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- 1Equilibria – How Much?In Kinetics we studied the rate of the reaction, i.e. how fast the product was produced. However kinetics tells us nothing about how much of the product will be produced, for that information we need to study Equilibria.
- 2A reversible reaction isone which can be made to go in either direction depending on the conditions.
- 3Demonstration of some reversible reactions CO2 dissolving and reacting – the Soda Stream CO2(g) ⇌ CO2(aq) CO2(aq) + H2O(l) ⇌ H+(aq) + HCO3-(aq)Providing the lid is on and the temperature is constant, an equilibrium is reached with few / no bubbles seen.
Releasing the pressure, opening the bottle allows CO2(g) to escape and the equilibrium shifts to the LHS to try and restore the original CO2 pressure. - 4Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 What colour is NO2 ?brown
- 5Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 N2O4?colourless
- 6Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 The flask contains an equilbium mixture of brown NO2 gas and colourless N2O4 gas. This mixture could have been produced either: by starting from 0.8 mol NO2
- 7Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 The flask contains an equilbium mixture of brown NO2 gas and colourless N2O4 gas. This mixture could have been produced either: or by starting with 0.4 mol N2O4
- 8Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 The equilibrium mixture reaches a steady state whenthe forward reaction forming N2O4 equals the rate of the reverse reaction.
Note the equilibrium position does not lie ‘in the middle’. In this case at equilibrium there are more N2O4 molecules. - 9Demonstration of nitrogen dioxide and dinitrogen tetroxide 2NO2 ⇌ N2O4 What colour would you expect it to go if the pressure was increased? Explain the changesGo much paler
Gas appears darker at first as pressure pushes molecules closer together (increases concentration).
Over the period of a few minutes the colour becomes paler than the original as the equilibrium becomes established. - 10Define Dynamic equilibrium.In a closed system in equilibrium
Rate of forward reaction = rate of reverse reaction
No overall change in concentration - 11What factors might affect the position of a chemical equilibrium?concentration.
temperature
pressure. - 12Le Chatelier’s Principle states:If a system is in a state of dynamic equilibrium, then a change in temperature, total pressure or concentration of a reactant or product will cause the equilibrium position to shift/change in order to counteract/cancel out the imposed change.
- 13In each case predict what effect changing the temperature and pressure and the removal or addition of a component will have on the system:- 1) 4C(s) + 3H2(g) + N2(g) ⇌ CH3CN(l) ∆Hforward = +31 kJmol-1 Decreasing TemperatureDecreasing the temperature shifts the equilibrium towards the LHS.
The forward reaction is endothermic so by shifting to the LHS the system will tend to ‘replace’ the lost heat energy so [CH3CN] decreases - 14In each case predict what effect changing the temperature and pressure and the removal or addition of a component will have on the system:- 1) 4C(s) + 3H2(g) + N2(g) ⇌ CH3CN(l) ∆Hforward = +31 kJmol-1 Decreasing PressureDecreasing the pressure shift the equilibrium towards the LHS.
4 moles of gas on LHS, none on the RHS so equlibrium shift will try to restore pressure to the original value by moving towards the side with more gaseous moles. so [CH3CN] decreases - 15In each case predict what effect changing the temperature and pressure and the removal or addition of a component will have on the system:- 1) 4C(s) + 3H2(g) + N2(g) ⇌ CH3CN(l) ∆Hforward = +31 kJmol-1 Removing the CH3CN (l) as it is formedEqulibrium shifts towards the RHS, to try to increase the [CH3CN] to the original value
- 16In each case predict what effect changing the temperature and pressure and the removal or addition of a component will have on the system:- 2) 2H2O2 (l) ⇌ 2H2O (l) + O2 (g) ∆Hforward H = -196 kJmol-1 Decreasing TemperatureDecreasing the temperature shifts the equilibrium towards the RHS.
The forward reaction is exothermic so by shifting to the RHS the system will tend to ‘replace’ the lost heat energy so H2O2 decreases - 17In each case predict what effect changing the temperature and pressure and the removal or addition of a component will have on the system:- 2) 2H2O2 (l) ⇌ 2H2O (l) + O2 (g) ∆Hforward H = -196 kJmol-1 Decreasing PressureDecreasing the pressure shift the equilibrium towards the RHS.
