chem 2019 Flashcards by Coral Dunk

An excess of aqueous potassium bromide was added to chlorine water and the
solution turned orange.
(i) Write an equation for this reaction. State symbols are not required.

Cl2+2KBr→Br2 +2KCl

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Silver nitrate solution was added to the mixture in (a) and excess dilute ammonia solution was then added to the precipitate formed. Only some of the precipitate dissolved.
(1)
Deduce why only some of the precipitate dissolved.

(M1)the precipitate is a mixture of silver chloride and bromide
(M2)silver chloride/AgCl dissolves in dilute ammonia
(M3)silver bromide/AgBr does not dissolve in dilute ammonia

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Aqueous potassium bromide was added to aqueous iodine, instead of chlorine water. There was no reaction.
Give a reason why no reaction occurred.

iodine is a weaker oxidising agent than chlorine

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Chlorine undergoes disproportionation when it reacts with hot aqueous sodium hydroxide solution.
(i) Complete the ionic equation for this reaction. State symbols are not required.

3Cl2 + 6OH− →5Cl− + ClO3− + 3H2O

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Explain, in terms of oxidation numbers, why this is a disproportionation reaction.

Oxidation number for chlorine changes from 0 to -1 so it is reduced
Oxidation number for chlorine changes from 0 to +5 so it is oxidised

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50.0 cm3 of propanoic acid solution, of concentration 1.00 mol dm–3, was added and thoroughly mixed with the sodium hydroxide solution in the polystyrene cup.
The maximum temperature rise was 6.5 °C.
Calculate the enthalpy change of neutralisation for propanoic acid, in kJ mol–1,
giving your answer to the nearest whole number.
[Assume density of the mixture = 1.00 g cm–3, specific heat capacity of the
mixture = 4.18 J g–1 °C–1]

Q=(100 x 4.18 x 6.5=) = 2717 (J) / 2.717 kJ
(1)
ΔH= 2.717 ÷ 0.05= (−)54.340
= −54 (kJ mol−1)

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Explain why the data book value for the standard enthalpy change of neutralisation of ethanoic acid with sodium hydroxide is –55.2 kJ mol–1 but the –1
value for hydrochloric acid is –57.1 kJ mol

ethanoic acid is a weak(er) acid / only partially ionised/dissociated
(some) energy is used to fully/completely ionise the ethanoic acid

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The glowing splint is used as a test for one of the gases given off in this experiment. Identify this gas and the positive result of the test.

(Identity of gas is) oxygen/O2

(test result is that the splint) relights

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Give the name and appearance of the other gas given off in this experiment when a Group 2 nitrate is heated.

(Identity of gas is) nitrogen dioxide

appearance is) brown (gas/fumes

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Write the equation for the decomposition if the Group 1 compound, sodium nitrate, was used in this experiment.
State symbols are not required.

2NaNO3 → 2NaNO2 + O2

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Describe the apparatus that would be used to compare the decomposition of metal carbonates. Include how the rate of decomposition would be compared.

use of a delivery tube to bubble gas into limewater

compare the time taken for the limewater to go cloudy

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Explain why magnesium carbonate decomposes much more readily on heating than barium carbonate.

The magnesium ion/cation is smaller than the barium ion/cation)
which polarises the (large) carbonate (ion)/anion
and weakens the carbon-oxygen bond (C-O/C=O) bonds

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Give a reason why the reaction between iodide ions and peroxodisulfate ions has a high activation energy and is therefore very slow without a catalyst.

the two negative ions repel each other

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Explain, with the aid of two equations, how the iron(II) ions catalyse this reaction. State symbols are not required.

2Fe2+ + S2O82−→ 2Fe3+ + 2SO42−
2Fe3+ + 2I− → 2Fe2+ + I2
(catalysis is possible because) variable oxidation state/iron has more than one oxidation state/number

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Deduce two ionic equations to show how cobalt(II) ions catalyse the reaction in acidic solution. State symbols are not required.

2Co2+ + 1⁄2O2 + 2H+ → 2Co3+ + H2O

2Co3+ + SO32− + H2O → 2Co2+ + SO42− + 2H+

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Give a possible reason why tungsten would also not be a suitable replacement for platinum in a catalytic converter. Refer to the mechanism of heterogenous catalysis in your answer.

(because) adsorption is too strong and so desorption would be too slow

Water might be expected to have a lower boiling temperature than hydrogen sulfide
but it actually has a higher boiling temperature.
Comment on this statement by referring to the intermolecular forces in both these substances.

(M1) (a lower boiling temperature is expected) because water has fewer electrons than hydrogen sulfide
(M2) water has weaker/less London forces
(M3) (a higher boiling temperature occurs because) water has hydrogen bonding
(M4) hydrogen bonding is stronger than London forces and requires more energy to break
(and results in a higher boiling temperature)

Explain why both water and carbon dioxide molecules have polar bonds but only water is a polar molecule.

