INFLUENCE OF GROWTH ZONES ON THE PHYSICAL CHEMICAL COMPOSITION OF THE FRUIT OF
PARTE EXPERIMENTAL Descripción de las zonas de estudio 5
a C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)
b 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
c C2H4(g) + H2(g) → C2H6(g)
d 2SO2(g) + O2(g) → 2SO3(g)
tHEOrY OF KnOWLEDgE
In your study of Chemistry so far you have gained an understanding of the different ways in which scientific knowledge is produced, in other words, ‘how scientists know’, and you are learning about ‘what chemists know’ about the nature of matter. As learners we rarely stop to think about the process by which this happens, its limitations and strengths. Reflect on your experiences as a knower in Chemistry by considering the following. • Give an example of chemical knowledge that has been
simultaneously reliable and tentative. What are the advantages of both existing simultaneously?
• What makes the knowledge claims made by scientists more believable (e.g. consider the use of empirical evidence, inductive reasoning, dealing with uncertainty, reproducibility).
• What similarities and differences have you found in the method used to gain knowledge in the natural sciences compared to the method used in the human sciences?
• How do creativity, imagination and risk taking assist in the production of new knowledge? Support with
• The process of science produces theories and laws. Distinguish between them and describe some of the characteristics of well-established scientific laws and theories.
• Why is it important for scientists from different countries and research organizations to collaborate and communicate with one another effectively during research? What is a peer review and why is it so important?
• Give some examples to illustrate how the social and cultural context of a scientist’s work and their experience, age and expectations might affect the outcome of a scientific endeavour.
• Describe a story from the history of science that shows scientific knowledge being produced from evolutionary or gradual changes in what we know, and an example from a revolutionary or sudden change in thinking. • Give an example to illustrate how the advancement of
scientific knowledge has impacted and is dependent on the development of new technologies.
CHAPTER 4
energetics
What is the driving force that makes a reaction spontaneous? Is it that heat energy is released as the reaction proceeds (an exothermic reaction) or perhaps that the system becomes more disordered as products are made (positive entropy)? Perhaps it is a combination of enthalpy and entropy?
Consider the reaction in which ammonia is produced. N2(g) + 3H2(g) Ý 2NH3(g)
∆H ʅrxn = ∑n∆H ̲(products) - ∑m∆H ̲(reactants) (see section 4.1)
∆H ʅrxn = [(2 mol)(-46.3 kJ mol-1)] - [(1 mol)(0 kJ mol-1) + (3 mol)(0 kJ mol-1)]
= -92.6 kJ mol-1 (i.e. the enthalpy change for the molar quantities in this reaction is -92.6 kJ)
The reaction is exothermic.
∆S ʅrxn= ∑n∆S ʅ(products) - ∑m∆S ʅ(reactants) (see section 4.3)
∆S ʅrxn = [(2 mol)(193 J K-1mol-1)] - [(1 mol)(191.5 J K-1mol-1)
+ (3 mol)(131 J K-1mol-1)] = 386 - 584.5
= -198.5 J K-1
The entropy of the reaction is not favourable to a spontaneous reaction. This can be seen qualitatively by noticing that the reaction has 4 mol of gaseous reactants and 2 mol of gaseous products—a significant increase in order (decrease in disorder).
So in the case of such a well-known reaction, enthalpy and entropy disagree as to whether the reaction should go ahead.
Gibbs free energy, G, reflects the balance between these two quantities. It
may be described as the energy available to do work or more clearly as the ‘driving force’ (although it is not really a force) of a chemical reaction. It is calculated as a combination of the enthalpy and entropy of a reaction with the temperature at which the reaction is carried out also being considered.
∆G = ∆H - T∆S
Gibbs free energy has two components: the change in enthalpy, ∆H, and the change in entropy, ∆S, which is combined with the temperature (in kelvin) at which the reaction is occurring.
• ∆H is the drive toward stability. When ∆H < 0 the products are more stable (have a lower enthalpy) than the reactants.
