7. Equilibrium
Show how changes in concentration, pressure, temperature, and the presence of a catalyst alter the position of an equilibrium. - Concentration
* if the concentration of the reactants increase, the concentration of the products will increase, the equilibrium will proceed more to the right. - The reactants will be able to produce more products as they are being "used". - Eventually the forward and reverse reaction rates are going to be equal and equilibrium will be restored * If you decrease the concentration of the reactants the equilb. will proceed more to the left. I f you increase the conc. of the products that is, you add more to the reaction, then the reaction will favour the left side of the equilibrium.
An equilibrium reaction has four particular features under constant conditions:
1. It is dynamic. 2. The forward and reverse (backward) reactions occur at the same rate. 3. The concentrations of reactants and products remain constant at equilibrium. 4. It requires a closed system.
Describe the key features of the Haber process (ammonia) and Contact process (sulfuric acid) as important industrial applications of reversible reactions, and the principles that control dynamic equilibrium. - Contact process
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -197 kJ mol-1 - Contact Process Le Châtelier's principle would suggest that the highest yield of sulfur trioxide would be obtained using high pressures and low temperatures. However, in practice, a very low temperature cannot be used, as the rate of achieving equilibrium would be so slow as to be uneconomic. Another factor here is that the vanadium (V) oxide catalyst only becomes effective at temperatures above 400°C. To improve the percentage conversion, the reacting gases are passed through a sequence of four separate catalyst beds. This repetitive technique produces a 99.5% conversion of sulfur dioxide to sulfur trioxide. Given this high rate of conversion, it is unnecessary to use such a high pressure to increase the yield of sulfur trioxide, SO3. The cost of using high pressures is uneconomic and a pressure of 1-2 atmospheres is used to ensure the gases circulate freely through the catalyst beds.
HL The value of the equilibrium constant for the following reaction: 2HI(g) ⇋ H2(g) + I2(g) is 0.25 at 440°C. What would the value of the equilibrium constant be for the reverse reaction at the same temperature? H2(g) + I2(g) ⇋ 2HI(g) 1. 4 2. 0.5 3. 2 4. 0.25
4.0 Explanation #1 is correct; the equilibrium constant for the reverse reaction is always the inverse of the value for the forward reaction; hence the value here is 1 / 0.25 or 4.0
Show how changes in concentration, pressure, temperature, and the presence of a catalyst alter the position of an equilibrium. - Catalyst
A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway of lower activation energy (Ea). Industrially, catalysts are of significance, as they allow reactions to occur at reasonable rates under milder and therefore more economic conditions. Consequently, the presence of a catalyst increases the rate of the forward and reverse reactions equally. There is no change in the position of the equilibrium, or the value of Kc. However, the advantage of using a catalyst is that its presence reduces the time required for the equilibrium to be established.
Explain why Equilibrium requires a closed system.
A closed system is one in which none of the reactants or products escape from the reaction mixture. In an open system, some matter is lost to the surroundings. Many chemical reactions can be studied without placing them in closed containers. They can reach equilibrium in open flasks if the reaction takes place entirely in solution and no gas is lost.
HL Calculate the solutions of homogeneous equilibrium problems using the expression for Kc for a particular reaction - ICE method
A slightly more difficult type of question centres on you being able to use the equation for the reaction to work out the concentrations of the various substances in the equilibrium mixture. The approach here is sometimes referred to as the 'ICE' method as it involves the tabulation of: - The initial number of moles, or concentration, I; - The change in number of moles, or concentration, C; - The equilibrium number of moles, or concentration, E, involved in the different stages of the reaction https://ieg.kognity.com/study/chemistry-hl-2016/equilibrium-ahl/equilibrium-law/calcualtions-homogeneous-equilibria/ https://youtu.be/sehNbIQbfCY
HL Understand and describe how Le Châtelier's principle for changes in concentration can be explained by the equilibrium law, Vet ej
According to Le Châtelier's principle , the equilibrium system responds to negate change by responding in the opposite way. That is, when a system in equilibrium is stressed, it moves to oppose the stress. The equlibrium law states that the equilibrium law can be used to explain and quantify the effect of concentration changes at a given temperature. Such explanations are based on the fact that the value of the equilibrium constant, Kc, itself is not affected by a change in concentration. The only thing that effects the value of Kc is a change in temperature.
