(a)(i) Mass of ethanoic acid = 0.04 x 60. = (.404 g) Volume of ethanoic acid =.404.049 =.97 =.3 (cm 3 ) Correct answer with no working () Ignore SF except only one 60.0 for molar mass which gives mass.4 and volume.88 =.3 cm 3 () First step.049 60/60. to find number of moles in cm 3 = 0.07 Then volume = 0.04 0.07 =.359 (cm 3 ) But note, if whole calculation done on calculator, 60 gives.879 and 6 gives.97. If units given, they must be correct, but penalise wrong units only once here. (a)(ii) Syringe Burette Gas syringe Biuret Graduated/adjustable pipette Just pipette
(a)(iii) To prevent evaporation/vapour escaping water vapour entering To maintain a closed system To maintain a closed environment To prevent: air oxidizing the alcohol reaction with air Due to volatility (of chemicals) IGNE gas escaping HCl escaping
(a)(iv) First and second mark Phenolphthalein Litmus/universal indicator 3 From colourless to (pale) pink/red Other indicators with pk in in range 7.5 0 Some examples are: Thymol blue ((base)) (yellow to blue) Phenol red (yellow to red) Thymolphthalein (colourless to blue) Pink to colourless Thymol blue (acid) Phenyl red Methyl red Second mark depends on correct indicator except bromothymol blue, which is incorrect but very close to range so allow colour yellow to blue. Third mark Sodium ethanoate is (slightly) alkaline Ethanoic acid is a weak acid Phenolphthalein ph range coincides with vertical section of the ph/titration curve Titration of weak acid with strong base Neutralisation/equivalence point is at 8-0/ any number between 8 and 0. pk in +/- lies within vertical region Third mark is independent
(b)(i) CH 3 COOH+CH 3 CH OH CH 3 COOCH CH 3 +H O Single arrow -CO H -C H 5 Displayed formulae IGNE state symbols even if incorrect (b)(ii) Volume of alkali reacting with ethanoic acid = 77.-.7 = 65.4 cm 3 Moles of ethanoic acid = 65.4 x 0.00 000 = 0.0308/.308x0 - (mol) Correct answer no working () Ignore SF except Allow internal TE for use of Moles of ethanoic acid = 77. x 0.00 000 = 0.054/.54x0 - (mol) max (b)(iii) of moles of ethanol = 0.0308/.308x0 (mol) TE same as (ii)
(b)(iv) of moles of ethyl ethanoate =0.0400 0.0308 = 0.069 (mol) Allow TE from (ii)/(iii) for example 0.054 gives 0.0458 (b)(v) K c = [CH 3 CO CH CH 3 ][H O] [CH 3 CO H][ CH 3 CH OH] = 0.069 x 0.069 0.0308 x 0.0308 = 4.3579 = 4.4 Ignore SF except one Allow TE from (ii), (iii) and (iv) for example 0.054 etc gives.54 No TE for incorrect expression of K c (b)(vi) The units cancel There are the same numbers of moles of reactants and products (b)(vii) (Concentrated) hydrochloric acid contains water
(c)(i) First test tube esterification addition/elimination Condensation Second test tube (acid) hydrolysis Two fully correct answers in wrong order ma Alkaline hydrolysis followed by acidification (c)(ii) The values are the same within experimental error Just...the same The values are concordant The values are similar The equilibrium can be approached from either direction The reaction is reversible Any comment relating equilibrium to reversibility IGNE Dynamic equilibrium Rate of reverse reaction = rate of forward reaction
(c)(iii) (Acid) catalyst (makes it faster) Provides H + (as a catalyst) Initiates Reacts with... Protonates... Protates Protonating agent... Donates protons Increases H + concentration
*(a) (A green solution) forms a yellow / orange / brown (solution) reddish-brown A grey / black precipitate silver ppt solid / crystals for precipitate Red Green(ish) with any other colour Silver mirror silver compound (b)(i) 0.05(00) (mol dm -3 ) (b)(ii) Amount of silver ion in 0 cm 3 = amount of thiocyanate = 5.6 x 0.000 = 0.000/. x 0-4 (mol) 000 So concentration of silver ion = 0.000 x 000 = 0.0/. x 0 - (mol dm -3 ) 0 (b)(iii) 0.0/. x 0 - (mol dm -3 ) Accept TE = answer to (ii) (b)(iv) 0.0500 0.0 = 0.0388/3.88 x0 - (mol dm -3 ) Accept TE = 0.05 - answer to (iii) Accept answer to (i) answer to (iii)