• ∆S is the drive toward disorder. When ∆S > 0, the products are more disordered than the reactants. Note that increasing T increases the influence of ∆S on the reaction’s ‘driving force.’
• Temperature is always quoted in kelvin for calculations of ∆G.
Gibbs free energy can be used to determine whether a reaction will go ahead on its own or not. That is, whether or not it is spontaneous. The change in Gibbs free energy is defined in the same way as the change in enthalpy, ∆H.
∆G = G(products) - G(reactants)
As for enthalpy and entropy, Gibbs free energy, G ʅ, is usually quoted under standard conditions (298 K and 1.01 × 102 kPa pressure) and the standard free energy change is ∆G ʅ.
4.4 Spontaneity
15.4.1
Predict whether a reaction or process will be spontaneous
by using the sign of ∆G ʅ.
A negative value for ∆G ʅ accompanies a spontaneous reaction, while a positive value means that the reaction is not spontaneous. Note that the value for ∆G° says nothing about the rate of the reaction. The reaction may proceed very slowly indeed, yet have a negative value of ∆G ʅ.
If ∆G ʅ < 0 the reaction is spontaneous. If ∆G ʅ = 0 the reaction is at equilibrium. If ∆G ʅ > 0 the reaction is not spontaneous.
For a chemical reaction ∆H ʅ and ∆S ʅ are independent of each other; that is, you cannot calculate one from the other. You can have situations where both are positive, both are negative, or one is positive and the other negative. Notice that if ∆H and ∆S have the same sign they are working against each other, and the larger quantity will win. Entropy values are usually much smaller than enthalpy values and are quoted in joules rather than kilojoules. However, if the temperature is increased the entropy term (T∆S) may become greater than the enthalpy term. Similarly, as the temperature decreases, the entropy term becomes less significant and the enthalpy term dominates.
1 in an exothermic reaction with an increase in entropy ∆H ʅ is negative, so if ∆S ʅ is positive, then
∆G ʅ = ∆H ʅ - T∆S ʅ
= (negative) - (positive) = negative
The reaction will be spontaneous, no matter what the temperature.
Worked example 1
For the following reaction, ∆H ʅ is negative and ∆S ʅ is positive: 2H2O2(l) → 2H2O(l) + O2(g)
Determine whether the reaction will be spontaneous at room temperature.
solution
There is no need to actually calculate ∆G ʅ. ∆G ʅ = ∆H ʅ - T∆S ʅ
= (negative) - (positive) = negative
This reaction will be spontaneous at any temperature. However, we cannot make any assumptions about the rate of the reaction. You may recognize this decomposition of hydrogen peroxide as one that requires a catalyst to proceed at a measurable rate.
15.4.2
Calculate ∆G ʅ for a
reaction using the equation
∆G ʅ = ∆H ʅ -T∆S ʅand by
using values of the standard free energy change of
formation, ∆G ̲. © IBO 2007
CHAPTER 4
energetics
2 in an exothermic reaction with a decrease in entropy being carried out at a low temperature
∆H ʅ is negative and ∆S ʅ is negative, therefore ∆G ʅ will only be negative if T∆S ʅ has a negative value that is smaller than the value of ∆H ʅ.
∆G ʅ = ∆H ʅ - T∆S ʅ
= (negative) - (smaller negative) = negative
The reaction will be spontaneous only at low temperatures. The actual temperatures depend upon the relative sizes of ∆H ʅ and ∆S ʅ.
Worked example 2
For the following example find ∆H ʅ and ∆S ʅ, then determine the highest temperature at which the reaction will be spontaneous.
NH3(g) + HCl(g) → NH4Cl(s)
solution
∆H ʅrxn = ∑n∆H ̲(products) - ∑m∆H ̲(reactants)
= [(1 mol)(∆H ̲(NH4Cl)] - [(1 mol)(∆H ̲(NH3) + (1 mol)(∆H ̲(HCl)]
= [(1 mol)(-315.4 kJ mol-1)] - [(1 mol)(-46.3 kJ mol-1) + (1 mol)(-92.3 kJ mol-1)]
= -315.4 - (-138.6) = -176.8 kJ mol-1 The reaction is exothermic.