The only thing that changes Kc is temperature!!!! What does an increase or decrease in temp do to the Kc of the reaction?
An increase of temp is going to favour the endothermic side (left side with reactants) meaning that the Kc is going to decrease. An decrease in temp is going to favour the exothermic side (right side with products) meaning that Kc is going to increase.
Kc value when the reaction has gone virtually to completion or hardly taken place at all?
As a general rule: If Kc > 1, then the reaction is said to have gone virtually to completion. There has been an almost complete conversion of reactants to products. If Kc < 1, the reaction has hardly taken place at all; very little of the reactants have been converted to products. Reaction hardly goes: Kc < 10^-10 'Reactants' predominate at equilibrium:Kc = 0.01 Equal amounts of reactants and products: Kc = 1 'Products' predominate at equilibrium: Kc = 100 Reaction goes virtually to completion: Kc > 10^10
¨HL Explain why the composition of an equilibrium mixture thus depends on the value of ΔGθ .
As a reaction proceeds the composition of the reacting mixture of reactants and products is constantly changing and the free energy will also be changing. The position of equilibrium corresponds to a maximum value of entropy and a minimum in the value of the Gibbs free energy change. Once this point has been reached the reaction will not proceed any further - the rate of the forward reaction and the reverse reaction are equal. The composition of an equilibrium mixture thus depends on the value of ΔGθ . A consideration of the relationship between ΔGθ, equilibrium composition and Kc will be returned to in Topic 17.
Explain the forward and backward reactions occur at the same rate.
At equilibrium, the rate of the forward reaction equals the rate of the backward (reverse) reaction. Molecules or ions of reactants are becoming products, and those in the products are becoming reactants at the same rate. One analogy to help understand this is shown in Figure 1. If the speed of the escalator is the same as that of the boy, then the position of the boy will not change.
Show how changes in concentration, pressure, temperature, and the presence of a catalyst alter the position of an equilibrium. - Pressure
For pressure you have to look for gas in the Eqb. *If you increase the pressure the eqb is going to try to remove some of the gas to reduce the stress(pressure/negate the change. *Increased pressure shifts the equilibrium position to the side of the equation with the least gas particles. *Decreased pressure shifts the equilibrium position to the side with most gas particles. *If a reaction has equal numbers of gas particles on each side, pressure has no effect on the reaction!*
Give examples of both physical and chemical instances of a dynamic equilibrium.
Hittade ej.
Show how changes in concentration, pressure, temperature, and the presence of a catalyst alter the position of an equilibrium. - Temperature
If you heat the reaction up it's goin to try to cool itself down and if you cool it down it tries to heat itself up. In a reaction the left side is endothermic (cooler) and the right side is exothermic. *An increase in temperature always favours the endothermic process. (forward) *A decrease in temperature always favours the exothermic process.(reverse)
HL Understand how Le Châtelier's principle enables us to predict the effect of various changes in experimental conditions on the position of equilibrium established for a reversible reaction
Le Châtelier's Principle is a descriptive statement of what happens when a dynamic equilibrium is disturbed by a change in conditions; it is not an explanation as to why the change happens. Put simply, the system responds to negate the change by responding in the opposite way. For instance, if we add more of a reactant, the system will react to remove it. The equilibrium expression for the general reaction: A + B ⇋ C + D is as follows (note again that the concentrations of the products are always on top in this expression): Kc = [C] [D] / [A] [B]
Why is Le Châtelier's principle of importance industrially?
