Chemistry, Third edition

ANSWERS TO EXERCISES AND REVISION QUESTIONS

12R

(i) 6 (ii) 2 .

12S

(i)A Sodium atom loses its outer shell electron when it

becomes an ion. The positive charge on the ion draws

the remaining electrons closer into the nucleus. A

chlorine atom gains an electron to its outer shell when it

becomes an ion. The negative charge on the ion makes

the electrons ‘spread out’.

(ii)The 2charge on the magnesium ion draws the

electrons closer to the nucleus.

(iii)The greater the negative charge on the ion, the more

the electrons repel each other and ‘spread out’.

Revision questions

12.1 (i)H (ii)X, L or Z (iii)Y or G (iv)X or D

(v)Z, M or K (vi)I (vii)J (viii)H (ix)D

(x)L.

12.2 (i)violently Ra(s) 2H 2 O(l)Ra(OH) 2 (aq)H 2 (g)

(ii)the most thermally stable and insoluble carbonate of

Group 2. The carbonate should decompose to the oxide

and carbon dioxide, on strong heating.

12.3Tin reverts to the crumbly allotrope at low

temperatures.

12.4CS.

12.5 (i)Li 2 O

Li 2 O(s)H 2 O(l)2LiOH(aq)

(ii)NO 2 (iii)BeO.

12.6 (i)ionic (ii)covalent.

12.7 (ii)C (network solid) and Al (strong metallic

bonding).

(iii)Any of H, He, N, O, F, Ne, Ar; London dispersion

forces between simple molecules or atoms.

(iv)Li, Na, Mg or Al. See Unit 5.

(v)Yes. There is a pattern of low boiling points then a

gradual rise to very high boiling point(s), then a drop.

12.8 (i) 6 (ii) 2 (iii)tetraaquadichloronickel(II).

12.9 (i)[Cr(H 2 O) 4 Cl 2 ]+ (ii)[PtBr 6 ]^2 –.

12.10 (i)[H 3 N–Cu–NH 3 ]+linear

(ii)

octahedral.

Unit 13

Exercises

13A

13B

(i)A balanced equation should be also written. Only then

do we know how much sodium and chlorine have

reacted.

(ii)Number of moles of carbon

120 g / 12 g mol–^1 10.

The equation shows that complete combustion of 1 mol

of carbon produces 393.15 kJ of heat. Therefore, the

complete burning of 10 mol of carbon produces

3931.5 kJ of heat energy.

13C

601.7 kJ is required to decompose 2 mol of MgO, so

300.9 kJ is required to decompose 1 mol of MgO.

13D

Eqn (13.14) is obtained by adding eqns (13.12) and

(13.13). By Hess’s law,

H^ —(13.14)H^ —(13.12)H^ —(13.13)

296.8(98.9)395.7 kJ mol–^1

13E

(i)To speed up the reaction so that the reaction was

complete in a short time.

(ii)The number of moles of lead(II) nitrate is

concentration volume

0.10 mol dm–^3 (100/1000)0.010

The number of moles of Zn used 0.007 7 moles. The

equation

Zn(s)Pb^2 +(aq)Zn^2 +(aq)Pb(s)

shows that 1 mol of solid zinc reacts with 1 mol of

Pb^2 +(aq), so that lead(II) nitrate is in excess.

13F

The solutions are dilute and may be assumed to be water

(C4.18 J g–^1 °C–^1 ). Therefore the total mass of water

(1 cm^3 1 g) is 50.0 g 50.0 g 100.0 g.

T23.6017.406.20 °C

Enthalpy change of reactants

mCT100.0 4.18 6.22600 J.

