2 How far? Equilibrium Answers

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1 How far? Equilibrium Answers ratie: pages Answer is D. Only a hange in temperature harges the value of the equilibrium onstant. Answer is D. [B] /[A] so [B] [A] or [B] [A] 1/ 3 Answer is B. Amounts at equilibrium are CO 0.6 mol, H O 0.6 mol, H 0.4 mol, CO 0.4 mol. 4 Answer is D. There are 7 mol on the right-hand side of the equation and mol on the left-hand side. So units are ka 7 /ka. 5 Answer is B. p is 600 ka. So 600 (0.5) /(0.5). Therefore, and 100 ka. 6 Answer is C. At equilibrium, [CO] 0.01/ mol dm 3, [CH 4 ] [H O] 0.008/ mol dm 3 and, from the equation whih shows that for every mol of CH 4, 3 mol of H reats, [H ] 0.001/ In pratie, this an be done without working out the detailed arithmeti. 7 Answer is A. The differene in amounts in mol between produts and reatants is 3. 8 Answer is A. All three statements are orret. 9 Answer is A. As the reation is endothermi an inrease in temperature gives more produt and the value of the equilibrium onstant will inrease. The volume inreases as the reation takes plae beause eah N O 4 (g) gives NO (g). An inrease in temperature also inreases the rate of the reation. 10 Answer is B. If the amount of produt dereases as the temperature rises the reation must be exothermi. SO 3 is breaking into SO (g) + O (g) so the total amount of gas in mol inreases. No dedution is possible about the strength of the sulfur oxygen bonds. 11 a) For eah of 1a, b and award 1 mark for the expression with the orret units: b) NO NO 4 N g O g NO ) Remember C(s) is not inluded in the equilibrium onstant. d) CO g O CO CO g O g Hodder & Stoughton Limited 015

2 How far? Equilibrium Answers 1 There is 80 mg (0.08 g) of ethanol in 100 m 3 of blood. So, onentration of ethanol 0.8 g dm 3 In mol dm 3 this is 0.8/ mol dm 3 CH5OH g CH5OH blood [C H 5 OH(g)]/0.017 Conentration of ethanol in the vapour in equilibrium with the blood mol dm 3 or g dm 3 13 a) This statement is partly true but hemial equilibrium is dynami so that there is onstant hange with opposing reations anelling eah other out. b) does not vary with onentration. However, some of the added hemial reats and more produts form until the system is again at equilibrium. ) Adding a atalyst does not affet the value of beause (at a onstant temperature) the yield of produt at equilibrium is not hanged by adding a atalyst. However, adding a atalyst may inrease the yield by enabling the reation mixture to reah equilibrium faster. d) This is true if the reation to make produts is endothermi. If the reation is exothermi, then the equilibrium yield of produt falls as the temperature rises. However, as with atalysts, raising the temperature an inrease the yield of a produt by allowing a reation mixture to reah equilibrium that would otherwise reat too slowly. 14 a) Before equilibrium is established the amounts (in moles) are: N O(g) N (g) + O (g) 1.00 mol 0 mol 0 mol One equilibrium is established, there is mol of N O present and, therefore, mol of the N O has been onverted to N and O. The equation indiates that 1.00 mol of N O reats to produe 1.00 mol of N and mol of O. So, mol of N O will produe mol of N and mol of O. Sine the volume of the ontainer is 1 dm 3, the onentration of N is mol dm 3 and the onentration of O is mol dm 3. b) To summarise, at equilibrium: N O(g) N (g) + O (g) mol mol mol mol dm NO ( 0.100) N g O g [ ] Hodder & Stoughton Limited 015

3 How far? Equilibrium Answers 15 a) ) The value of will not hange. It is unneessary to redo the alulation, hanging the onentrations to half their previous values to obtain a new value for. However, hanging the volume of the ontainer hanges the amount of moles of eah omponent in suh a way that the value of stays the same. Remember, only a hange in temperature auses a hange in the value of. Cl g Cl If the value of inreases when the temperature is raised, this must be beause more Cl atoms are produed. Using Le Chatelier s priniple, this means that the forward reation of the equilibrium Cl (g) Cl(g) is endothermi. (This is preditable beause heat is neessary to break the Cl Cl bond in Cl.) b) i) The reason is that the pressure in the stratosphere is very muh lower and, by Le Chatelier s priniple, a lower pressure will ause the equilibrium to move to the side of higher volume (greater number of moles) or the hlorine bond is broken readily by the UV. (The tempting answer is to suggest that it is beause of an inreased temperature. However, the temperature in the stratosphere is onsiderably lower than that at the Earth s surfae.) ii) Chlorine radials attak the ozone layer by the mehanism: Cl + O 3 ClO + O ClO + O Cl + O 16 NO H N H O Initial amount (mol) a b d Equilibrium amount (mol) (a x) (b x) ( + x) (d + x) Initial amount Equilibrium amount 0.14 The initial amount of NO 0.00 mol and the equilibrium amount of NO 0.14 mol therefore, x mol. The onentration of the NO in the dm 3 ontainer at equilibrium is 0.14/ 0.06 The equilibrium amount of eah reatant and produt an now be alulated using the equation: NO(g) + H (g) N (g) + H O(g) 1 mol of NO reats with 1 mol of H to form 1 mol of N (g) and mol of H O. Hodder & Stoughton Limited 015

