Chemistry Scholarship

Assessment Schedule 2007 Scholarship Chemistry (93102)

Evidence Statement

Question Evidence Mark Allocation

1(a) (i) "Chemical reactions occur to enable atoms to obtain full outer electron shells or octets".

eg hydrogen and chlorine gases react spontaneously to form

hydrogen chloride.

Discussion should include recognition that:

* 'Full outer shells' are not always octets.

The term 'outer shell' refers to a property of an atom. As such the number of electrons needed to 'fill' the 'outer shell' of an atom depends on the period of the Periodic Table in which the atom is found. Only those in the second period have a 'full outer shell' of 8 electrons, eg if a hydrogen atom has 2 electrons its 'outer shell' is filled. Using the same argument chlorine would need 18 electrons

to fill its outer shell.

* 'Obtaining full outer shells' is not the reason why chemical

reactions occur.

In the example given the outer shell arrangement of the H and Cl atoms are the same whether they are found in reactants or products, ie H has 2 e-s in H2 and in HCl and chlorine has 8 e-s in Cl2 and HCl. Chemical reactions are better explained by considering the energy changes that occur during reactions. Many spontaneous chemical reactions (at room temperature) occur because the energy of the products is lower than the energy of the reactants and the activation energy of the reaction is small. As a consequence the products will have stronger bonds than the reactants. 7-8 marks

Shows understanding of principles of chemistry by

(a) recognising misconceptions and providing correct explanations for the observations given

AND

(b) carrying out thermochemical calculations and linking the answers to the observations.

5-6 marks

Shows understanding of principles of chemistry by

(a) identifying misconceptions and providing a reasonable explanation for the observations given

AND

(b) using the correct procedure for the thermochemical calculations.

1-4 marks

Answers include:

* identification of a misconception (s) with limited explanation

* correct method for themochemcal calculation.

(ii)

"Successive ionisation energies for any atom increase because when an electron is removed from an atom or ion, the remaining electrons receive an extra share of the attraction from the nucleus."

Eg: the first three ionisation energies for sodium are:

Na(g) → Na+(g) + e- ΔH = 502 kJ mol-1

Na+(g) → Na2+(g) + e- ΔH = 4569 kJ mol-1

Na2+(g) → Na3+(g) + e- ΔH = 6919 kJ mol-1

Discussion should include recognition that:

The attraction of the nucleus for an electron is dependant on the distance of the electron from the nucleus and the size of the nuclear charge. The energy needed to remove the outermost electron from a gaseous atom (ionization energy) will be greater for electrons in energy levels closest to the nucleus. When electrons are removed from a filled or partially filled energy level the number of electron-electron repulsions are reduced which allows the electrons to move in closer to the nucleus. This accounts for the increase in energy needed to remove successive electrons from the same energy level.

For Na, the electron arrangement is 2, 8, 1. The first electron to be removed is from the energy level furtherest from the nucleus. The second electron is removed from a closer energy level so is held more strongly hence ionization energy is higher. The third ionization energy is similar to second because the electron is removed from the same energy level but there is one less electron meaning that there will be fewer electron-electron repulsions. This allows the electrons to be closer to the nucleus so they are more strongly held than in Na+.

(b) ΔcHo(CH4) = -812 kJ mol-1

ΔcHo(SiH4) = -1428 kJ mol-1

ΔcHo for both reactions is negative so both are thermochemically favorable. However, only the SiH4 reaction is spontaneous which suggests that the activation energy for this reaction is lower than that for reaction of CH4, ie a smaller energy input is needed to get the reaction started.

2(a)

70% Fe is 0.140 g

n(Fe) = 0.140 g / 55.9 g mol-1 = 0.00250 mol

[Fe3+] = 0.00250mol / 0.200 L = 0.0125 mol L-1

10% Ni is 0.0200 g

n(Ni) = 0.020 g / 58.7 g mol-1 = 3.41 × 10-4 mol

[Ni2+] = 3.41 × 10-4 mol / 0.200 L = 0.00170 mol L-1

For Fe(OH)3 to precipitate when the concentration of Fe3+ is reduced to 0.1% of original conc then

[Fe3+] = 0.001 × 0.0125 mol L-1 = 1.25 × 10-5 mol L-1

Ks(Fe(OH)3)) = 2.79 × 10-39 = [Fe3+] × [OH-]3

= 1.25 × 10-5 × [OH-]3

[OH-]3 = 2.79 × 10-39 / 1.25 × 10-5 = 2.23 × 10-34

[OH-] = 6.07 × 10-12 and [H3O+] = 0.00165 mol L-1

and pH = 2.78

To determine the pH at which Ni(OH)2 will precipitate:

Ks(Ni(OH)2)) = 5.48 × 10-16 = [Ni2+] × [OH-]2

= .00170 × [OH-]2

[OH-]2 = 5.48 × 10-16 / 0.00170 = 3.22 × 10-13

[OH-]