Edexcel Jan 2016 (IAL) Paper 4 Q18 (with explained solutions) |

Skip to content

Edexcel Jan 2016 (IAL) Paper 4 Q18

Answers available below

18 Methanol is synthesised by the following reaction. CO(g) + 2H2(g) (cid:85) CH3OH(g) A mixture of 39.5 mol of carbon monoxide and 77.5 mol of hydrogen was allowed to reach equilibrium at 500 K and 50 atm pressure. Under these conditions, the equilibrium mixture contained 38.5 mol of methanol. (a) Write the expression for the equilibrium constant in terms of pressure, Kp, for this reaction. A E R A S I H T N I E T I R W T O N O D *(b) Complete the table below. Hence calculate the value of Kp under these conditions. Give your answer to three significant figures and include the units. mol at start mol at equilibrium CO 39.5 H2 77.5 CH3OH Total mol 38.5 A E R A S I H T N I E T I R W T O N O D (1) (5) A E R A S I H T N I E T I R W T O N O D *P46940A01728* 17 Turn over<br />
*(c) When the reaction is carried out at 700 K and 50 atm pressure, the value of Kp is smaller. Use this information to deduce the sign of (cid:168)Ssurroundings for the forward reaction. Justify your answer. (2)D O N O T W R I T E I N T H I S A R E A (d) One industrial use of methanol is in the production of bio-diesel from vegetable oils. A component of these oils is shown below. CH2OOCC15H31 CHOOCC15H31 CH2OOCC15H31 Complete the equation for the reaction of methanol with this compound. (2) CH2OOCC15H31 CHOOCC15H31 + 3CH3OH (cid:111) CH2OOCC15H31 (Total for Question 18 = 10 marks) 18 *P46940A01828* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />

Show answer

$Acceptable Answers Reject Kp = p(CH3OH) p(CO) (x) p(H2) ALLOW Lower or upper case p/ pp Expression without brackets p2H2 PCO etc Acceptable Answers Mark Square brackets Expressions without p/ pp/ P/PP to show partial pressure Reject Mark Question Number 18a Question Number 18b CO 39.5 H2 77.5 CH3OH total 1.0 0.5 38.5 40(.0) Mol at start Mol at eqm Mol H2 (1) Mol CO and total mol ALLOW Total mol shown in working below (1) ALLOW TE on mols of hydrogen CO 0.025 H2 0.0125 CH3OH 0.9625 Mole fraction pp 1.25 0.625 48.125 Kp = ((48.125)/ (1.25) x (0.625)2 = 98.56 atm-2 ) = 98.6 atm-2 All three partial pressures (1) Correct value for Kp to 3SF TE on partial pressures and expression for Kp in table (1) Units TE on expression in table (1) Correct answer with no working scores last 3 marks<br /> CH3OH total If candidate incorrectly use the ratio 1 CO to 1 H2 then first mark is lost but the remaining four can be achieved as shown below Mol at start Mol at eqm 38.5 78.5 H2 77.5 1.0 39(.0) CO 39.5 Mol H2 (0) Mol CO and total mol (1) ALLOW TE on mols of hydrogen CO 0.0127 H2 0.497 CH3OH 0.490 Mole fraction pp 0.637 24.8 24.5 Kp = ((24.5)/ (0.637) x (24.8)2 = 0.06259 atm-2 = 0.0626 atm-2 If candidate does not approximate to 3 SF during the calculation allow 0.0623 to 0.0625 All three partial pressures (1) Correct value for Kp to 3 SF TE on partial pressures and expression for Kp in 18(a) (1) Units TE on expression in 18(a) (1) Correct answer with no working scores last 3 marks<br /> Question Number 18c Question Number 18d Acceptable Answers Reject Absence of Ssurroundings = - H/T (Kp is smaller so reaction does not go as far to right) reaction is exothermic/ H is negative (1) Ssurroundings = -H/T so is positive ALLOW If in explaining Ssurroundings is +ve, the expression Ssurroundings =H/T is quoted, then the mark can be awarded (1) IGNORE References to Stotal = RlnK Endothermic reaction scores 0 Acceptable Answers Reject CH2OH CHOH Mark Mark CH2OH (1) + 3 C15H31 COOCH3 /CH3OOC C15H31 / CH3OCOC15H31 (1) ALLOW partially displayed or skeletal formulae (Total for Question 18 = 10 marks)<br />$

Question 16 17 18 19 20 21

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.