4 How far? Equilibrium Answers In making the 0.14 mol of NO, the H will have lost mol of its original amount so, the amount (in mol) of H at equilibrium is mol and its onentration is 0.01 mol dm 3. The amount (in mol) of N formed is ½ mol and its onentration is mol dm 3. The amount of H O obtained is mol and its onentration is mol dm 3. N g HO g NO H ( 0.019)( 1.038) ( 0.06) ( 0.01) dm mol + (units) 3 (0.07 mol dm ) 17 a) (0.01 mol dm 3) (0.01 mol dm 3) 49 no units b) The quantity of iodine that has reated mol dm 3 This reats with mol dm 3 hydrogen and produes mol dm 3 HI. So the new onentrations are: [H (g)] mol dm 3 and [HI(g)] mol dm 3 18 For eah of 18 a, b and award 1 mark for the expression with the orret units. a) p NO NO 4 units are ka b) N g O g p NOg units are ka ) Remember C(s) is not inluded in the equilibrium onstant. CO p O no units CO d) p ( CO ) O units are ka 1 Hodder & Stoughton Limited 015

5 How far? Equilibrium Answers 19 p 3 O (g) 3 (Neither ClO 3 (s) or Cl(s) are inluded in the expression for p as they are solids.) Therefore, the partial pressure of O 3 is the same as the total pressure 36 ka And p (36) ka 3 for answer, for units 0 4 NO O p NO NO 5 N O / and N O ka 1 Remember C(s) is not inluded in the equilibrium onstant. CO CO The equation shows that if 1.4 mol of CO is formed then 0.7 mol CO must have reated. Therefore the amount of CO remaining at equilibrium is mol Total amount in mol at equilibrium is mol partial pressure mol fration total pressure CO(g) (1.4/1.7) ka CO (g) (0.3/1.7) ka (0.59) / ka SO3 p SO O g The equation shows that if 0.98 mol of SO 3 is formed then 0.98 mol of SO must have reated and therefore ( ) 0.0 mol remains mol of oxygen must have reated and ( ) 0.01 mol remains. Total amount in mol at equilibrium is mol partial pressure mol fration total pressure SO 3 (g) (0.98/1.01) 0.97 atm SO (g) (0.0/1.01) atm O (g) (0.01/1.01) atm Therefore, ( 0.97) ( ) ( ) 4 p /( )(0.0099) 0.94/( )(0.0099)( ) 8.1 (8.08) atm Hodder & Stoughton Limited 015

6 How far? Equilibrium Answers 3 4 (g) (g) Therefore when 1 mol of 4 is reated, mol of is formed. At equilibrium, 0.9 mol of 4 remains, so 0.1 mol has reated. Therefore 0. mol of must be present at equilibrium. The total amount in mol present at equilibrium must be mol partial pressure mol fration total pressure 4 (0.9/1.1) (0./1.1) And p (7.3) / ka + units Challenge 4 a) mass (in g) volume (in m 3 ) density (in g m 3 ) amount in mol mass/moleular mass the amounts in mol are therefore: C H 5 OH ( )/ (3) mol CH 3 COOH ( )/ mol CH 3 COOC H 5 ( )/ (4) mol H O ( )/ (6) mol b) The amount, in moles, of sodium hydroxide in the m 3 is ( )/ mol This will have neutralised the sulphuri aid atalyst and also the ethanoi aid. The equation for the neutralisation of the sulphuri aid is: NaOH(aq) + H SO 4 (aq) Na SO 4 (aq) + H O(l) 10.0 m 3 of 0.5 mol dm 3 ontains mol and this will be ontained in the 80 m 3 of the equilibrium mixture. However, only 1 m 3 was titrated so the amount in mol ontained in this will be 0.005/ mol. Hene this will reat with mol NaOH The equation for the neutralisation of the ethanoi aid is: CH 3 COOH(aq) + NaOH(aq) CH 3 COO Na + (aq) + H O(l) Therefore 1 mol of NaOH neutralises 1 mol of CH 3 COOH. So the amount, in moles, of ethanoi aid in the 1 m 3 is mol ) The amount, in moles, in 80 m 3 is mol d) Using the equation: C H 5 OH(l) + CH 3 COOH(l) CH 3 COOC H 5 (l) + H O(l) The amount of ethanoi aid used up in reating the equilibrium is mol Hodder & Stoughton Limited 015

7 How far? Equilibrium Answers e) Ethanol present in the equilibrium mixture is mol Ethyl ethanoate in the equilibrium mixture is mol Water present in the equilibrium mixture mol C H OH( l) CH3COOCH5 l HO l CH COOH l Therefore f) The total number of moleules on either side of the equation is the same and, therefore, the volume would anel out in the equilibrium expression. g) H ( ) ( ) 9.5 kj mol h) The differene in the enthalpy between the reatants and produts is quite small so that neither side of the equation is likely to dominate within the mixture. (Note, however, that this does not mean the equilibrium will neessarily form quikly as this depends on the ativation energy.) i) The forward reation to form ethyl ethanoate and water is exothermi and a derease in temperature would favour their formation. Therefore at 0 C the equilibrium onstant would be higher. So, the differene between the value obtained and the value quoted in data books is not explained by the experiment being arried out at 0 C. The differene is far more likely to result from either a failure of the mixture to reah true equilibrium or experimental error. Hodder & Stoughton Limited 015