(M1) oxygen is more electronegative than hydrogen and carbon
(M2) which results in a polar bond with oxygen δ− so carbon and hydrogen δ+
(M3) carbon dioxide is a symmetrical/linear molecule and so the dipole moments/vectors cancel
(M4) the lone pairs of electrons of oxygen/ the V-shape of the water molecule mean that the dipole moments/vectors do not cancel

The value of Kw at 310 K is 2.40 × 10–14 mol2 dm–6 (i) Calculate the pH of water at 310 K.
Give your answer to two decimal places.

[H+] = (√2.40 x 10−14) = 1.549..x10−7 (mol dm−3)
pH = (−log 1.549..x10−7)
= (6.809894379) = 6.81

Predict, with a reason, whether water is acidic, alkaline or neutral at 310K

neutral

because [H+(aq)] = [OH−(aq)] /equal amounts of H+ and OH− ions

Predict, with a reason, the sign of the enthalpy change for the ionisation of water.

positive / + sign because Kw increases as the temperature increases

colour is often used in chemistry to identify substances.
Compare and contrast the origin of the colour of a copper(II) complex with the origin
of the colour of the copper(II) ion in a flame test. You do not need to state any specific colours.

Similarities
 (IP1) the differences in energy levels determines the colour of the flame test and complex ion
Differences Flame test
 (IP2) heat (energy) results in electron promotion
 (IP3) return of an (excited) electron to a lower (energy) state
Complex ion
(IP4) d orbitals are split (in energy by the ligands)
(IP5) light (energy) is needed for electron promotion
(IP6) the colour not absorbed is the colour seen

Identify and correct the two errors in this Born-Haber cycle.

Error 1 – arrow for enthalpy change of formation should go down/be reversed
Error 2 – the word ‘half ’ should be deleted from the enthalpy change of atomisation of hydrogen

Calculate the first electron affinity, in kJmol–1, of hydrogen, using the values given in the cycle.

1st EA= −(218+496+107)−56 +804

= −73 (kJ mol−1)

Deduce the feasibility of this reaction at 298 K by calculating the free energy change, G.

ΔG = −56 –(298 x −76.5/1000) =−33.203 (kJ mol−1) ΔG is negative/ <0 and so reaction is feasible

Calculate the temperature at which G = 0.

``` ΔG = 0, so ΔH = TΔS(system) T = 56/0.0765 = 732 K ```

Deduce the reduction half-equation for the alkaline fuel cell.

1⁄2O2 + H2O + 2e−→ 2OH−

A buffer solution always

resists changes in pH if small quantities of acid or base are added

A buffer solution with a pH of 3.90 is required. Calculate the mass, in grams, of sodium ethanoate that should be added to 50.0 cm3 of an ethanoic acid solution of concentration 0.800 mol dm–3 to form this buffer solution.

[H+] = 10−3.9= 1.2589..x 10−4 (mol dm−3) [CH3COO−] = (1.74 x 10−5 x 0.800) = 1.2589.. x 10−4 =0.11057.. (mol dm−3) n=(0.11057.. x 0.05 =) 5.5285..x 10−3 (mol) m=(5.5285..x 10−3 x 82=) 0.453339 m=0.45/ 0.453 (g)

Explain how this buffer system helps to control the pH of blood when extra carbon dioxide is present due to strenuous exercise.

Carbon dioxide dissolved in the blood forms carbonic acid (and so this concentration increases) The equilibrium will shift to the right and produces more H+/acid ions The (high) concentration of hydrogencarbonate ions suppress the ionisation of carbonic acid (to help to control pH)

Draw a titration curve

curve starts at pH of 3.0 and ends at pH of 12-13 vertical part of the curve at 40 cm3 Allow buffering area to be labelled anywhere between ~5 and 35 cm3 inclusive

Describe, without calculation, how you would use your curve to determine the value of Ka for propanoic acid.

determine the pH at the point when half of the acid is neutralised Ka = 10−pH

Calculate the value of Kp at this temperature.

XSO2 = 0.0160 ÷ 0.8 = 0.02 XO2 = 0.0120 ÷ 0.8 =0.015 XSO3 = 0.772 ÷ 0.8 =0.965 ``` PSO2 = 0.02(0) x 2.40 = 0.048 PO2 = 0.015 x 2.40 = 0.036 PSO3 = 0.965 x 2.40 =2.316 ``` atm−1

Calculate the number of sulfur dioxide molecules present in this equilibrium mixture.

N=(n x L = 0.0160 x 6.02 x 1023) | = 9.632 x 1021

Deduce, by referring to Kp , how the number of sulfur dioxide molecules will change if more oxygen is added to the equilibrium mixture.

to ensure that Kp stays the same/ quotient stays the same the number of (sulfur dioxide) molecules decreases because the equilibrium shifts to the right

Explain any change in the position of equilibrium if a few drops of sodium hydroxide solution are added to this equilibrium system.

equilibrium position shifts to the left | (because) the hydroxide ions combine with/neutralise the H+ ions to remove them from the equilibrium