Module 5 (Remember to do questions on equilibrium graphs week 3 and write syllabus points each flashcard covers)
Examples of irreversible reactions
Combustion of fossil fuels
Reaction between Mg and O2
Define dynamic equilibrium
The rates of forward and reverse reactions are equal and simultaneous but non-zero. Called dynamic since there is always movement of particles in both f.r and r.r. This is the type of equilibrium more often seen in chemical reactions. On a macroscopic level, the concentration of reactants and products are constant but on the microscopic level, reactions are continually occurring.
F.r proportional to reactants
Define static equilibrium
The rates of forward and reverse reaction are equal and zero (or practically zero). There is no reaction in either direction once equilibrium has been reached, due to the forward and reverse reactions having extremely large activation energies.
Values of ΔH, ΔS and ΔG in forward and reverse reactions
They will be the negative of that for the forward reaction meaning that it the forward reaction is exothermic, then the reverse reaction is endothermic and vice versa
Colours of CoCl2.6H2O originally, when heated, when water is added, when evaporated
The cobalt (II) chloride is a dark pink colour. When it is heated, it turns into a purple colour, and then after a while, it turns sky blue in colour. When water is added, the solution is a dark pink. When evaporated, same sequence of colour changes repeat.
Cause of the colours of CoCl2.6H2O originally, when heated, when water is added, when evaporated
The hexahydrate is pink in colour, the dihydrate is purple and the anhydrous form is sky blue. Different forms of cobalt (II) chloride can interconvert by hydration or dehydration.
CoCl2.6H2O(s)⇌CoCl2.2H2O(s) + 4H20(l)
CoCl2.2H2O(s)⇌CoCl2 + 2H2O(l)
When the hexahydrate is heated, the waters of crystallisation are gradually vaporised, and the purple dihydrate form is observed before the anhydrous form. When cobalt (II) chloride is dissolved in water, it produces a pink solution similar to the colour of the hexahydrate solid. This is because all the waters of crystallisation, would go into solution so the appearance of cobalt (II) chloride solution is the same regardless of the hydration number of the solid that was dissolved.
If the water is then allowed to evaporate without heating, CoCl2.6H2O(s) will be reformed. However reverse and forward reactions don’t occur simultaneously.
Observations in combustion of magnesium strip and steel wool
White solid forms in crucible as magnesium strip is heated. When placed in an ice bath, no changes occurs.
A reddish-brown solid forms when steel wool is heated. When placed in an ice bath, no changes occur.
Describe an closed system
A constant number of particles in the system (no matter transfer)
Energy can be transferred, (dependent on the wall of the system, either allowing the transfer of heat energy (rigid) or work (adiabatic).
Example is saucepan with lid, lid can transfer energy to the environment via radiation and conduction. However, no matter enters or exits the saucepan.
This is required for static and dynamic equilibria
Describe an open system
System can interact with the surroundings allowing the exchange of matter and energy
Types of static equilibrium
Irreversible reaction at completion
Initially, the rate of forward reaction starts high. It reduces over time as the concentration of reactants of reactants reduces
The reverse reaction doesn’t occur, so has a rate of zero
Eventually, the rate of forward reaction reaches zero, and static equilibrium is reached
An example of this type is the dissolution of salt in an unsaturated solution
NaCl(s) –>Na+(aq)+Cl-(aq)
Irreversible reaction before initiation
If insufficient energy is inputted, then neither the forward nor reverse reactions will occur so their rates will be equal to zero
An example of this is the combustion of a fuel without an initial spark
C8H18(l) +25/2 O2(g) –>8CO2(g) + 9H2O(l)
Without an initial spark, there is insufficient energy to overcome the activation energy barrier, so the fuel doesn’t combust
Reversible reaction with insurmountable activation energy
An example of this is the diamond-graphite equilibrium
This reaction has ΔG=-2.90 kj mol^-1, so it should be spontaneous. However, it demands an almost insurmountable activation energy, especially at room temperature and hence the reaction doesn’t happen. Hence, f.r and r.r of reaction are almost 0.
Advantages of modelling dynamic eq for the experiment involving cylinder A and B where water in the straws in cylinder A exchange water to cylinder B
It models how the fr is proportional to the amount of reactants, and the rate of reverse reaction is proportional to the amount of products.
