Suggested answers to George Facer A2 equilibrium questions



Suggested answers to George Facer A2 APPLICATION of equilibria 4.6 questions.doc p86-7

Q1

a)

For 2SO2 (g) + O2(g) ( 2SO3(g), Kc = [ SO3(g) ]2

[ SO2(g) ]2 [ O2(g) ]

If eqm then the initial moles for each reagent (as given in the question) will be the equilibrium moles. So calc their concn and test the quotient to see if it agrees with the given.

Kc value.At 425 0C, Vol = 20 dm3, Kc= 1.7 x 106

| |Reactants |Products |

| |2SO2(g)) |O2(g) |2SO3(g) |

|Initial moles |0.2 |0.1 |2.0 |

|Eqm moles |0.2 |0.1 | 2.0 |

|20 dm3 container |0.2/20 = |0.1/20 = | 2/20 = |

|so, eqm concn |0.01 |0.005 |0.1 |

-----------------------------------------------------

Q = [ SO3(g) ]2 ( (0.1)2 ( 0.01 = 20000 = 2.0 x 104

[ SO2(g) ]2 [ O2(g) ] (0.01)2 (0.005) 5 x 10-7

Q < Kc so the system is NOT at equilibrium.

Note: We were testing the quotient so we used the symbol Q. We didn’t know it was at eqm so we didn’t use the symbol Kc

The reaction will move in favour of the products, increasing the value of the quotient until it increases sufficiently to be the same as Kc. This happens irrespective of whether a catalyst was used or not, but the equilibrium will be achieved over a shorter period of time.

Q2

a)

One mole of solid produces two moles of gas. (S(sys = ((S prods) - ((S reacts) (

(S(sys = big (positive) number – smaller (positive) number ( (S(sys is likely to be positive.

(S(sys being positive is obvious to see as the system becomes more disordered due to more disordered gas states being produced.

b)

For NH4Cl(s) ( NH3(g) + HCl(g), Kp = ((NH3) x ((HCl) Kp = 50 atm-2 at 700 K

c)

The (H( = +176 kJ mol-1

(S(TOT = (S(sys + (S(surr The (H( and the (S(sys remain almost constant at different temperatures. In general, the change in (S(TOT depends on (S(surr and (S(surr = -((H( / T)

So at any Temp, (S(surr will be –‘ve, but at higher Temp. it will become less –‘ve

Therefore (S(TOT becomes less negative (i.e. more positive).

(S(TOT = R lnK so the value of K must increase.

Note: The value of the quotient at the higher temp will have a greater partial pressure of NH3 and HCl

d)

(S(TOT = R lnK

(S(TOT = 8.314 x ln(50) = 8.314 x 3.912 = + 32.5 J K-1 mol-1

(S(surr = -((H( / T) = -(176,000 / 700) = - 251.4 J K-1 mol-1

Remember: convert kJ values of (H into J when calculating (S values !!

(S(sys = (S(TOT - (S(surr = (+ 32.5) – (- 251.4) = 283.9 J K-1 mol-1

NB: Data book for (S(sys = NH4Cl = +94.4, NH3 = +192.3, HCl = +186.6.

(S(sys =((S prods) - ((S reacts) ( { (+192.3) + (+186.6) } – (+94.4) = 378.9 - 94.4 = +284.5

Q3

CO(g) + 2H2(g) ( CH3OH(g) (H( = -90 kJ mol-1

Kp at 573 K = 5 x 10 - 4 atm-2

Conditions employed are: Temp=573 K, ZnO catalyst, pressure 300atm, Temp = 573 K (i.e. 300 oC)

Q = ( (CH3OH(g) )

( (CO(g) ) x ( (H2(g) )2

Temp:

An elevated temp (relative to room temp) is used to increase the rate of reaction by making more molecules with energy ( the activation energy, but as the reaction is exothermic, use of elevated temperature gives a lower equilibrium value due to the fact the (S(surr will be less positive as temperature increases, (S(sys and (H stay relatively constant so (S(TOT decreases. A smaller value of the equilibrium constant will result causing in a lower yield, so the Temp used isn’t too high or yield would be too low. A lower temp is not used as although a greater equilibrium yield will result, the rate at which it occurs will take too long. Hence the temperature used gives the best trade-off between rate and yield.

