Mastery questions
The student repeats Experiment 1 using 0.020moldm-3 methanoic acid, HCOOH (aq) (pKa =3.75) instead of 0.020moldm-3 HCl (aq) as a source of H+ (aq). Determine the initial rate in this experiment (3 marks)
(Ka =) 10-3.75 OR 1.78x0-4 (mol dm-3) [H+] = (square root) 1.78x10-4 x 0.0200 = 1.89x 10-3 (mol dm-3) initial rate = 6.7x108 x 0.01 x 0.0152 x (1.89 x 10-3)^2 = 5.33x10-3 to 5.38x10-3 (mol dm-3 s-1) OR 5.3x10-3 to 5.4x10-3 (mol dm-3 s-1) Actual value will depend on amount of acceptable rounding
4H2(g) + CS2(g) --> CH4(g) + 2H2S(g) Substance | CS2(g) | CH4(g) | H2S (g) S/ JKmol-1 | 238 | 186 | 206 Calculate the standard entropy for H2. (2 marks)
-164 = (186 + 2 × 206) - (4 × S + 238) OR 4 S = 164 + (186 + 2 × 206) - 238 S = (+)131 (J K-1 mol-1)
The student prepares a buffer solution containing propanoic acid, C2H5COOH, and propanoate ions, C2H5COO-. The concentration of C2H5COOH and C2H5COO- are both 1.00 mol dm-3. The following equilibrium is set up: C2H5COOH (aq) ⇌ C2H5COO- (aq) + H+ (aq) Ka = 1.35x10-5 mol dm-3 The student adds 6.075g Mg to 1.00 dm-3 of this buffer solution. Calculate the pH of the new buffer solution. Give your answer to two decimal places. (4 marks)
ALLOW HA and A- throughout Amount of Mg (1 mark) n(Mg) = 6.075 24.3 = 0.25(0) mol Moles/concentrations (2 marks) n(C2H5COOH) = 1.00 - (2 × 0.25) = 0.50 (mol) n(C2H5COO-) = 1.00 + (2 × 0.25) = 1.50 (mol) [H+] and pH (1 mark) [H+] = 1.35x10-5 / (0.50/1.50) = 4.5x10-6 pH = -log (4.5x10-6) = 5.35
The reverse reaction of the dissociation of water is called neutralisation. Plan an experiment that a student could carry out to measure the enthalpy change of neutralisation. In your answer, you should explain how the enthalpy change of neutralisation could be calculate from experimental results. (6 marks)
Acid and alkali mixed Amount of acid AND alkali stated Temperature taken at start AND finish energy, Q = mc∆T OR in words AND meaning of m, c AND ∆T given Energy scaled up to form 1 mol of water ∆Hneut = -energy change
Explain how this buffer solution controls pH when an acid or an alkali is added (2CH3COOH + CaCO3 --> (CH3COO)2Ca + H2O + CO2) In your answer you should explain how the equilibrium system allows the buffer solution to control pH. (5 marks)
CH3COOH ⇌ H+ + CH3COO- CH3COOH reacts with added alkali OR CH3COOH + OH- OR added alkali reacts with H+ OR H+ + OH- Equilibrium shifts to right OR Equilibrium shifts to CH3COO- CH3COO- reacts with added acid Equilibrium shifts to left OR Equilibrium shifts to CH3COOH
Explain why the 2nd ionisation energy of calcium is more endothermic than the first ionisation energy of calcium (2 marks).
Ca+ is smaller than Ca Greater attraction from nucleus
The 'magic tang' in many sweets is obtained by use of acid buffers. A sweet manufacturer carried out tasting tests with consumers and identified the acid taste that gives the 'magic tang' to a sweet. The manufacturer was convinced that the 'magic tang' would give the company a competitive edge and he asked the company's chemists to identify the chemicals needed to generate the required taste. The chemists' findings would be a key factor in the success of the sweets. The team of chemists identified that a pH of 3.55 was required and they worked to develop a buffer at this pH. The chemists decided to use one of the acids in Table 4.1 (page 8) and a salt of the acid to prepare this buffer. • Deduce the chemicals required by the chemists to prepare this buffer. • Calculate the relative concentrations of the acid and its salt needed by the chemist to make this buffer. • Comment on the validity of the prediction that the pH of the sweet would give the sweets the 'magic tang'. (6 marks)