Le Châtelier's principle is of importance industrially, as it allows chemists to alter the reaction conditions to produce an increased amount of the product and, therefore, increase the profitability of a chemical process.
Describe the key features of the Haber process (ammonia) and Contact process (sulfuric acid) as important industrial applications of reversible reactions, and the principles that control dynamic equilibrium. - Haber process
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = -92 kJ mol^-1 - Haber process The conditions used industrially in the main reaction vessel are arrived at by consideration of both Le Châtelier's principle and kinetic factors. The aim is to achieve a satisfactory yield of ammonia at a reasonable and economic rate. The Haber process is markedly affected by changes in pressure. If the pressure is increased in accordance with Le Châtelier's principle, the position of the equilibrium will shift to the right - there are fewer molecules on that side of the equation. On the left-hand side of the equation there are four moles of gas; on the right-hand side there are two moles of gas. This means that higher pressures will move the position of the equilibrium to the right, producing more ammonia gas In the Haber process, the forward reaction is exothermic (ΔH = -92 kJ mol-1). This means that the production of ammonia will be favoured by lower temperatures. Increased temperature will result in less ammonia in the equilibrium mixture. By this consideration, it would follow that the Haber process should be carried out at low temperatures. Industrially, however, a temperature of 450°C is actually used. This is an optimum temperature - a compromise between producing sufficient ammonia and producing it at a reasonable rate of reaction. If too low a temperature is used, the reaction is too slow, even though you get more ammonia in the mixture. The Haber process uses a catalyst of freshly produced, finely divided iron. Remember that a catalyst doesn't alter the equilibrium position - it speeds up the forward and reverse reactions equally.
Explain how Le Châtelier's Principle predicts the effect of a change in conditions on a physical or chemical equilibrium.
Put simply, "the system responds to negate the change by responding in the opposite way. " (lanterna) or "When a system in equilibrium is stressed, it moves to oppose the stress" (R.Thornley)
Explain how the reaction quotient, Qc, is related to the equilibrium constant and gives an indication of the progress of a reaction towards equilibrium.
Qc is the reaction quotient of the concentrations of reactants and products when they are not in equilibrium. When Q < Kc --> equilibrium proceeds to the right When Q > Kc --> equilibrium
examiner tip
Remember that a very large Kc value does not mean that the reaction goes to total completion as there is still an equilibrium situation, but that only there is a relatively far, far more products than reactants. Some argue that no reaction goes to total completion.
Attention! What's the difference between ∆G and ∆Gθ ?
The 'standard' symbol makes a world of difference. The standard free energy change, ∆Gθ , has a single value for a particular reaction. The other quantity, ∆G , represents the total free energies of all the substances (products - reactants) in the reaction mixture at any particular composition. In contrast to ∆Gθ which is a constant for a given reaction, ∆G varies continuously as the composition changes, finally reaching zero at equilibrium.
HL understand that the Gibbs free energy change of a reaction and the equilibrium constant can both be used to measure the position of an equilibrium for a reaction and are related by the equation, ΔGθ = −RT lnK.
The chemical equilibrium constant, K, is directly related to the Gibbs free energy change, ∆Gθ, (Topic 15) by the following equation (van't Hoff's equation): ∆Gθ = − RT lnK Where R represents the molar gas constant and T the absolute temperature in kelvin. For reactions in solution or the liquid phase the value of K derived is that of Kc. If ∆Gθ is negative, Kc is greater than 1 and the products predominate in the equilibrium mixture. Alternatively, if ∆Gθ is positive, Kc will be less than 1 and the reactants will predominate in the equilibrium mixture.
Explain the concentrations of reactants and products remain constant at equilibrium.