∆S ʅrxn = ∑n∆S ʅ(products) - ∑m∆S ʅ(reactants) (see section 4.3)
= [(1 mol)(∆S ʅ(NH4Cl)] - [(1 mol)(∆S ʅ(NH3) + (1 mol)(∆S ʅ(HCl)]
= [(1 mol)(95 J K-1mol-1)] - [(1 mol)(193 J K-1mol-1) + (1 mol) (187 J K-1mol-1)]
= 94.6 - 380 = -285 J K-1
The entropy of the reaction is not favourable to a spontaneous reaction. This can be seen qualitatively by noticing that the reaction has
2 mol of gaseous reactants and 1 mol of solid products—an increase in order (decrease in disorder).
∆G ʅ = ∆H ʅ - T∆S ʅ
To find the temperature at which the reaction ceases to be spontaneous, substitute ∆G ʅ = 0 as this is the greatest value at which the reaction is not spontaneous.
0 = -176 800 - T × -285 T × -285 = -176 800 T = - - 176800 285 = 620 K
The reaction is spontaneous at temperatures below 620 K (347°C).
Figure 4.4.2 The formation of ammonium chloride is clearly demonstrated when cotton wool balls soaked in HCl and NH3 are placed at either end of a tube and the vapours diffuse towards each other.
Note that units of ∆Hʅ have been changed to J (multiplied by 1000) to agree with units of ∆Sʅ .
3 in an endothermic reaction with an increase in entropy at high temperatures.
∆H ʅ is positive, therefore ∆G ʅ will be negative only if T∆S ʅ has a positive value that is larger than the value of ∆H ʅ.
∆G ʅ = ∆H ʅ - T∆S ʅ
= (positive) - (larger positive) = negative
The reaction will be spontaneous only at high temperatures.
1 in an endothermic reaction with a decrease in entropy
∆H ʅ is positive, therefore if ∆S ʅ is negative ∆G ʅ will always be positive. ∆G ʅ = ∆H ʅ - T∆S ʅ
= (positive) - (negative) = positive The reaction will never be spontaneous.
Worked example 3
Explain why oxygen gas does not change spontaneously into ozone, O3(g), at room temperature (298 K).
3O2(g) → 2O3(g)
solution
∆H ʅrxn = ∑n∆H ̲(products) - ∑m∆H ̲(reactants)
= [(2 mol)(∆H ̲(O3)] - [(3 mol)(∆H ̲(O2)]
= [(2 mol)(+142.2 kJ mol-1)] - [(3 mol)(0 kJ mol-1)] = +284.4 kJ mol-1 = 2.844 × 105 J mol-1
∆S ʅrxn = ∑n∆Sʅ(products) - ∑m∆Sʅ(reactants)
= [(2 mol)(237.6 J K-1 mol-1)] - [(3 mol)(205.0 J K-1 mol-1)] = 475.2 - 615.0 = -139.8 J K-1 Since ∆Gʅ = ∆Hʅ - T∆Sʅ ∆Gʅ = +2.844 × 105 - (298 × -139.8) = +2.844 × 105 - (-4.17 × 104) = 3.26 × 105 J mol-1 = +326 kJ mol-1
The Gibbs free energy for this reaction is a large positive value, so oxygen gas will not change spontaneously into ozone at room temperature.
Note that the subtraction of a negative number from a positive number will always result in a positive value, so not only is this reaction not spontaneous at room temperature, it will never be spontaneous.
CHAPTER 4
energetics
2 in an exothermic reaction
∆H ʅ is negative, therefore if ∆S ʅ is negative and the temperature is high, then T∆S ʅ may have a larger negative value than ∆H ʅ and ∆Gʅ will be positive.
∆G ʅ = ∆H ʅ- T∆Sʅ
= (negative) - (larger negative) = positive
The reaction will not be spontaneous at high temperatures.