Chemicals (1 mark) Lactic acid (sodium) lactate Concentrations (4 marks) EITHER [H+(aq)] = 10-3.55 OR 2.8X10-4 OR 2.82X10-4 (mol dm-3) separate marking point Ka = 10-3.86 OR 1.4X10-4 OR 1.38X10-4 (mol dm-3) separate marking point [HA] / [A-] = [H+] / [Ka] OR [A-] / [HA] = [Ka] / [H+] [HA] / [A-] = 2.8X10-4 / 1.4X10-4 OR 2/1 = 2 OR [A-] / [HA] = 0.5 / 1 OR 0.5 Comment (1 mark) Magic tang/taste could come from other chemicals/substances in the sweet OR The buffer would have the dame taste/tang as the magic tang
Lattice enthalpies are exothermic. Explain why it is difficult to predict whether the lattice enthalpy of magnesium bromide would be more or less exothermic than the lattice enthalpy of sodium chloride. (3 marks)
Comparing cation size AND charge (ORA based on Na+) Mg2+ is smaller AND Mg2+ has a greater charge OR Mg2+ has a greater charge density Comparing of anion size (ORA based on Cl- ) Br- is larger OR Br- has a smaller charge density Comparing cation to anion attraction Mg2+ has stronger attraction AND Cl- has stronger attraction IGNORE 'nuclear' attraction
The chemists adds CH4 to 4dm3 container Equilibrium mixture contains 9.36x10-2 mol CH4 and 0.168 mol C2H2 Amount of H2 = 0.504 mol Calculate equilibrium constant, Kc, at this temperature including units. Give your answer to three significant figures. (3 marks)
conc. CH4 = 9.36 X10-2/4 = 0.0234 conc. C2H2 = 0.168/4 = 0.042 conc. H2 = 0.504/4 = 0.126 Kc = [0.042][0.126]3/[0.0234]2 = 0.153 mol2dm6
Explain, with a calculation, the significance of temperatures above 1154℃ for this reaction. (2 marks)
∆G = ∆H - T∆S (∆G =) -234 - 1427 × −164 1000 = 0 (calculator 0.028(kJ) OR 28 (J) The reaction is not feasible because ∆G is positive
The chemist increases both temperature and pressure of the equilibrium mixture. The mixture is left to reach equilibrium again. Explain why it is difficult to predict what would happen to the position of equilibrium after these changes in temperature and pressure 2CO(g) + 2NO(g) ⇌ 2CO2(g) + N2 (g) (2 marks)
Effect of T and P on equilibrium (increased) temperature shifts equilibrium to the left AND (increased) pressure shifts equilibrium to the right AND fewer (gaseous) moles on right hand side Overall effect on equilibrium Difficult to predict relative contributions of two opposing factors.
The chemist prepares 1dm3 of a buffer solution by mixing 200cm3 of 0.200 mol dm-3 HNO2 with 800cm3 of 0.0625 mol dm-3 sodium nitrite, NaNO2. Calculate the pH of the buffer solution. Give your answer to two decimal places. (4 marks)
Expression: Ka x acid/base ratio Use of Ka x [HNO2] / [NO2 - ] OR 4.69x10-4 x [HNO2] / [NO2 - ] Using correct concs/mol in expression [H+ ] = 4.69x10-4 x (0.0400 / 0.0500) Subsumes previous mark Calculation of [H+ ] [H+ ] = 3.752x10-4 (mol dm-3 ) pH to 2 DP (From 3.42573717) pH = -log 3.752x10-4 = 3.43
A biochemist plans to make up a buffer solution with a pH of 5.000 The biochemist adds solid sodium ethanoate, CH3COONa, to 400cm3 of 0.200 mol dm-3 ethanoic acid Calculate the mass of sodium ethanoate that the biochemist needs to dissolve in the ethanoic acid to prepare this buffer solution Assume that the volume of the solution remains constant at 400cm3 on dissolving the sodium ethanoate. (5 marks)
H+] = 10-5 (mol dm-3 [CH3COO- ] = (1.75x10-5 / 10-5) x 0.200 = 0.350 mol dm-3 n(CH3COONa/CH3COO-) in 400 cm3 = 0.350 x \(400 / 1000) = 0.14(0) (mol) mass CH3COONa = 0.140 x 82.0 = 11.48 OR 11.5 (g)
Explain why the first ionisation energy of calcium is endothermic (1 mark)
Ionisation energy refers to removing electrons that are attracted to the nucleus/energy needed to overcome the force of attraction between outer electrons and nucleus.
