OCR A Jan 2013 Paper 2 Q5 (with explained solutions) |

Skip to content

OCR A Jan 2013 Paper 2 Q5

Answers available below

5 Chloroethene, CH2CHCl, is made from 1,2-dichloroethane, CH2ClCH2Cl. 12 CH2ClCH2Cl CH2CHCl + HCl equation 5.1 (a) A chemical plant uses 19.80 tonnes of 1,2-dichloroethane to make 11.25 tonnes of chloroethene. Calculate the percentage yield of chloroethene. (1 tonne = 1.00106 g) percentage yield =% [3] (b) The table below shows some average bond enthalpies. Bond Average bond enthalpy / kJ mol1 CH CC C=C CCl HCl +413 +347 +612 +346 +432 Using the average bond enthalpies in the table above, calculate the enthalpy change of reaction, Hr, for equation 5.1.OCR 2013 Hr =kJ mol1 [3]<br />
(c) Chloroethene is used to make poly(chloroethene), PVC. 13 Write an equation, using displayed formulae, for the reaction to form poly(chloroethene) from its monomer. [3] (d) Waste polymers are often disposed of by combustion. The heat released can be used to generate electricity. (i) Hydrogen chloride, HCl, is a toxic waste product formed by combustion of PVC. Suggest how the HCl can be removed from the gases produced during the combustion of PVC.[1] (ii) Some waste polymers are put into landfill sites. This uses up a valuable resource. State two ways, other than landfill and combustion, of processing waste polymers. 12[2] (iii) Chemists are trying to minimise the environmental damage caused by the disposal of waste polymers by developing new types of polymers. Give an example of a type of polymer being developed to minimise environmental damage.[1] [Total: 13]OCR 2013 Turn over<br />

Show answer

F322/01 Mark Scheme January 2013 Question (a) Answer FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 90% award 3 marks Marks amount of dichloroethane = OR 200000 (mol) OR 2105 (mol) (cid:57) amount of chloroethene = OR 180000 (mol) OR 1.8105 (mol) (cid:57) Calculates percentage yield = = 90 % (cid:57) Guidance IF there is an alternative answer, check to see if there is any ECF credit possible using working below. ALLOW approach based on mass for 2nd and 3rd marks Theoretical mass of chloroethene = 20000062.5 OR 12500000 (g) OR 1.25107 (g) (cid:57) Calculates percentage yield = -------------------------------------------------------------------------------- ALLOW approach based on grams rather than tonnes: n(dichloroethane) = OR 0.2 (mol) (cid:57) = 90 % (cid:57) n(chloroethane) = OR theoretical mass chloroethane = 0.262.5 OR 12.5 g (cid:57) OR 0.18 (mol) = 90 % (cid:57) = 90 % OR % yield = -------------------------------------------------------------------------------- ALLOW ECF throughout from wrong Mr value(s) with final % yield to 2 or more significant figures DO NOT ALLOW final mark for an answer above 100% -------------------------------------------------------------------------------- Note: If this is the only working seen award no marks ie = 56.81% 16<br />
F322/01 Question 5 (b) (c) Mark Scheme January 2013 Answer FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = (+)62 award 3 marks H for bonds broken = 2691 (kJ mol1) (cid:57) H for bond formed = 2629 (kJ mol1) (cid:57) H = (+)62 (kJ mol1) (cid:57) Displayed formulae of monomer and polymer required for the marks. Cl Cl Only chloroethene on left hand side (cid:57) Only the correct polymer on right hand side (cid:57) A correctly balanced equation using displayed formulae for any monomer and matching polymer including the correct use of n (cid:57) Marks Guidance IF there is an alternative answer, check to see if there is any ECF credit possible. IGNORE sign ALLOW 1106 (CCl, CC and CH bonds) IGNORE sign ALLOW 1044 (HCl and C=C bonds) ECF based on bonds brokenbonds formed ALLOW 2 marks for 62 Polymer must have side links (do not have to cut through bracket and can be dotted lines) ALLOW a correct section of the polymer with side links as below would score two marks as the equation is not balanced Cl Cl Cl DO NOT ALLOW ECF from wrong monomer n on LHS can be at any height to the left of formula AND n on the RHS must be a subscript (essentially below the side link) The equation below would be worth 1 mark for balancing 17<br />
F322/01 Question (d) (i) React with an alkali Answer Mark Scheme Marks OR react with a base/carbonate OR Bubble through water (to make HCl(aq)) OR dissolve in water (cid:57) (ii) Sort and recycle (cid:57) Organic feedstock OR cracked (cid:57) January 2013 Guidance ALLOW react with a named alkali or base eg calcium carbonate, calcium hydroxide, magnesium oxide, ammonia ALLOW an appropriate chemical formula IGNORE use of gas scrubbers ALLOW separate and recycle or sorting and remoulding ALLOW use for the production organic compounds OR synthesis gas ALLOW the production of plastics or monomers or new polymers IGNORE a named polymer if degradable DO NOT ALLOW any addition polymer eg PTFE (iii) (Bio) degradable (polymers) OR compostable (polymers) OR soluble (polymers) OR photodegradable (polymers) (cid:57) Total 13 18<br />

Question 1 2 3 4 5 6 7 8

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.