Worked example 4
Find the maximum temperature at which this reaction is spontaneous. N2(g) + 3H2(g) → 2NH3(g)
∆H ʅ = -92.6 kJ mol-1 = -9.26 × 104 J
∆S ʅ = -198.5 J K-1
solution
∆G ʅ = ∆H ʅ - T∆S ʅ
A value of ∆G ʅ = 0 is the greatest value at which the reaction is not spontaneous. 0 = -9.26 × 104- T × -198.5 T = - - 92600 198 5. = 466 K
The reaction is spontaneous at temperatures up to 466 K (193°C). 3 in an endothermic reaction with an increase in entropy at low temperatures
∆H ʅ is positive, therefore ∆G ʅ will be positive if T∆S ʅ has a positive value that is smaller than the value of ∆H ʅ.
∆G ʅ = ∆H ʅ - T∆S ʅ
= (positive) - (smaller positive) = positive
The reaction will not be spontaneous at low temperatures.
Worked example 5
NH4NO3(s) water→ NH4+(aq) + NO3-(aq)
Find the minimum temperature at which this reaction will be spontaneous. ∆H ʅ = +28.05 kJ mol-1 = +2.805 × 104 J
∆S ʅ = 108.7 J K-1
solution
∆G ʅ = ∆H- - T∆S ʅ
A value of ∆G ʅ = 0 is the lowest value at which the reaction is spontaneous. 0 = +2.805 × 104 - T × 108.7
T = 258 K
In theory this reaction will be spontaneous down to a temperature of 258 K (-15°C); however, it involves dissolution in water, whch would be very difficult below 0°C as the water would be frozen!
table 4.4.1 suMMary of effects of signs of ∆H ʅ and ∆S ʅ on ∆G ʅ ∆H ʅ
positive (+) negative (-)
∆S ʅ
positive (+) ∆G ʅ is negative at high temperatures
and positive at low temperatures. e.g. H2(g) + I2(g) → 2HI(g)
∆G ʅ is always negative.
These reactions are always spontaneous. e.g. 2H2O2(l) → 2H2O(g) + O2(g)
negative (-) ∆G ʅ is always positive.
These reactions are never spontaneous. e.g. 3O2(g) → 2O3(g)
∆G ʅ is positive at high temperatures and negative at low temperatures.
e.g. NH3(g) + HCl(g) → NH4Cl(s)
The standard Gibbs free energy of a reaction can also be calculated from tabulated values of standard free energies of formation, if these are available. Note that, as for standard enthalpies of formation, the standard free energy
change of formation of any element in its most stable state is equal to zero.
The same method of calculation is applied for standard free energy change
of reaction as was used for calculating standard enthalpy of reaction and
standard entropy of reaction.
∆G ʅrxn= ∑n∆G ̲(products) - ∑m∆G ̲(reactants)
table 4.4.2 soMe values of ∆G
(Some of these values can a lso be found in Appendix 3 and in table 11 of the IB Data booklet. © IBO 2007)
substance formula state ∆G ̲(kJ mol−1) substance formula state ∆G ̲(kJ mol−1)
Ammonia NH3 g −17 Methane CH4 g –51
Ammonium chloride NH4CI s −204 Methanol CH3OH l −166
Butane C4H10 g −16 Nitrogen N2 g 0
But-1-ene C4H8 g 72 Nitrogen dioxide NO2 g 52
Carbon dioxide CO2 g −394 Nitrogen monoxide NO g 87
Chlorine CI2 g 0 Oxygen O2 g 0
Ethane C2H6 g −33 Ozone O3 g 16
Ethene C2H4 g 68 Propane C3H8 g –24
Ethanol C2H5OH g −169 Propene C3H6 g 63
Ethanol C2H5OH I −175 Tetrachloromethane CCI4 g −65
Ethyne C2H2 g 209 Water H2O g −229
Hydrogen H2 g 0 Water H2O l –237
Hydrogen chloride HCI g –95
Worked example 6
Determine the standard Gibbs free energy of the following reaction. 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
CHAPTER 4