No moles of gas on LHS, one on the RHS so equlibrium shift will try to restore pressure to the original value by moving towards the side with more gaseous moles. so H2O2 decreases - 18Industrial application of the principles of equilibria – The Haber Process From your knowledge of equilibria predict what will be the most favourable conditions for the production of ammonia:- Temperature: Pressure:temp: High for rate but low for equilibrium position
pressure: high - 19The actual conditions used in the modern Haber process are 400-450 oC and 130-300 atm Are these the conditions you found gave either highest production rate or lowest cost above? Cost. What other factors might have to be considered?1) Cost of high pressure system plant costs of heavy duty / thick pipes and containers.
Also cost of running compressors.
2) Safety - risk analysis of running high pressure systems does not justify their extra yield
3) Environmental implications of any gas release. - 20In most industrial processes the conditions used are a compromise between:-approaching an equilibrium quickly (i.e. the rate of both forward and backward reactions)
the position of equilibrium
the capital costs to build the plant
running costs
safety - 21catalystis often used so that the reaction can be run efficiently at a lower temperature.
Industrially systems do not reach equilibrium as the products are drawn off as they are formed, thus shifting the equilibrium position in favour of products. - 22High pressure systemsrarely used because of the cost of running the compressors (electricity), cost of the high pressure vessels and pipe-work, and safety issues (risk of explosions).
Low pressure systems are also rarely used because the vessels would need to be too large and it would be difficult to move gases under low pressure around the plant. - 23Stored methane in equilibrium – a major environmental issue. methane hydrate(s) ⇌ methane(g) + water(l)There are huge amounts of methane trapped as solid methane hydrate in ice structures deep in the oceans and within the permafrost. Methane hydrate is in equilibrium with gaseous methane and water:-
- 24methane hydrate(s) ⇌ methane(g) + water(l) The forward reaction is endothermic, if the temperature of the earth rises the equilibrium position will shift to theright.
- 25The reaction could come to the equilibrium position (at which significant amounts of both reactants and products are present) either starting frompure reactants or products. What this implies this that going from either pure reactants or pure products results in a decrease in the Gibbs energy. This can be shown on a plot of the Gibbs energy of the reaction mixture as a function of the composition.
- 26From the plot we can see that the Gibbs energy goes to?a minimum at some point intermediate between pure reactants and pure products. This minimum corresponds to the equilibrium position. The shape of the plot funnels the composition towards the minimum Gibbs energy. From this point any change in the composition would result in an increase in the Gibbs energy, which is not permitted. So, once at it’s equilibrium value the composition does not change, provided the temperature does not change.
- 27If we take two liquids which do not mix (are immiscible), such as water, and volasil (or hexane) they will form:two separate layers, one on top of the other. Both these liquids can act as a solvent for iodine, a solute. You may remember from the halogen packs that iodine has quite different and distinct colours in these two solvents.
- 28Colour of iodine in aqueous solution:brown
- 29Colour of iodine in organic solvent ispurple:
- 30If iodine is shaken with the mixture of the two solvents, they will both compete for it. The concentration of iodine in each layer will be shown by the colour. What do you think is really happening when the concentration of iodine in each layer is apparently constant (the colour no longer changes)?Rate at which iodine passes into the aqueous layer is the same as the rate at which iodine passes back into the volasil layer.
- 31I2(aq) ⇌ I2 (volasil) Partition Coefficient,
- 32reaction between iodine and sodium thiosulphate: Reduction half equation:I2 + 2e- ----> 2I-
- 33reaction between iodine and sodium thiosulphate: Oxidation half equation:2S2O32- ----> S4O62- + 2e-
- 34reaction between iodine and sodium thiosulphate: Total redox reaction:2S2O32- + I2 ----> S4O62- + 2I-
- 35Where square brackets [ ] =molar concentration of I2
- 36Partition Coefficient Do you think Kc has units?No
Explain your answer:
same number of concentration expressions on top and bottom of Kc expression so the units cancel
- 37The Equilibrium Law For a reaction aA + bB ⇌ cC + dD For a closed system at constant temperature, the following expression can be written The equilibrium constant, Kc =The subscript ‘c’ indicates that Kc is expressed in molar quantities
The products of the forward reaction always appear on the top line,
and the backward reaction on the bottom line.
