpy change (∆ H^{o}) = - 613 n – 195 (using bond energy values) ……. Equation-1

Here, n is the number of C atoms in the alkanes.

**Deduction of general expression of enthalpy change of combustion of alkanes using values of enthalpy of formation**

C_{n}H_{2n+2} + \((n\ +\ \frac{2n+2}{4})\) O_{2 }-------🡪 n CO_{2} + \(\frac{2n+2}{2}\) H_{2}O

Enthalpy change (∆ H^{o})

= [ (∆H^{0}_{f} of CO_{2} (g)× n ) + (∆H^{0}_{f} of H_{2}O (g) ) × \((\frac{2n+2}{2})\) ] - ∆H^{0}_{f} of alkanes) ]

= [ (-393.50 × n ) + ( -241.8 ) \((\frac{2n+2}{2})\) ] - (∆H^{0}_{f} of alkanes)

= [ -393.5 n -241.80 n – 241.8 ] - (∆H^{0}_{f} of alkanes)

= [ - 635.30 n – 241.80 ] - ∆H^{0}_{f} of alkanes

Enthalpy change (∆ H^{o}) = [ - 635.30 n – 241.80 ] - (∆H^{0}_{f} of alkanes)…. (using enthalpy of formation)

As the values of enthalpy of formation of elements at standard state is zero, O_{2} is ignored.

Formulas used:

Enthalpy change (∆ H^{o})= - 613 n – 195 (using bond energy values) ……. Equation-1

Enthalpy change (∆ H^{o}) = [ - 635.30 n – 241.80 ] - (∆H^{0}_{f} of alkanes)…. (using enthalpy of formation)

The values of enthalpy of combustion have been directly used from the Chemistry Data Booklet

Calculation of entropy change:

C_{n}H_{2n+2} + \((n\ +\ \frac{2n+2}{4})\) O_{2} -------🡪 n CO_{2} + \(\frac{2n+2}{2}\)H_{2}O

Entropy change (∆S^{0})

=[ (S^{0}of CO_{2}× n) + (S^{0} of H_{2}O) × \((\frac{2n+2}{2})\) ]- [ (S^{0} of alkane) + (S^{0} of O_{2}) × (n + \(\frac{2n+2}{4}\)) ]

= [ (213.80 × n) + (188.80) \((\frac{2n+2}{2})\) ]- [ (S^{0}of alkane) + (161.1) × (n + \(\frac{2n+2}{4}\)) ]

= [ 213.80n + 188.8 n + 188.8 ] – [ (S^{0} of alkane) + 241.65 n + 80.55)]

= [402.6 n + 188.8 – 241.65 n – 80.55 ] - S^{0}of alkane

= (160.95 n + 108.25) - S^{0}of alkane

Alkane

Number of C atoms (n)

S^{0} of alkane in J mol^{-1} K^{-1}

Entropy change (∆S^{0}) in J mol^{-1} K^{-1}

Methane

1

+186

83.20

Ethane

2

+230

200.15

Propane

3

+270

321.10

Butane

4

+310

442.05

Pentane

5

+347.82

565.20

Gradient = \(\frac{∆ G° }{T}\) = 0.0833 (considering absolute value)

Gradient = \(\frac{∆ G° }{T}\) = 0.2001 (considering absolute value)

Gradient =\(\frac{ ∆ G° }{T}\) = 0.3211 (considering absolute value)

Gradient = \(\frac{∆ G° }{T}\) =0.422 (considering absolute value)

Gradient = \(\frac{∆ G° }{T}\) = 0.5625 considering absolute value

Figure - 6 to Figure - 10 shows are scattered plots of free energy change in kJ mol^{-1} against temperature (T). As the value of enthalpy change (∆H) is negative and the entropy change (∆S) is positive,

Alkane

Number of C atoms

\(\frac{∆ G°}{T}\) in kJ mol^{-1} K^{-1} / 10^{-2}

Methane

1

8.33

Ethane

2

20.11

Propane

3

32.11

Butane

4

42.20

Pentane

5

56.25

The graph above plots the values of \((\frac{∆ G°}{T})\) along the y axes and the number of C atoms along the x axes. Quantitatively, it shows that as the chain length increases from one C in methane to five C in pentane, the values of \((\frac{∆ G°}{T})\) is also increasing from 8.33 X 10^{-2} kJ mol^{-1} K^{-1} to 56.25 X 10^{-2} kJ mol^{-1} K^{-1}. It indicates that as the alkane gets longer or we go down the homologous series, the dependence of the free energy change on temperature gets stronger and more effective.

Higher the magnitude of the \(\frac{∆ G°}{T}\) , it means that the same increase of temperature will make the reaction more spontaneous. Figure - 6 to Figure - 10 clearly shows that in all cases, the reaction is spontaneous at all temperature. As temperature increases, the value of ∆G^{0} increases and becomes more negative. This means that all the combustion reactions are spontaneous at all values of temperature and the spontaneity increases as the temperature increases. However, the extent in which spontaneity increases with temperature is not the same in all cases. As the chain length increases, the magnitude of \(\frac{∆ G°}{T}\) increases and that means that the increase of spontaneity with the rise of temperature is more for a higher homologue of alkane than the lower homologue of it.