Initially, the fr rate is high since the amount of water in cylinder A is high
Over time, the rate of fr decreases and the rate of r.r increases as more water is transferred to cylinder B
Shows that dynamic eq is reached when the lvls of water in cylinder A and B reach a constant lvl, the rates of fr and r.r=
The eq can be reestablished if its disturbed where the eq can be disturbed by adding more water to either cylinder
In this scenario, the rates of the f.r and r.r will instantaneously become different. Over time, a new eq will be reached.
Disadvantages of modelling dynamic eq for the experiment involving cylinder A and B where water in the straws in cylinder A exchange water to cylinder B
System isn’t closed since water can evaporate or be spilled
This model doesn’t deal with concentrations; it only deals with volumes
This system only models one reactant and one product
How does gibbs free energy impact equilibriums
Chemical reactions proceed towards a point of lowest G i.e Gibbs free energy
The extent of a reaction is used to describe the proportion of reactants that have been converted to products
Extent of reaction at lowest G Definition Implications
0% No reaction has occurred This reaction doesn’t occur spontaneously
x% (0-100%). x% of reactants by moles This reaction reaches a dynamic equilibrium
have converted to products
100% All reactants have converted to products This reaction occurs
spontaneously to completion
Difference between ΔG° and ΔG
ΔG° is the standard free energy change which has a single value for a particular reaction at given temperature and pressure.
It is the change in Gibbs free energy for a reaction that goes to completion.
ΔG varies with the extent of reaction and used to describe reactions that can reach dynamic eq. It measures the distance in free energy terms of a particular reaction mixture from reaching dynamic eq.
Key things to remember about gibbs free energy
ΔG°<0 doesn’t mean it is spontaneous always as this only works for reactions where ΔG=0 at either 0% or 100% extent of reaction (static equilibria)
Some dynamic equilibrium reactions can have ΔG°>0 but still proceed to equilibrium since ΔG=0 somewhere between 0% and 100% extent of reaction.
If a reaction reaches ΔG=0 at any extent of reaction that isn’t 0% or 100%, then the reaction will reach dynamic equilibrium. ΔG° doesn’t have to be 0 for dynamic equilibrium to be possible.
However, if a reaction has ΔG°=0, then it has ΔG=0 at every extent of reaction (0%-100%), so the reaction will definitely reach dynamic equilibrium.
Define enthalpy
The internal heat energy of a system. measured in Jmol^-1 pr kjmol^-1. If ΔH<0, the reaction has forward enthalpy drive hence exothermic reactions
If ΔH>0, the reaction has backward/reverse enthalpy drive hence endothermic reactions
How to determine entropy drive
Measured in jmol^-1K^-1 or kjmol^-1K^-1, if ΔS>0, the reaction has forward entropy drive
If ΔS<0, the reaction has backward/reverse entropy drive
What are the law of thermodynamics
Zeroth law: if two systems are in thermal equilibrium with a third, then they are in thermal equilibrium with each other.
1st law: Energy movement into or out of a system is in accordance with the law of conservation of energy
2nd law: The entropy of an isolated system will increase over time, approaching a maximum value at equilbrium. Another interpretation is that the entropy of the universe must be increasing i.e ΔS(universe)=ΔS(system)+ΔS(surrounding)>0
3rd law: The entropy of a system approaches a minimum as temperature approaches zero.
Entropy changes in combustion
The combustion of solid sulfur has a positive entropy change. The temperature increases as the reaction proceeds with more heat energy needing to be dispersed thus producing a greater degree of disorder. General combustion reactions have forward entropy drive for example the combustion of octane (C8H18).
C8H18(l)+25/2 O2(g) ->8CO2(g) +9H20(l)
When octane is combusted at reaction conditions with a spark or flame as the initial source of energy, the temperature is much higher than the bp of water, so gaseous water is formed as a product first
There are 25/2 moles of gaseous O2 on the reactants side, and 17 moles of gas initially on the products side. This means that the entropy drive is forward,
However, we write this equation with water in liquid state to comply with the convention of writing combustion equations at standard conditions.
Entropy changes in photosynthesis
Photosynthesis reduces the entropy of the system by creating ordered glucose molecules from free carbon dioxide and water molecules.
However, we also know that photosynthesis is endothermic, so this reaction has reverse entropy and enthalpy drive and will never occur spontaneously.