The concentrations remain constant because, at equilibrium, the rates of the forward and backward reactions are equal. The equilibrium can be approached from two directions. For example, in this reaction... H2(g) + I2(g) ⇌ 2HI(g)
HL Understand how the position of equilibrium can be quantified by the equilibrium law and that the equilibrium constant for a particular reaction only depends on temperature,
The equilibrium expression for the general reaction: Aa + Bb ⇋ Cc + Dd is as follows (note again that the concentrations of the products are always on top in this expression): Kc = [C]^c [D]^d / [A]^a [B]^b The equilibrium law can be used to explain and quantify the effect of concentration changes at a given temperature. Such explanations are based on the fact that the value of the equilibrium constant, Kc, itself is not affected by a change in concentration. The only thing that effects the value of Kc is a change in temperature. e.g. Using the Haber process reaction: N2(g) + 3H2(g) ⇋ 2NH3(g) we can illustrate the changes when more nitrogen is added at constant volume. The concentration increases and using arguments based on the equilibrium law we can make the following statements regarding the subsequent changes: H2 reacts with the added N2 to form NH3. [NH3] increases a lot; [N2] is initially increased and [H2] is decreased since Kc remains unchanged. The position of equilibrium shifts from left to right (favouring products). Both forward and backward rates will be greater than before. The effect of the addition of more nitrogen on the rates of the forward and reverse reactions can be described as follows. The adding of nitrogen to the mixture increases its concentration, which causes an increase in the rate of forward reaction. This uses up nitrogen and hydrogen, so forward rate decreases, and it makes more ammonia so the backward rate increases. The decreasing forward rate and increasing backward rate finally become equal when a new equilibrium is re-established (
HL Understand that the position of equilibrium corresponds to a maximum value of entropy and a minimum in the value of the Gibbs free energy for a reaction,
The equilibrium mixture always has a lower Gibbs free energy (and higher entropy) than either the pure reactants or the pure products (a mixture has higher entropy than pure substances); therefore the conversion of either reactants or products into the equilibrium mixture results in a process in which ∆Gθ is negative. Note that the overall Gibbs free energy of a system (Gθ ) depends on how much of each substance is present, and the equilibrium mixture represents the composition that gives the minimum value of the Gibbs free energy (maximum value of entropy). When the system is at equilibrium, the Gibbs free energy of the remaining amount of reactants present is the same as that of however much of the products has been formed. Thus ∆G is zero at this point and there is no tendency to spontaneously move in either direction away from equilibrium. Any shift away from the equilibrium position results in an increase in G and therefore a process for which ∆G is positive, i.e. non-spontaneous The value and sign of ∆G give us information about the position of equilibrium. If ∆G is negative then the position of equilibrium will be closer to the products than the reactants (Figure 2). The more negative the value of ∆G, the closer the position of equilibrium lies awards the products. If ∆G is numerically very large and negative then the position of equilibrium lies very close to pure products, which corresponds to the idea discussed above (see Tables 1 and 2) - a reaction for which ∆G is negative proceeds spontaneously from reactants to products. If ∆Gθ is positive then the position of equilibrium lies more towards the reactants (Figure 3) - the more positive the value, the closer the position of equilibrium lies towards pure reactants. In summary, ∆Gθ can be used to predict the spontaneity of a reaction and position of equilibrium as follows: A reaction that has a value of ∆Gθ that is both large and negative takes place spontaneously and reaches an equilibrium position to the right, favouring the products. The equilibrium mixture contains a large proportion of products. A reaction with a value of ∆Gθ that is large and positive does not take place spontaneously and reaches an equilibrium position to the left favouring the reactants. The equilibrium mixture contains predominantly reactants with only a limited amounts of the products formed. https://www.youtube.com/watch?v=YITesmOK4u8&list=PL816Qsrt2Os394LaHIen_URp8ZFC90Kpg&index=10
How does the magnitude of Kc gives us a useful indication of how far a reaction has gone towards completion under the given conditions?
