CHEMISTRY 2813/01 How Far, How Fast?

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1 THIS IS A LEGACY SPECIFICATION ADVANCED SUBSIDIARY GCE CHEMISTRY 2813/01 How Far, How Fast? *CUP/T57245* Candidates answer on the question paper A calculator may be used for this paper OCR Supplied Materials: Data Sheet for Chemistry (Inserted) Other Materials Required: Scientific calculator Friday 9 January 2009 Afternoon Duration: 45 minutes * * INSTRUCTIONS TO CANDIDATES Write your name clearly in capital letters, your Centre Number and Candidate Number in the boxes above. Use black ink. Pencil may be used for graphs and diagrams only. Read each question carefully and make sure that you know what you have to do before starting your answer. Answer all the questions. Do not write in the bar codes. Write your answer to each question in the space provided, however additional paper may be used if necessary. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 45. You may use a scientific calculator. FOR EXAMINER S USE A copy of the Data Sheet for Chemistry is provided as an insert with this question paper. You are advised to show all the steps in any calculations. This document consists of 12 pages. Any blank pages are indicated. Qu. 1 Max 5 Mark TOTAL 45 [A/100/3425] SP (CW/CGW) T57245/4 OCR is an exempt Charity Turn over

2 2 Answer all the questions. 1 Students carried out an experiment to investigate the rate of a reaction. The students added dilute hydrochloric acid to marble chips. They collected the gas and measured the volume, at regular intervals, until the reaction was complete. They obtained the graph shown in Fig. 1.1 below. volume of gas/ cm time/s Fig. 1.1 (a) (i) How does the rate of reaction change as the reaction proceeds?... [1] (ii) Explain the changes in the rate of the reaction in terms of collisions.... [2] (b) The experiment was repeated using crushed marble in place of the marble chips. On Fig. 1.1, sketch the graph the students obtained from this experiment. The quantities and the temperature remain the same. [2] [Total: 5]

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4 4 2 Bond enthalpies can be used to calculate enthalpy changes of reactions. Some bond enthalpies are given in Table 2.1. bond bond enthalpy / kj mol 1 Cl Cl +242 H H +436 H F +568 H Cl +432 Table 2.1 (a) (i) Define the term bond enthalpy.... [2] (ii) Why is the sign of bond enthalpy positive?... [1] (b) Hydrogen halides can be made from their elements. Although the reaction between hydrogen and fluorine is explosive, the enthalpy change for the reaction can be found. H 2 (g) + F 2 (g) 2HF(g) H = 542 kj mol 1 Use this value of the enthalpy change for the reaction, and the data in Table 2.1, to calculate the bond enthalpy of the F F bond. bond enthalpy of the F F bond =... kj mol 1 [3]

5 5 (c) Use the data in Table 2.1 to calculate the enthalpy change of formation of hydrogen chloride, HCl. enthalpy change of formation of HCl =... kj mol 1 [2] (d) Hydrogen fluoride and hydrogen chloride dissolve in water to give hydrofluoric acid and hydrochloric acid. Hydrofluoric acid is a weak acid but hydrochloric acid is a strong acid. (i) What is meant by the term weak acid?... [2] (ii) Use the data in Table 2.1 to suggest why hydrofluoric acid is a weaker acid than hydrochloric acid.... [1] (iii) Write the equation for the reaction between hydrofluoric acid and solid sodium carbonate, Na 2 CO [1] (iv) Write the ionic equation for the reaction between hydrofluoric acid and solid sodium carbonate, Na 2 CO [1] (v) Describe how the reaction with Na 2 CO 3 can be used to show that hydrochloric acid is strong and hydrofluoric acid is weak.... [2] [Total: 15] Turn over

6 6 3 Hydrogen is needed in large quantities for the manufacture of ammonia and margarine. Methane and steam can be used to manufacture hydrogen. This is a reversible reaction. CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) H = +210 kj mol 1 equation 3.1 (a) Describe the conditions that would produce a high yield of hydrogen at equilibrium. Explain your answers. temperature... pressure [4] (b) Describe the conditions that would produce hydrogen at a fast rate. Explain your answers. temperature... pressure [4] (c) Use your answers to (a) and (b) to suggest why a temperature of 800 C and a pressure of 30 atm might be used for the manufacture of hydrogen from methane and steam.... [3]

7 (d) The carbon monoxide produced in the reaction can be used as a fuel. 7 Hess s Law can be used to calculate the enthalpy change of formation, H f, of carbon monoxide. (i) State Hess s Law.... [1] (ii) The table below shows some enthalpy changes of formation, H f. substance H f / kj mol 1 CH 4 (g) 75 H 2 O(g) 242 Use these data, and the enthalpy change of equation 3.1, to determine the enthalpy change of formation of carbon monoxide. H f =... kj mol 1 [3] [Total: 15] Turn over

8 8 4 Catalysts are often used to speed up chemical reactions. (a) Explain how a catalyst increases the rate of a reaction. Use an enthalpy profile diagram and a Boltzmann distribution in your answer. Enthalpy profile diagram Boltzmann distribution