A continuous supply of energy in the form of sunlight is required to drive this reaction in plants
Does photosynthesis violate the 2nd law of thermodynamics
Endothermic reactions typically have ΔSsurrounding<0. Then theoretically, this would mean that photosynthesis decreases the entropy of the universe since ΔSuniverse=ΔSsystem +ΔSsurrounding and both seem to be less than 0. This would violate this law. However, since this reaction also uses the UV energy from the sun thus it cannot be classed as an isolated system. If we consider the Sun as part of the system, then the energy released from the Sun is actually absorbed and reflected off the plant into a state of greater disorder. Indeed, the continuous photo excitation from the Sun excites the pigment molecules in the plant including chlorophyll and increases the entropy of their electrons. This amount of disorder is greater than the order gained from organisation of the molecular energy into gluose, i.e ΔS(surrounding), is actually positive, and thus the overall entropy of the universe increases.
Relationship between ΔG° and reversibility
In general, reversible reactions tend to have competing entropy and enthalpy drives.
A chemical reaction with ΔG°>0 for all temperatures is non-spontaneous and likely to be an irreversible chemical reaction. One example is photosynthesis, where ΔH°>0 and ΔS°<0
A chemical reaction with ΔG°<0 for al temperatures is spontaneous and likely to be an irreversible chemical reaction an example being combustion where ΔH°<0 and ΔS°>0.
In general reversible reactions tend to have either:
1. ΔH°<0 (forward enthalpy drive) and ΔS°<0 (reverse entropy drive) OR
2. ΔH°>0 (reverse enthalpy drive) and ΔS°>0 (forward entropy drive)
2NH3(g)⇌N2(g) +3H2(g) ΔH=+92 kjmol^-1
Explain why this reaction is more likely to reach dynamic eq than static eq
This reaction has reverse enthalpy drive as it is endothermic. It has forward entropy drive as ΔS°>0. This is because 2 moles of gas on the LHS are being converted to 4 moles on the RHS, which is an increase in entropy. Therefore, the sign of ΔG° depends on the temperature, so it is not always spontaneous or non-spontaneous. These properties means that it is more likely to reach dynamic eq than static eq.
2NO2(g)⇌N2O4(g). This reaction reaches dynamic eq.
Explain whether the forward reaction is more likely to endothermic or exothermic.
This reaction has a reverse entropy drive i.e ΔS°<0. This is because there are 2 moles of gas on the LHS and 1 mole of gas on the RHS, so there is a net reduction in entropy. Since the reaction reaches dynamic eq, the enthalpy drive is likely forward be an irreversible non-spontaneous reaction, and instead reach static eq. Therefore, the f.r. is likely to be exothermic i.e ΔH<0.
What does collision theory explain
How to increase rate of reaction
Collision theory explains that chemical reactions occur when molecules with sufficient energy collide at a correct orientation. This is known as a successful collision.
Frequency of collision need to increase of any individual collision must have a hgher chance of being successful.
Relevancy of collision theory to type of equilibria
In any reaction, initially there is a high concentration of reactants. Due to this high concentration, the reactant particles will collide at a high frequency, by collision theory, and thus the rate of reactants being converted to products is high.
The concentration of reactants decreases as they are being converted to products, hence the rate of the forward reaction reduces over time. Conversely, as products are being formed, the increased concentration of products increases the collision frequency of the reverse reaction and thus the rate of reverse reaction increases. This continues until the forward and reverse reactions are at the same rate thus making the system in dynamic equilibrium. This point is rarely when the concentration of products and reactants are equal. This position is dependent on the activation energies and the heat of reactions in both the forward and reverse direction.
What does LCP state
Le Chatelier’s principle states that if a system at dynamic eq is disturbed, then the system will shift so as to minimise the change until a new eq is reached. If the system shifts to the right, then the rate of forward reaction begins to exceed the rate of reverse reaction. This will increase the concentration of products and decrease the concentration of reactants until eq is re-established. If the system shifts to the left, then the rate of reverse reaction begins to exceed the rate of forward reaction. This will increase the concentration of reactants and decrease the concentration of products until eq is re-established.
How does concentration impact the point of equilibrium
If the concentration of a substance in equilibrium is changed, the system will shift to favour the reaction that counteracts the initial change
For example Cl2(g) + H2O(l)⇌Cl-(aq)+Clo-(aq)+2H+(aq)
If KCL is added to the system, [Cl-] increases meaning the system shifts to favour the reverse reaction that removes Cl- (by LCP) i.e equilibrium shifts to the side of the reactants (left). This lowers [Cl-], thereby minimising the overall change in [Cl-]. Concurrently, the [ClO-] and [H+] will decrease and [Cl2] will increase
If H+ is removed, [H+] decreases where the system shifts to increase production of H+, i.e the equilibrium shifts to the side of the products (right).