The higher the value of Kc, the further to the right the equilibrium position will lie at that temperature. This relationship arises from the structure of the equilibrium expression - a consequence of product concentrations being on the top! If the value of Kc is high, this shows that there is a high proportion of products compared to reactants at equilibrium. In this case, we say that the equilibrium lies well over to the right, as the equation is written. If the Kc value is low, then this indicates that only a small fraction of the reactants have been converted into products and the equilibrium lies over to the left.
The key features of an equilibrium expression that you should be aware of are:
The key features of an equilibrium expression that you should be aware of are: 1. Concentrations used are those of the reactants and products in the equilibrium mixture. 2. Concentrations of the products are always on top in the expression. 3. Powers to which the concentrations are raised are the stoichiometric coefficients in the equation for the reaction. 4. The only factor which can change the value of the equilibrium constant is a change of temperature.
Explain dynamic equilibrium
The phrase dynamic equilibrium means that the molecules or ions of reactants and products are continuously reacting. Reactants are continuously being changed to products, and products are continuously being changed back to reactants.
What is the equilibrium constant?
We can write an equilibrium expression for Kc for a general reaction: aA + bB ⇌ cC + dD where a, b, c and d represent the stoichiometric coefficients in the balanced equation. Kc=[C]^c [D]^d / [A]^a [B]^b *It is important to note that the concentration values fed into the expression must be those that occur at equilibrium - not the starting values. *It should be noted that the equilibrium constant, Kc, is constant for a given temperature. If the temperature changes, then the value of Kc will change. *The key features of an equilibrium expression that you should be aware of are: - Concentrations used are those of the reactants and products in the equilibrium mixture. - Concentrations of the products are always on top in the expression. - Powers to which the concentrations are raised are the stoichiometric coefficients in the equation for the reaction. - The only factor which can change the value of the equilibrium constant is a change of temperature.
What happens if you reverse the equilibrium?
When discussing a reversible reaction it is important to give the equation for the reaction clearly. Even though the reaction is reversible, the species on the right of the equation are often referred to as 'products', while those on the left are still termed the 'reactants'. The structure of the equilibrium expression depends on the direction in which the equation is written, as does the value of Kc that we derive from it. Reversing the equation: if we know the equilibrium constant for an equation written in one direction, then if we carry out the reaction bringing it to equilibrium starting from the opposite side of the equation the value of Kc will be the inverse (or reciprocal) of the known value.
Examiner tip HL
When solving for homogeneous equilibrium problems, don't forget to ensure that you convert a given number of moles into concentrations (mol dm-3). The easiest situation is where the volume of the container is 1 dm -3. However, this will obviously not always be the case. If you are not given a value then use V dm3 as in the example above. The V's will then usually cancel out. The change in concentration line in the above method must always reflect the coefficients of the balanced chemical equation.
HL Calculate the solutions of homogeneous equilibrium problems using the expression for Kc for a particular reaction - Standard question
Worked example 1: Nitrogen(II) oxide, NO, is a pollutant released into the atmosphere from car exhausts. It is also formed when nitrosyl chloride, NOCl, dissociates according to the following equation: 2NOCl(g) ⇋ 2NO(g) + Cl2(g) To study this reaction, different amounts of the three gases were placed in a closed container and allowed to reach equilibrium at two different temperatures 503 K and at 738 K. a Write the expression for the equilibrium constant,Kc, for this reaction. b Calculate the value of Kc at each of the two temperatures given. c Is the forward reaction endothermic or exothermic? (base your answer on ideas we covered in Topic 7) Solution a Kc = [NO]2 [Cl2] / [NOCl]2 b at 503 K; Kc = (1.46 x 10-3)2 (1.15 x 10-2) / (2.33 x 10-3)2 = 4.5 × 10−3 at 738 K; Kc = (7.63 x 10-3)2 (2.14 x 10-4) / (3.68 x 10-4) = 9.2 × 10−2 c The value of Kc is greater at 738 K; Kc increases with temperature, so the forward reaction is being favoured to increase the proportion of products in the equilibrium mixture. This suggests that the forward reaction is endothermic.