In simple words, if the temperature of the combustion of methane and ethane is increased by the same amount, both of them will become more spontaneous. But, the increase in spontaneity for the combustion of propane would be more than the increase in spontaneity for the combustion of ethane. Quantitatively, we can say that, for methane if the value of standard free energy change decreases by 8.33 kJ mol^{-1} with the rise of temperature by 1 K, for ethane it decreases by 20.11 kJ mol^{-1} for the rise of temperature by 1 K.

Figure - 12 depicts a linear trend line with a positive gradient and has a value of R^{2} = 0.998; it shows that there is a strong correlation between the magnitude of temperature dependence of free energy change \((\frac{∆ G°}{T})\) of combustion of straight chain alkanes (methane – CH_{4}, ethane – C_{2}H_{6}, propane – C_{3}H_{6}, butane – C_{4}H_{10} and pentane – C_{5}H_{12}) in presence of excess oxygen and the chain length of the alkane.

The aim of the investigation was to answer the question

Is there a correlation between the temperature dependence of free energy change \((\frac{∆G°}{T})\) of combustion of straight chain alkanes (methane – CH_{4}, ethane – C_{2}H_{6}, propane – C_{3}H_{6}, butane – C_{4}H_{10} and pentane – C_{5}H_{12}) in presence of excess oxygen and the chain length of the alkane?

Figure 12 - depicts a linear trend line with a positive gradient and has a value of R^{2} = 0.998; it shows that there is a strong correlation between the magnitude of temperature dependence of free energy change \((\frac{∆G°}{T})\) of combustion of straight chain alkanes (methane – CH_{4}, ethane – C_{2}H_{6}, propane – C_{3}H_{6}, butane – C_{4}H_{10} and pentane – C_{5}H_{12}) in presence of excess oxygen and the chain length of the alkane.

Quantitatively, it shows that as the chain length increases from one C in methane to five C in pentane, the values of \((\frac{∆G°}{T})\) is also increasing from 8.33 X 10^{-2} kJ mol^{-1} K^{-1} to 56.25 X 10^{-2} kJ mol^{-1} K^{-1}. It indicates that as the alkane gets longer or we go down the homologous series, the dependence of the free energy change on temperature gets stronger and more effective.

As the chain length increases, the magnitude of \(\frac{∆G°}{T}\) increases and that means that the increase of spontaneity with the rise of temperature is more for a higher homologue of alkane than the lower homologue of it.

A possible scientific justification for this may be the fact that as the chain length increases, the number of bonds increases. Thus, when we move from combustion of methane to ethane, a greater number of bonds are broken and a greater number of bonds are formed. This results in release of more energy which makes the reaction more exothermic. The values in Figure - 3 indicates that. As we are moving from methane to pentane, the value of enthalpy change increases and the reaction becomes more exothermic. Similarly, in Figure - 4, the value of entropy change is increasing which means that as we are moving from methane to pentane, the increase of disorder-ness increases. Thus, as the chain length increases from methane to pentane, the combustion becomes more exothermic and the disorder-ness also increases. Both of these factor drives the reaction more towards spontaneity. Thus, as we move from methane to pentane, the combustion becomes more exothermic, more disordered and consequently more spontaneous.

- Source of data is a major concern in a secondary data IA. Here in this investigation barring a few, almost all the data has been taken from the IB Chemistry Data Booklet. Thus, any questions related to the discrepancy of data can be easily averted.
- During the exploration, generic expressions has been derived to calculate the values of enthalpy change and entropy change instead of calculating them individually for each of the reactions. These equations deduced using the general equation is a sole personal input and makes the table more convenient and definitely makes the exploration more precise.
- Enthalpy change of a reaction can be calculated in multiple ways. In this investigation, enthalpy change has been calculated using three different ways and an average value has been considered. This makes the result more accurate and reliable.

- To ensure that the data used is accurate and reliable use of a justified secondary source is essential. The data used in the investigation has been taken from the IB Chemistry Data Booklet. Considering other reliable sources would have been an appropriate venture as that may make the data set more accurate.
- Enthalpy change of a reaction can be computed in multiple ways- using calorimetry, using enthalpy of formation, using energy cycle, using enthalpy of formation and many more. The values obtained are not same in all cases. Thus, the method used to calculate the value of enthalpy change may have an impact on the reliability of the conclusion made. To optimize this, the enthalpy change was calculated in three different sources and an average value has been used.
- Data processing based on secondary data involves the appropriate use of mathematical methods. The magnitude of \(\frac{ ∆ G°}{T}\) has been calculated here as a gradient of the curve obtained on plotting free energy change against temperature. This curve was obtained based on an approximation that the magnitude of entropy change and enthalpy change does not change with temperature.

As an extension of this investigation, I would like to calculate the enthalpy change of neutralization of organic carboxylic acids with NaOH using calorimetry and compare the values of enthalpy change obtained against the chain length of the organic carboxylic acid used. To do this, the acid and the base can be taken in two separate beakers and the temperature can be monitored against time before mixing them and after mixing them. A scatter plot of temperature against time can then be used to compute the difference of temperature and then the heat change can be calculated using the equation, q = -mC∆t. Following this, the enthalpy change of the reaction can be calculated using the equation, ∆H= \(\frac{q}{n}\) where n is the number of moles of the acid used. Following this, we can plot a graph of ∆H against the number of C and deduce a trend of how the enthalpy of neutralization changes with chain length.

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