The value of Kc gives us an idea of the position of equilibrium. The greater the value, the further the equilibrium position to the right and the greater the yield.
The power to which you raise the concentration of the substance is the same as the number of moles which appears in front of the balanced equation,
The concentrations are multiplied together because this represents all possible collisions of particles which might result in a reaction.
E.g. If there are 3 particles of A and 3 particles of B,
The number of possibilities of the particles colliding and reacting = 3 x 3 =9
- 38The Equilibrium Law For a reaction aA + bB ⇌ cC + dD For a closed system at constant temperature, the following expression can be written The equilibrium constant, Kc = continuedIf the concentration composition is changed, the reactants and products will arrange themselves until equilibrium is re-established and Kc is unchanged.
Similarly, if the total pressure composition is changed Kc is unchanged.
If the temperature of the reaction is changed, then Kc is changed.
All chemical reactions are equilibrium reactions. Even reactions that seem to go to completion actually have a little of reactant left in equilibrium with the product.
Do not include in the equilibrium expression:- solids
reactants/ products also present as a solvent
because the concentrations of these will remain virtually constant. - 39What are the units of Kc?The units of Kc vary!
You can see that the units of Kc vary from reaction to reaction, and need to be worked out from the Kc expression. When you quote a value for Kc you must always show a balanced equation for the reacti - 40Equilibrium quantities questions 1) Work out the actual number of moles of each species that are present at equilibrium following the worked example a). You are given the circled figures.
- 41What use is the equilibrium law?It enables us to calculate the composition of a reaction mixture which has reached equilibrium and is therefore useful if we are preparing a 'target compound’ via a reversible reaction.
- 42Consider the reaction of ethanol and ethanoic acid: Imagine that Kc = 4 at the temperature of our experiment. Suppose we want to know how much ethyl ethanoate we could produce by mixing 1 mol of ethanol and 1 mol of ethanoic acid. Set out the information as shown below:
- 43Consider the reaction of ethanol and ethanoic acid: continuedWe do not know how many moles of ethyl ethanoate will be produced, so we call this x. The equation tells us that x mol of water will also be produced and in doing so x mol of both ethanol and ethanoic acid will be used up. So the amount of each of these remaining at equilibrium is (1-x)mol.
- 44Consider the reaction of ethanol and ethanoic acid: continued These figures are in moles, but we need concentrations in mol dm-3 to substitute in the equilibrium law expression. Suppose the volume of the system at equilibrium was V dm3. Then:
- 45Consider the reaction of ethanol and ethanoic acid: continued These figures may now be put into the expression for Kc:
- 46The Vs cancel, so we do not need to know the actual volume of the system. (This will happen for all systems with equal numbers of moles of product and reactant, so V is sometimes omitted. It is always better to include V, so you will not forget it for systems where the Vs do not cancel out.)
- 47Taking the square root of both sides, we get:So ⅔ mol of ethyl ethanoate and ⅔ mol of water are produced if the reaction reaches equilibrium, and the composition of the equilibrium mixture would be:-
ethanol ⅓ mol,
ethanoic acid ⅓ mol,
ethyl ethanoate ⅔ mol,
water ⅔ mol
- 48Determination of the equilibrium constant for an esterification reactionThe practical determination of the equilibrium constant Kc for a reaction presents some difficulties. To find Kc for a reaction it is necessary to measure the concentration of reactants and products at equilibrium. Most methods for doing this involve removal of reactant or products, which changes their concentration and shifts the equilibrium. It is therefore ideal to use a method of measuring concentration that does not involve removing any component - colorimetry, for example.
- 49The reaction under consideration here is the esterification of ethanoic acid by ethanol: CH3COOH + CH3CH2OH ⇌ CH3COO CH2 CH3 + H2OThe reaction is catalysed by H+ ions. The advantage of studying this particular reaction is that it is kinetically slow, so it is possible to measure the concentration of a reactant by titration (which removes the reactant), without significantly disturbing the equilibrium in the relatively short time it takes to carry out the titration.