Pressure:

At the elevated compromise temperature, moving to an elevated pressure of 300 atm will not affect Kp but will give a higher yield because of the greater power terms for the partial pressures in the denominator of the quotient can only by offset to give the same Kp by the partial pressure in the numerator increasing. The high pressure will have an extra effect on the rate, although marginal in the presence of a heterogeneous catalyst, by causing the gas molecules to experience greater frequency of successful collisions, leading to a greater rate of reaction. – But it should be remembered that the rate from the catalysed reaction is far more significant. A higher pressure than 300 atm is not used as although would it would increase the very low yield (indicated by the small value of Kp), the costs of operating at higher pressures as well as maintenance is very expensive, making the manufacture of methanol too costly and leading to a poor economical process. It may also cause the gases concerned to condense in the active sites of the catalysr reducing the ability of the ZnO to catalyse the reaction.

Catalyst: A catalyst is used to provide an alternative pathway with a lower activation energy so a greater proportion of molecules has an energy ( the new lower activation energy which leads to an increase in rate.

A heat exchanger can be used to cool the post-reaction mixture to liquefy the methanol (which has the highest boiling point) and therefore remove it from the reaction mixture, recycling the unreacted reactant gases back over the catalyst being heated from the heat exchanger.

Q4

a)

Kp = (CO2(g)

b)

Units depend on how we measure the pressure. If the pressure is measured in atm, then the units for Kp in this reaction are atm. If the pressure is measured in ‘mm Hg’ then the value for Kp will be mm Hg, if the pressure is measured in Pascals (Pa), then the Kp units will be in Pa

c)

CaCO3(s) ( CaO(s) + CO2(g) (H( = +177 kJ mol-1

At 700 K

(S(TOT = (S(sys + (S(surr

Using the data book (!) the S( values for CaCO3(calcite), CaO and CO2 are +92.9, +39.7 and +213.6 J K-1 mol-1 respectively.

(S(sys =((S prods) - ((S reacts) ( { (+39.7) + (+213.6) } – (+92.9) = 253.3 - 92.9 = +160.4 J K-1 mol-1

(S(surr = -((H( / T) = - (177,000 / 700 ) = - 252.9 J K-1 mol-1

(S(TOT = (+160.4) + (- 252.9) = -92.3 Note: The reaction is not spontaneous (feasible at this temp)

(S(TOT = R lnK ( K = e^( (S(TOT / R ) ( e^( -92.3 / 8.314 ) ( e^(-11.10) = 1.51 x 10-5 atm

(note units are assumed to be atmospheres)

At 1200 K,

(S(sys and (H( remain effectively constant, therefore:

(S(surr = -((H( / T) = - (177,000 / 1200 ) = - 147.5 J K-1 mol-1

(S(TOT = (+160.4) + (- 147.5) = +12.9 J K-1 mol-1 Note: At this Temp, the reaction is now spontaneous.

(S(TOT = R lnK ( K = e^( (S(TOT / R ) ( e^( 12.9 / 8.314 ) ( e^(1.55) = 4.72 atm

(note units are assumed to be atmospheres)

d)

The Temp at which the reaction is in the verge of becoming feasible is when (S(TOT = 0

Therefore (S(sys + (S(surr = 0 ( +160.4 – (+177,000) / T) = 0 ( –177,000 / T = -160.4

( –177,000 / -160.4 = T So T = 1103.49 K

So at T > 1103.49 K the reaction changes from being disfavourable to favourable

Q5

Atom economy for N2(g) + 3H2(g) ( 2NH3(g) is high because ammonia is removed from the reaction mixture by cooling it to liquefaction. The unreacted remaining gases (N2 and H2) are constantly recycled and passed through the reaction changer for further reaction. The continual process of withdrawal of the product ammonia, means the system never gets to equilibrium. This is advantageous in terms of rate as the rate of product formation as equilibrium approaches is very small and tends towards zero. By preventing the attainment of equilibrium, the rate of NH3 production is sufficiently high to compensate for the relatively low continuous yield.

