F.6/Chem II/Mock Exam/12-13/P.1

S.K.H.LAMKAUMOWSECONDARY SCHOOL

MOCK EXAMINATION (2012-2013)

MARKING SCHEME

Form: 6

ChemistryPaper: 2

1. (a)

(i)Colorimetry/Change in colour intensity/absorbance[1]

Br2 is brown in colour. When the reaction proceeds, the concentration of bromine decreases and hence the colour intensity of the mixture also decreases. [1]

(ii)Plot a graph of conc. of Br2 against time,[1]

Find the slope of the tangent of the curve at time = 0, which is the initial rate of the reaction.[1]

(iii)Since, rate = k [Br2]x [CH3COCH3]y

If [CH3COCH3] > [Br2], the rate equation becomes rate = k’[Br2] [1]

The variation of reaction rate throughout the experiment is due to the change in the concentration of Br2 only. [1]

(iv)This makes sure the volumes of the mixture in all experiments are the same, and hence the volume ratio of the bromine (or the propanone) used represents the concentration ratio. [1]

(v)Compare the result of experiments 1 and 2,

The reaction rate doubles when [Br2] is doubled.[1]

The order of reaction w.r.t. Br2 is 1.[1]

(vi)Repeat the experiment but keeping the volume of Br2 constant and using different volumes of propanone. [1]

By comparing the initial rates and the conc. of propanone used in each experiment, the order of reaction to propanone can be determined. [1]

1.(b)(i)As the product side has less overall gaseous mixture volume/ no. of molecules, [1]according to Le Châtelier’s principle, the equilibrium shifts to the right with higher pressure to reduce the stress. [1]

(ii)Because maintaining such extreme pressure is so expensive that even a higher productivity cannot compensate for the cost. [1]

(iii)

(I)Because the reaction is exothermic, [1]a higher temperature shifts the equilibrium back to the left. [1]

(II)Although a lower temperature can shift the equilibrium to the right, [1] it takes longer time for the reaction to attain equilibrium, which will increase the running cost of the plant. [1]

(iii)A positive catalyst increases rate of a chemical reaction. [1]

Because the activation energy of the uncatalyzed process is too high for it to proceed in the provided conditions, an alternative pathway of lower activation energy provided by a positive catalyst. [1]

2. / (a) / (i) / Between oxalate ion and permanganate ion:
2 MnO4-(aq) + 5 C2O42-(aq) + 16 H+(aq)  2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) / [1]
(ii) / Potassium permanganate solution is not stable on standing / To determine the concentration of potassium permanganate solution. / [1]
(iii) / Potassium permanganate is purple in colour. When end point is reached, the solution in the conical flask will change from pale yellow to purple. / [1]
(iv) / For Part I,
No. of moles of C2O42-(aq) = mol / [1]
No. of moles of MnO4-(aq) = mol / [1]
Concentration of KMnO4(aq) = / [1]
(v) / For Part II,
No. of moles of KMnO4(aq) used = mol / [1]
No. of moles of Fe2+(aq) in 25 cm3 = mol / [1]
No. of moles of Fe2+(aq) in 250 cm3 = mol
Mass of Fe2+ in an iron tablet =
=1.269g / [1]
Percentage by mass of Fe2+ in an iron tablet = / [1]

2. (b)

(i)In 1 mole (74.0 g) of compound X,

[2]

The molecular formula of compound X is C4H10O. [1]

(ii)The board band at around 3 360 cm–1 represents O–H bond. [1]

The peaks ataround 2 860 cm–1 to 2 970 cm–1 represents C–H bond. [1]

(iii)

[1] [1]

(iv)

Butan-2-olor[1]

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