2SO2 (g) + O2 (g) ⇌ 2S03 (g) - Mixed together 1 mol SO2 and 0.5 mol O2 with a catalyst at room temperature -compressed the gas mixture to a volume of 250cm3 -allowed the mixture to reach equilibrium at a constant temperature and without changing the total gas volume At equilibrium, 82% of SO2 had been converted to SO3 Determine the conc of SO2, O2, and SO3 present at equilibrium and calculate Kc for this reaction. (6 marks)
Kc = [SO3]^2 / [SO2]^2[O2] = Kc = 3.28^2 / 0.720^2 x 0.360 Kc = 57.6 dm3mol-1
The research chemist doubled the pressure of the equilibrium mixture. All other conditions were kept the same. As expected, the yield of ammonia increased. Explain, in terms of Kc, why the equilibrium yield of ammonia increased. (3 marks)
Kc doesn't change Increased pressure increases concentration terms on bottom of Kc expression more than the top OR system is no longer in equilibrium Top of Kc expression increases and bottom decreases until Kc is reached
Ion: Na+ | K+ | Rb+ | Cl- | Br-| I- Radium/pm: 95 | 133 | 148 | 181 | 195 | 216 Predict the order of melting points for NaBr, KI and RbCl from lowest to highest. Explain your answer. (3 marks)
Lowest melting point: KI RbCl Highest melting point: NaBr Mark 2nd and 3rd marking points independently Attraction and ionic size linked: Greater attraction from smaller ions/closer ions/larger charge density Comparison needed Energy AND attraction/breaking bonds linked: More energy/heat to overcome attraction (between ions) OR More energy/heat to break (ionic) bonds
The student adds 50.0 cm3 of 0.250 mol dm-3 butanoic acid to 50.0 cm3 of 0.0500 mol dm-3 sodium hydroxide. A buffer solution forms. Calculate the pH of the buffer solution. The Ka of butanoic acid is 1.51 × 10-5 mol dm-3. Give your answer to two decimal places. (5 marks)
Moles (2 marks) amount CH3(CH2)2COOH = 0.0100 (mol) amount CH3(CH2)2COO- = 0.0025 (mol) Concentration (1 mark) [CH3(CH2)2COOH] = 0.100 mol dm-3 AND [CH3(CH2)2COO-] = 0.025 mol dm-3 [H+] and pH (2 marks) [H+] = 1.51x10-5 x (0.100 / 0.025) = 6.04x10-5 (mol dm-3) pH = -log 6.04x10-5 = 4.22 pH to 2 DP
Suggest why the second electron affinity of oxygen is positive (2 marks).
Oxide ion, O- and electron are both negative Hence energy is required to overcome repulsion
The student measures out 35.0 cm3 of 2.40 mol dm-3 NaOH and 35.0 cm3 of 2.40 mol dm-3 HCl. The temperature of each solution is the same. The student mixes the two solutions. The temperature rises by 16.5 °C. The specific heat capacity of the mixture is 4.18 J g-1 K-1. Assume that the density of the mixture is 1.00 g cm-3. Calculate the enthalpy change of neutralisation, in kJ mol-1. (3 marks)
Q = mc∆T energy change = 70.0 x 4.18 x 16.5 = 4827.9 (J) OR 4.8279 (kJ) amount of H2O formed = 2.4(0) x (35.0 / 1000) = 0.084(0) mol ΔHneut = - 4.8279 / 0.084(0) = -57.475 OR -57.48 OR -57.5 (kJ mol-1)
A reaction is not feasible at low temperatures but is feasible at high temperatures. Deduce the signs of ∆H and ∆S for the reaction and explain why the feasibility changes with temperature. (3 marks)
Signs of ∆H and ∆S ∆H is positive AND ∆S is positive T∆S and temperature 'Value of' T∆S increases with temperature Feasibility At high temperatures, ∆G is -ve OR ∆G < 0 AND At low temperatures, ∆G is +ve OR ∆G > 0 OR ∆H - T∆S decreases with (increasing) temperature OR ∆H - T∆S from +ve to -ve with (increasing) temperature OR the idea: As temperature increases, T∆S outweighs ∆H to make ∆G <
Composing a mechanism Overall equation: H2O2 + 2I- + 2H+ --> I2 + 2H2O Rate equation: rate = k[H2O2][I-] (3 marks)
Step 1: H2O2 + I- --> H2O + IO- Step 2: IO- + H+ --> HIO Step 3: HIO + I- --> I2 + OH- Step 4: OH- + H+ --> H2O
The rate equation is rate= k[H2O(aq)][I-(aq)] -Show that the student's results support this rate equation -Calculate the rate constant, k, for this reaction (6 marks)
When comparing experiments 1 and 2, the concentration of H2O2 x2 and the concentrations of I- and H+ stays constant. The initial rate x2 therefore H2O2 is a 1st order reactant When comparing experiments 2 and 3, the conc of H2O2 AND I- stays constant and the conc of H+ x2. The initial rate stays constant therefore H+ is a 0 order reactant When comparing experiments 2 and 4, the concentration of H2O2 x2, the concentration of I x2 and the concentration of H+ stays constant. H2O2 is a 1st order reactant which causes the initial rate to x2, going from 1.14x10-5 to 2.28x10-5. The concentration of I- x2 and the initial rate x2 therefore I- is a 1st order reactant. k= rate/[H2O2][I] k= 5.70x10-6/0.0010x0.020 k= 0.0285 k=2.85x10-2 dm-3mol-1s-1
Calculate the pH of a 0.1250 mol dm-3 solution of Ba(OH)2. Give your answer to two decimal places. (3 marks)
[OH- ] = 2 × 0.1250 = 0.25(0) (mol dm-3 ) [H+ ] = 1.00x10 -14 0.25(0) OR 4(.00) × 10-14 (mol dm-3 ) Subsumes 1st mark pH = -log 4.00 × 10-14 = 13.40