Q6

a)

The reaction, 2SO2(g) + O2(g) ( SO3(g) has a (H of -196 kJ mol-1

If the gases were passed through the catalyst bed at 900 K, as the reaction is exothermic, the yield at this higher Temp would cause an unacceptably lower value of K resulting a poor conversion conversion factor into SO3(g).

Also, at a higher temperature, the catalyst may no longer be efficient. Catalysts have what is known as an operating temperature.

b)

To remove the SO3(g) product from the equilibrium, which if still present on passing through the catalyst bed a second time, would cause a much lower conversion factor.

Q7

8C(s) + 9H2(g) ( C8H18(l) (H( = -250 kJ mol-1, (S(sys = -860 J K-1 mol-1

The decrease in entropy of the system is large!

The turning point for feasibility comes when (S(sys + (S(surr = 0

( -860 – (-250,000) / T) = 0 ( +250,000 / T = +860

( 250,000 / 860 = T So T = 290.7 K, which is below RT, so from that aspect, the reaction seems very favourable. i.e. at 290.7K, the entropy change in the surroundings is compensates the entropy change of the system. The Activation energy might be high as it uses H2 and C(s) but once the reaction is initiated, the high exothermic value per mole of product is likely to sustain the reaction. The reaction should be cooled to prevent the heat produced from increasing the temp of the reaction as this would lower the value of the equilibrium constant. It should be possible to remove the product quite easily from the reaction mixture to it having a unique phase compared to all the other species involved. The H2(g) could be continuously recycled and passed over the C(s) so a high atom economy could be achieved. Although non-renewable, C in the form of coal, is still quite plentiful but H2 which has to be generated is costly requiring high energy conditions to produce it. The process would be carbon neutral which some people believe is something ‘good’ as it wouldn’t contribute to the greenhouse effect (assuming CO2 actually has a significant effect). So overall the suggestion looks very favourable.

Q8

a)

i)

[Cu(H2O)6 ]2+(aq) + 4 Cl-(aq) ( [CuClH4]2-(aq) + 6H2O(l)

At equilibrium, when conc HCl is added, The concentration term, Q, is no longer at equilibrium.

Q = [ [CuCl4]2- ]

[ [Cu(H2O)6 ]2+] x [ Cl-]4

Note: H2O is a solvent so it is not shown in the concentration term equation.

The [Cl-]4 term will increase substantially, and Q, the value of the quotient, will therefore be less than Kc

The system will act in such a way so that Q = Kc. This means the [ [CuCl4]2- ] will increase. The solution will become more yellow and less green in appearance.

ii) The reaction is exothermic, so heating it up will cause the Kc value to drop to a lower number. The quotient Q = [ [CuCl4]2- ]

[ [Cu(H2O)6 ]2+] x [ Cl-]4

Will be equal to the previous higher Kc value so Q > Kc at the higher temp. Consequently, the [ [CuCl4]2- ] term will decrease and the [ [Cu(H2O)6 ]2+] x [ Cl-]4 terms will increase. The solution will look more blue.

iii) Ag+ ions will remove the Cl- ions from the equation. The denominator will the lowered and the [ [Cu(H2O)6 ]2+] x [ Cl-]4 will increase in value until Q=Kc. The solution will once again become more blue in appearance.

b) The answer: “ The solution will become more yellow and less green in appearance.” Is already given in part i)

Q9

Because it’s power term, i.e. 3, is higher than that of N2 which is simply 1. So doubling the partial pressure of N2 will only have ¼ the effect of altering the position of the eqm than would be the case for doubling the partial pressure of H2(g).

