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OCR A Jun 2017 Paper 3 Q3
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![This question is about reactions of hydrogen peroxide, H2O2. (a) Hydrogen peroxide, H2O2, iodide ions, I, and acid, H+, react as shown in the equation below. H2O2(aq) + 2I(aq) + 2H+(aq) I2(aq) + 2H2O(l) A student carries out several experiments at the same temperature, using the initial rates method, to determine the rate constant, k, for this reaction. The results are shown below. Experiment Initial concentrations [H2O2(aq)] / mol dm3 [I(aq)] / mol dm3 [H+(aq)] / mol dm3 Rate / 106 mol dm3 s1 1 2 3 4 0.0100 0.0100 0.0200 0.0200 0.0100 0.0200 0.0100 0.0100 0.100 0.100 0.100 0.200 2.00 4.00 4.00 4.00 (i) Determine the rate equation and calculate the rate constant, k, including units. k =units[3] (ii) The rate constant, k, for this reaction is determined at different temperatures, T. Explain how the student could determine the activation energy, Ea, for the reaction graphically using values of k and T.[3]<br />
(b) Solutions of hydrogen peroxide decompose slowly into water and oxygen: 2H2O2(aq) 2H2O(l) + O2(g) This reaction is catalysed by manganese dioxide, MnO2(s). Standard electrode potentials are shown below. O2(g) + 2H+(aq) + 2e MnO2(s) + 4H+(aq) + 2e H2O2(g) + 2H+(aq) + 2e Eo = +0.70 V H2O2(aq) Mn2+(aq) + 2H2O(l) Eo = +1.51 V Eo = +1.78 V 2H2O(l) Using the electrode potentials, explain how MnO2 is able to act as a catalyst for the decomposition of hydrogen peroxide. You answer should include relevant equations.[4]<br />
(c) Peroxycarboxylic acids are organic compounds with the COOOH functional group. Peroxyethanoic acid, CH3COOOH, is used as a disinfectant. (i) Suggest the structure for CH3COOOH. The COOOH functional group must be clearly displayed. (ii) Peroxyethanoic acid can be prepared by reacting hydrogen peroxide with ethanoic acid. This is a heterogeneous equilibrium. [1] A 250 cm3 equilibrium mixture contains concentrations of 0.500 mol dm3 H2O2(aq) and CH3COOOH(aq) + H2O(l) Kc = 0.37 dm3 mol1 H2O2(aq) + CH3COOH(aq) 0.500 mol dm3 CH3COOH(aq). Calculate the amount, in mol, of peroxyethanoic acid in the equilibrium mixture. amount =mol [3]<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_QP_-_Paper_3_OCR_Chemistry_A_A-Level_3-0.jpg)
![Mark Scheme June 2017 Answer Marks H432/03 Question (a) (i) (rate =) k [H2O2] [I]k = rate [H2O2] [I] units: dm3 mol1 s12.00106 0.01000.0100 = 0.02(00)(a) (ii) Plot graph using ln k AND 1/T(Measure) gradientIndependent mark Ea = ()Rgradient OR ()8.314gradientIndependent mark, even if variables for graph are incorrectSubsumes gradient mark Guidance Square brackets required IGNORE any state symbols IGNORE [H+]0 ALLOW ECF from incorrect rate equation BUT units must fit with rate equation used ALLOW mol1 dm3 s1 OR in any order NOTE Kc expression with calculation and units 0 marks Unless otherwise stated, assume, that ln k is on y axis and 1/T is on x axis IGNORE intercept ALLOW gradient = () Ea R ------------------------------------------------------------------ NOTE: ALLOW Inverse graph (special case) Plot graph of 1/T against ln k(Measure) gradientIndependent mark Ea = () gradient OR () 8.314 gradient OR gradient = () R Ea Subsumes gradient mark 12<br />
H432/03 Question (b) Marks Answer ALLOW equilibrium sign in equations provided reactants on left Reaction of H2O2 with MnO2: H2O2 + MnO2 + 2H+O2 + Mn2+ + 2H2OReaction of H2O2 with Mn2+: H2O2 + Mn2+MnO2 + 2H+Use of E data Use of E data to support equation(s) above or half direction of provided half equations (one including MnO2)Also look for evidence around half equations MnO2 regenerated/reformedMust be linked to an equation showing MnO2 as reactant and an equation showing MnO2 as product (c) (i) ALLOW skeletal OR displayed formula OR mixture of the above as long as non-ambiguous, e.g. 13 Mark Scheme June 2017 Guidance ALLOW correct multiples IGNORE state symbols ----------------------------------------------------------------- ALLOW uncancelled H2O and H+ H2O2 + MnO2 + 4H+O2 + Mn2+ + 2H2O + 2H+ H2O2 + Mn2+ + 2H2O + 2H+MnO2 + 4H+ + 2H2O ExamplesMore negative E moves to left ORAReduction half equation to the right ORAMost positive E is reduced ORACalculated E cell = +0.81 V (from top 2) OR +0.27 V (from bottom 2) ALLOW combining of equations above to show that MnO2 is used and reformed ALLOW OR Structure must include OH as part of COOOH group ALLOW O H+ in structure<br />
H432/03 Question Answer Marks (c) (ii) FIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = 0.023(125) (mol) award 3 marks for calculation ------------------------------------------------------------------------------- Kc expression (Kc =) [CH3COOOH] [H2O2] [CH3COOH] [CH3COOOH] = 0.370.5000.500 = 0.0925 (mol dm3)Subsumes Kc expression n(CH3COOOH) = 0.0925= 0.023(125) (mol)250 1000 Mark Scheme June 2017 Guidance If there is an alternative answer, check for any ECF credit ------------------------------------------------------------------ ALLOW 0.37 = [CH3COOOH] 0.5000.500 ALLOW ECF but ONLY if 0.37 AND 0.50.5 have been used Common errors 0.076 2 marks Use of [CH3COOOH]2 0.675 0.169 0.338 2 marks Use of 0.5 for [H2O] on Kc 2 marks Inverted Kc 1 mark Inverted Kc AND 0.5 for [H2O] 5.78103 2 marks 250 1000 before [CH3COOOH] Total 14 14<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_MS_-_Paper_3_OCR_Chemistry_A_A-Level_3-0.jpg)