9 9... [8] (b) Explain the effect that adding a catalyst has on the position of equilibrium in a reversible reaction.... [2] [Total: 10] END OF QUESTION PAPER

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12 12 PLEASE DO NOT WRITE ON THIS PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

13 2813/01 Mark Scheme January /01 How Far? How Fast?/Experimental Skills 1 Written Paper Question Expected Answers Marks Additional Guidance 1 (a) i reaction slows 1 Not becomes constant ii because there are less particles per unit volume (as the reaction proceeds)/particles further apart/ the concentration decreases 2 marks are stand alone (rate) of collision decreases allow successful collisions (b) sketch to show graph starting more steeply finishing at same level 2 Total 5 19

14 2813/01 Mark Scheme January 2009 Question Expected Answers Marks Additional Guidance 2 (a) i energy/enthalpy change to break 1 mole of a (covalent) bond do not allow first mark: if energy released if break and make if ionic if heat in the gaseous state 2nd mark is stand alone ii energy is put in to break the bond/ energy is needed to overcome the attraction (between electrons and nuclei) in the bond 1 (b) bonds broken = 1(H H) + 1(F F) = (F F) and bonds made = 2(H F) = (F F) 1136 = 542 bond enthalpy = 158 (kj mol -1 ) 3 (c) enthalpy change is ΔH for ½ H 2 + ½ Cl 2 HCl 2 ignore under standard conditions not attraction between oppositely charged ions endothermic alone does not score. ecf possible bonds broken = = 339 and bonds made = 432 ΔH = 93 (kj mol -1 ) 2 (d) i a weak acid is partially dissociated allow partially ionised to form H + ions/ protons 2 can be shown in an equation ii the H F bond is stronger ora 1 accept harder to break the bond. iii Na 2 CO 3 + 2HF 2NaF + CO 2 + H 2 O 1 ignore state symbols; allow H 2 CO 3 iv 2 + CO 3 + 2H. CO2 + H 2 O/ ignore state symbols Na 2 CO 3 + 2H + 2Na + + CO2 + H 2 O 1 do not allow the inclusion of spectator ions v hydrochloric acid bubbles faster / solid disappears faster concentrations (of acids) must be the same 2 allow conditions must be the same Total 15 20

15 2813/01 Mark Scheme January 2009 Question Expected Answers Marks Additional guidance 3 (a) high temperature needed (to send equilibrium to RHS) since (forward) reaction is endothermic low pressure needed (to send equilibrium to RHS) all stand alone marks since fewer moles of gas on LHS 4 (b) high temperature (gives a fast rate of reaction) because a higher proportion of collisions exceed Ea all stand alone marks accept because a higher proportion of collisions have enough energy to cause a reaction do not accept more collisions or more energetic collisions high pressure (gives a fast rate of reaction) because molecules are closer together and collide at a faster rate/ because particles are more concentrated and collide at a faster rate 4 (c) would use high temperature for rate and yield but compromise on pressure because high pressure gives a fast rate but poor yield 3 (d) i the (total) enthalpy change for a reaction is the same whichever route is taken owtte 1 not answer based on E a compromise must be on pressure not temperature both rate and yield needed ii cycle/ ΣΔH (products) - ΣΔH (reactants) = ΔH f ΔH f = (kjmol -1 ) 3 Total 15 cycle need not be drawn correctly/ drawn at all 107 scores 3 21

16 2813/01 Mark Scheme January 2009 Question Expected Answers Marks Additional Guidance 4 (a) enthalpy profile diagram y axis labelled energy/enthalpy and one curve drawn to include either horizontal lines or reactants and products labelled 1 max for enthalpy profile diagram if diagrams are reversed second curve included to start and finish at same energy level and one curve labelled (catalysed or uncatalysed) Boltzmann distribution axes labelled x as energy and y as number/ fraction or % of particles/ molecules/atoms shape of curve catalysed and uncatalysed E a shown as vertical lines touching or crossing the curve and labelled for x axis allow kinetic energy/ KE/ velocity/ enthalpy do not allow activation energy/ E a must start at 0,0 and at higher energies must not touch or cross the x axis if 2 graphs are drawn, this mark does not score explanation to increase the rate of reaction more collisions/particles/molecules have to exceed E a /have enough energy to react a catalyst acts by lowering E a by allowing the reaction to proceed via a different route 8 (b) equilibrium position unchanged if candidate says catalyst gives molecule more energy, this mark does not score rate of forward and reverse reaction increased by same amount 2 Total 10 22

CHEMISTRY. How Far, How Fast? THURSDAY 11 JANUARY 2007 ADVANCED SUBSIDIARY GCE 2813/01. Morning. Time: 45 minutes

CHEMISTRY. How Far, How Fast? THURSDAY 11 JANUARY 2007 ADVANCED SUBSIDIARY GCE 2813/01. Morning. Time: 45 minutes ADVANCED SUBSIDIARY GCE 83/0 CHEMISTRY How Far, How Fast? THURSDAY JANUARY 007 Morning Additional materials: Scientific calculator Data Sheet for Chemistry (Inserted) Time: 45 minutes INSTRUCTIONS TO CANDIDATES