(b)(v) K c = [Fe 3+ (aq)] [Fe + (aq)] [Ag + (aq)] [Ag] in numerator 4 K c = [Fe 3+ ] [Fe + ] [Ag + ] = = 0.0388 0.0 309.3 = 309 dm 3 mol - Value Unit (any order) Three SF Accept TE from (iii) and (iv): ( use of 0. from (i) gives 708 dm 3 mol - ) If [Ag] is included in the numerator and taken as =[Fe 3+ (aq)], then allow unit and SF marks ONLY, but must either state no units or show working (c)(i) ΔS o total = 8.3 x ln 309 = + 47.6(4) / +47.6(5)J mol - K - = 8.3 x ln 309.3 = +47.6(5) J mol - K - Accept TE : 8.3 x ln(answer from b(v)) Value Sign and Unit (any order) IGNE sf except
(c)(ii) First Mark: One of the products is a solid Two moles going to two moles but one of them is a solid Two moles of solution react to form one mole of solution / liquid and one mole of solid Second Mark (Hence) RHS more ordered / LHS less ordered (c)(iii) ΔS o surroundings = ΔS o total - ΔS o system = +47.6 (-08.3) = (+)55.9 (J mol - K - ) Accept TE on c(i) IGNE sf except (c)(iv) Because ΔS o surroundings = - ΔH o ΔS o total = - ΔH o 3 T T ΔH= -98 x 55.9 = -7658 (J mol - ) = -76.58 (kj mol - ) Units if given must be correct Correct answer with or without working scores marks IGNE SF except As T increases ΔS o surroundings becomes less positive / decreases therefore ΔS total becomes less positive / decreases more negative for less positive
*(d) No change in the titre No significant change Stand alone mark (though silver solid was removed the equilibrium constant remains the same so) the equilibrium concentration(s) would remain the same Second mark dependent on first IGNE references to temperature
3 (a)(i) (K p =) pch 3 CO H pch 3 OH (x) pco Partial pressure symbol can be shown in various ways, eg pp, p CO, (CO)p, etc p in upper or lower case, round brackets IGNE units [ ] State symbols given as (l) + in botto line 3 (a)(ii) P CH 3 OH = 4.9 (atm) P CO = 4.9 (atm) mark for recognition that pressures are equal IGNE units 3 (a)(iii) K p = ((.)/(4.9) ) = 0.95 atm stand alone mark but must match expression used in (a)(iii) 9.5 x 0 4 Pa / 9.5 kpa () TE from (a)(i) if inverted and/or (a)(ii) Answers to other than 3 significant figures
3 (b)(i) CH 3 OH: 3. CO : 3. for both values CH 3 CO H: 46.8 TE for moles of ethanoic acid based on numbers of methanol and carbon monoxide used, as long as moles of methanol and carbon monoxide are equal and moles ethanoic acid + moles methanol = 50 3 (b)(ii) 46.8 x 3 = 8. / 8.504 (atm) 53. IGNE sf except Value = 8.6 if mol fraction rounded TE from (b)(i) 8. 46.8 x 3 = 50 9.95 (atm) 3 (b)(iii) exothermic as yield / pp of ethanoic acid / conversion of reactants/ K p is higher at lower temperature / as equilibrium moves (right) at lower temperature if partial pressure of ethanoic acid <. atm in (b)(ii), endothermic as yield / pp of ethanoic acid / conversion of reactants/ K p is lower at lower temperature
3 (c)(i) No effect and other concentrations change to keep K p constant / K p is only affected by temperature/ as equilibrium moves (right) to keep K p constant / change in pressure does not change K p As K p is a constant 3 (c)(ii) Yield increased to restore fraction / quotient / partial pressure ratio back to K p (equilibrium moves) to use up the methanol /answers based on entropy or Le Chatelier Correct prediction in (c)(i) and (c)(ii) with inadequate explanations scores mark in (c)(ii) Just equilibrium moves to the right 3 (d) Mark independently Reaction can occur at lower temperature / has lower activation energy / requires less energy less fuel needed / fewer emissions (from fuels) / fewer raw materials needed / less natural resources used Answer based on car exhaust emissions Enables use of an alternative process with higher atom economy fewer raw materials needed / less natural resources used