Known amounts of acid, alcohol and/or ester are mixed with dilute hydrochloric acid in a stoppered bottle and allowed to reach equilibrium at room temperature. After reaching equilibrium, which takes several days, the mixture is titrated with alkali in order to find the total amount of acid present. After allowing for the presence of the hydrochloric acid catalyst, the amount of ethanoic acid present at equilibrium can be calculated. From this the amounts of the other reagents can be calculated, and hence a value for the equilibrium constant can be found. - 50ΔH for the esterification reaction is +17.5 kJ mol-1. Would you expect the value of Kc´ to be greater or smaller at 80oC than at room temperature?Hydrolysis ΔH = -17.5 kJ mol-1 so the value of K will decrease, the numerator, [CH3COOH] and [CH3CH2OH] will decrease and so equilibrium will shift to the LHS
Increasing the temperature will favour the esterification and Kc’ will increase - 51The yield of ester formed in the reaction between ethanoic acid and ethanol is considerably increased if concentrated sulphuric acid is added to the reactants. Suggest a reason for this.Conc. H2SO4 is a dehydrating agent and will ‘remove’ the water produced by esterification so the equilibrium position will be shifted towards the esterification. Addition of CH3CH2OH would have the same effect.
Also conc. acid acts as a catalyst (no effect on Kc) by protonating the organic acid facilitating subsequent nucleophilic attach by the alcohol: - 52Heterogeneous equilibria. So far the substances involved in equilibria have all been in the same phase. Phase is a region of space throughout which all physical properties of a material are essentially uniform. The term phase is sometimes used as a synonym for state of matter, however there are important differences. State this difference.States of matter describe the interrelationship between particles i.e. particles are close in fixed positions in a solid, close but freely moving in a liquid and widely spaced with little attraction in a gas.
Phase is a region of space throughout which all physical properties of a material are essentially uniform.
If you look at a mixture and can see a boundary between two of the components, those substances are in different phases. - 53A heterogeneous mixture iswhen there is more than one phase present
- 54Complete the table
- 55when we consider Kc equilibria whose components are expressed in terms of concentrations (mol dm-3). The concentration of a solid would not change, it is determined byits density, so it would not appear in a Kc expression. Similarly a non-volatile solvent would not exert any vapour pressure so would not appear in a Kp expression.
- 56In a system where one of the reactants or products acts as a solvent for the system thenthis is left out of the Kc expression.
- 57The basis of equilibrium
- 58Temperature affects the value of KA change in T cause a change in ΔSsurr and hence ΔStotal. This causes a change in K which alters the position of equilibrium.
An increase in temperature makes - ΔH/T a smaller number, hence ΔSsurroundings a smaller positive number (if exothermic) or a smaller negative number (if endothermic)
This cause ΔStotal and hence K to get smaller for exothermic reactions and bigger for endothermic reactions.
If K decreases, the position of equilibrium moves to the left and if K increases, the position of equilibrium moves to the right
NOTE position moves, so K alters. - 59Pressure has no effect on the value of KIf there are a different number of gas moles on each side, the concentration / partial pressure term will alter.
K no longer equals the concentration / partial pressure term
The position of equilibrium moves until the concentration / partial pressure term once more equals the unchanged K - 60A method for the calculations of:- Kc and Kp expressions
- 61For the equilibrium reaction between hydrogen and iodine H2(g) + I2(g) ---> 2HI(g) increasing the pressure of the systemincreases the rate but does not affect the position of equilibrium.
- 62The equation for the equilibrium reaction between hydrogen and iodine may also be written as ½H2(g) + ½I2(g) ------> HI(g) This change to the equation, compared to that in part (a),square roots the value of the equilibrium constant.
- 63as Kc is the ratio of products to reactants what does the value tell us?which way the equilibrium lies
- 64a Kc value bigger than 1?tends to favour / equilibrium pushed over to product side
- 65a Kc value smaller than 1?tends to favour / equilibrium pushed over to reactants side
- 66heterogeneous equilibria catalysts don't affect the equilibrium constant
- 67you might need to work out equilibrium constant
- 68Kc can be used to find concentrations in an equilibrium mixture
- 69the equilibrium constant can be calculated from experimental data
- 70the equilibrium constant can be calculated from experimental data example
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- A level Chemistry Karteikarten am Kc, erstellt von megan trent am 30/01/2019.
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