Q10

[CH4(H2O)6](s) ( CH4(g) + 6H2O(s) (H( = exo Kp = ((CH4)1

a)

If P(, Kp will remain the same. As there is only one species in the equation and the mole fraction will always be ‘1’, then when the P( Kp can only remain the same if the partial pressure If the p There will be no change in the yield of CH4(g) produced.

Note: I find myself disagreeing with the “correct” answer here. I cannot mathematically justify why a decrease in pressure would cause a greater yield of CH4(g) because it is the only species appearing in the Kp expression.

Le Châtelier would give you the “correct answer” : At lower P, more gaseous moles would be produced, but like I said. I am unable to justify it. Sorry!

b)

As the reaction is exo, increasing the Temp will make Kp smaller therefore ((CH4)1 will decrease.

c)

Kp = ((CH4)1 = PTOT x (mole fraction of CH4) Kp = 0.86 * 1 = 0.86 atm

NOTE: Given what I said in Q10 part a) I am not so sure if this answer would be “correct” either. Eeek!

Q11

4NH3(g) + 5O2(g) ( 4NO(g) + 6H2O(g) (H = -906 kJ mol-1

Ammonia oxidised by air at 900 oC over a Pt catalyst at 2 atm.

Kp = ( (NO )4 x ( (H2O )6

( (NH3 )4 x ( (O2 )5

a)

The reaction is highly exothermic. Increasing the temperature will cause a decrease in the positive value of (S(surr which in turn causes a decrease in (S(TOT therefore a decrease in K. The new equilibrium will now lie more towards the left hand side, than at a lower Temp.

b)

As P(, the Kp is unaffected but the yield of NO2 (and H2O) is lower because the sum of the coefficients of the products is greater than the sum of the coefficients in the denominator. As P( therefore, the position of the eqm will shift (albeit by only a small degree) to the left.

c) Adding excess air only will increase the pressure of the same number of moles in the reaction chamber. The result therefore is the same as part b)

d) No effect. The catalyst has no effect on the value of Kp or the position of equilibrium. (assuming the volume of the catalyst is negligible)

Q 12

a) Le Châteliers principle gives us the correct result : The yield of NH3 will increase.

Discussion: However the Kp will not change as the temp was constant. If on re-establishment of eqm, the yield of NH3(, then the eqm amount of N2 and H2 must decrease (a greater amount of N2 and H2 must have been consumes to produce the extra NH3), yet Kp must remain the same as the temperature remained constant. So the partial pressures on the denominator must increase to balance out the NH3 increase in partial pressure (remember we now have a greater mole fraction value of the NH3, so the partial pressure of NH3 must have increased as the pressure increased). How does the denominator term increase? Via the pressure term. The mole fractions of N 2 and H2 do indeed decrease but the pressure terms throughout the quotient when cancelled, leave a residual multiplicative term on the denominator which when multiplied by the smaller N2 and H2 mole fractions will lead to the an actual overall increase in the value of the denominator which balances out the increase in the numerator. Hence the Kp remains constant. {Note in a reaction where the sum of the coefficients for the gaseous reactants is the same as the sum of the coefficients for the gaseous reactants, then adding an inert gas like argon at eqm would have no effect on the equilibrium amount of either reactants or products, e.g. H2(g) + I2(g) ( 2HI(g) }

b) At constant P and constant temp, the volume of the system must increase. The mole fractions would change as there is not a greater total of gaseous moles. The mole fractions in the denominator of the quotient, would become smaller (as they are raised to more significant powers). Hence the value of the denominator would fall therefore to maintain the same Kp value, the numerator must also fall. The eqm yield of NH3 would be less.

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