Do pure substances affect concentration
In general, if a pure solid or pure liquid is removed or added from a system, there will be no effect on the equilibrium. This is because solid and liquids are virtually incompressible and their concentrations are essentially constant. For instance, the concentration of liquid water is 1000g/L or 55.5 molL^-1. This value won’t change regardless of how much water there is in a system.
How does adding or removing a solvent affect the equilibrium
However, if a solvent (which may be liquid) is added or removed from a system, changes to the equilibrium will be observed. Changing the volume of solvent will change the concentrations of all species dissolved in the solvent.
AgCl(s)⇌Ag+(aq)+Cl-(aq)
Addition of liquid water (the solvent in this case), will cause a reduction in the concentrations of Ag+(aq) and Cl-(aq), since c=n/V
Hence by Le Chatelier’s Principle, the system will shift to favour to increase the concentrations of the ions and hence the forward reaction is favoured and equilibrium shifts to the side of the products.
How does temperature impact the point of eq
N2(g)+3H2(g)⇌2NH3(g) ΔH<0
If the temperature is increased, the system shifts to favour the backward, endothermic reaction that removes heat i.e shifts to the left. Thus the temperature is lowered minimising the initial change. Concurrently, [N2] and [H2], while [NH3] decreases. Thus, in exothermic equilibria, increasing the temperature decreases yield (the output of products). This can be disadvantageous in industrial and commercial contexts. If the temperature is decreased, the system shifts to favour the forward exothermic reaction that produces heat, i.e shifts to the right.
Define partial and total pressure.
The partial pressure of a gas is its hypothetical pressure if it were the only gas occupying the volume of the mixture, under the same conditions.
The total pressure of an ideal gas mixture is equal to the sum of the partial pressures of each of the component gases
Adding an inert gas doesn’t change any of the partial concentration only the total concentration
How does pressure/volume impact the point of eq
2NO2(g) +H20(l)⇌HNO3(aq) + HNO2(aq)
If NO2 is pumped in, the partial pressure of NO2 increases, as the number of moles of NO2 per unit volume has increased. The system shifts to favour the forward reaction, which consumes NO2, and so lowers the number of moles of NO2 per unit volume. This decreases the partial pressure of NO2, counteracting the initial change. The system has shifted to the RIGHT.
N2(g) + 3H2(g)⇌2NH3(g). ΔH<0
If the total pressure if the system is increased (e.g. by shrinking the volume of the reaction chamber):
The no of moles of gas per unit volume increases (same no of moles, smaller volume)
Since the molar ratio of gases for reactants and products is 4:2, then the system will favour the forward reaction which decreases the no of moles present in the system and thus counteracts the initial change. Therefore, the eq shifts to the right,
How does an inert gas change the point of equilibrium
N2(g) + 3H2(g)⇌2NH3(g). ΔH<0
If an inert gas is added to the system, provided the volume is held constant, the addition of an inert gas will raise the total pressure of the system. However despite the increase in total pressure, the partial pressure for each of the reactants and products are unchanged. This is because the no of moles of each gas and the volume of the container are unchanged. For example, if the concentration of nitrogen gas was 5molL^-1 before the addition of the inert gas, it will still be 5 molL^-1 after the inert gas is added, provided the volume of the container is constant. Therefore, there will be no effect on equilibrium positions.
How does catalysis affect the point of equilibrium
For the catalysed reaction, the reactants will be depleted more rapidly and the products will be produced more rapidly. Catalysts will have no effect on eq position. Instead they hasten the attainment of eq. This means that at eq, the ratio of reactants to products is identical for both the catalysed and uncatalysed reactions.
Explain the overall observations about eq in terms of collision theory for concentration
In a system equilibrium, when the concentration of the reactants are increased, there is an increased number of collision between reactant particles. Thus, the rate of forward reaction increases as per the collision theory. Initially, since there is no change to the product concentration, the rate of R.r. remains constant. Thus, the system would no longer be in eq. However, as new product particles form due to the increased rate of f.r, the rate of reverse reaction also increases as the concentration of products increases. Over time, a new eq will be re-established where the r.r and f.r are equal. Overall, the eq will have shifted to the right in response to an increase in concentration of reactant molecules. Importantly, since the rate of reverse reaction started out at its original value and then climbed as more product molecules formed, the reverse rate at the new eq is higher than the reverse rate at the old eq. This means that the new forward rate must also be higher than its original value, which requires a greater reactant concentration than the original. Therefore, although the system shifted to remove the added reactant, not all of it was removed.