Numbers of moles of HCl initially present is

50.01.000.0500

1000

The number of mol of NaOH initially present is double

this. The reaction

H+(aq)OH–(aq)H 2 O(l)

(H+(aq) from hydrochloric acid, OH–(aq) from sodium

hydroxide) shows that 1 mol of water is produced when 1

mol of acid and 1 mol of alkali react. Therefore, HCl is

the limiting reagent, and the production of 0.0500 mol of

water is accompanied by the production of 2591 J of

heat. When onemole of water is made, the enthalpy

change is

H^1

0.0500^2591  51820 J, or 51.8 kJ mol

• 1

The accepted value is H—^ f57.3 kJ per mole of water

produced. The most important source of error is caused

by the loss of heat from the calorimeter, which results in

an underestimation of the temperature rise.

13G

To complete the equations, add the elements in their

standard states. The physical states of the elements

and compounds at 298 K should also be included

because we are assuming that H—^ frefers to this

temperature.

H 2 (g)S(s)2O 2 (g)H 2 SO 4 (l)

6C(s)6H 2 (g)3O 2 (g)C 6 H 12 O 6 (s)

13H

(i)2CO(g)O 2 (g)2CO 2 (g)

Using the standard enthalpy data in Table 13.2,

H^ —f(CO 2 (g))393.51 kJ mol–^1 ,

H^ —f(CO(g))110.53 kJ mol–^1 and

H^ —f(O 2 )(g))0 kJ mol–^1. Therefore,

H^ —f[2Hf^ —(CO 2 (g))][2H—^ f(CO(g))H^ —f(O 2 (g))]

[ 2(393.51)][ 2(110.53)0]

565.96 kJ mol–^1

(ii)Fe 2 O 3 (s)3CO(g)2Fe(s)3CO 2 (g)

H^ —f(CO 2 (g))393.51 kJ mol–^1 ;

H^ —f(CO(g)110.53 kJ mol–^1 ;

H^ —f(Fe 2 O 3 )(s)824.2 kJ mol–^1 ;

remember that

H—^ f(Fe(s))0 kJ mol–^1

H—^ f[3H^ —f(CO 2 (g)) 2 H^ —f(Fe(s)]

[3H—^ f(CO(g)H—^ f(Fe 2 O 3 )]

1180.5(1155.8)24.74 kJ mol–^1

(iii)NH 3 (g)^12 – N 2 (g)^3 – 2 H 2 (g)

H—^ f(NH 3 (g))46.11 kJ mol–^1 , and

H—^ f(H 2 (g))0 kJ mol–^1

H—^ f(N 2 (g))0 kJ mol–^1

H—f^ [ 21 – H—f^ (N 2 (g))^32 – H^ —f(H 2 (g))] [H—f^ (NH 3 (g))]

(46.11)46.11 kJ mol–^1

The positive sign shows that energy is absorbed in order

to decompose the molecule.

13I

(i)M(C 12 H 22 O 11 )342 g mol–^1. 5644 kJ of heat is

given out when 342 g of sucrose are completely burned

in oxygen.

Therefore, 1 g of sucrose gives out 5644/342 

16.5 kJ; energy value 16.5 kJ g–^1.

(ii)1 g of H 2 produces 142 kJ of heat. Therefore, the

mass of H 2 that produces 100 kJ of heat is

(^100)  1 0.704 g
142
0.704 g of H 2 contains (0.704/2.00) 0.352 mol of H 2 ;
1 mol of H 2 occupies 24 dm^3 at room temperature and
pressure. Therefore, 0.352 mol of H 2 occupies
0.352 24 8.5 dm^3
i.e. 8.5 dm^3 of gas burns to give 100 kJ of heat.
13J
20 g produces 120 Cal (120 kcal) when burned. The
energy value of the chocolate bar is
4.18 120
500 kJ. The energy value of 1 g of
chocolate is
(1/20) 500
25 kJ g–^1 (compare with the materials in
Table 13.3).
13K
Consumption of O 2 0.27 60  24 390 dm^3 day–^1.
Energy requirement per day  390  20 7800 kJ
(1870 Cal).
13L
2237 0.1223.7 kJ (endothermic)
13M
(i)The process Na(s) Na+(g)e–is the sum of the
atomisation and ionisation steps in the Born–Haber
cycle. Therefore, the energy requirement is
H—^1 H— 2  109  494 603 kJ mol–^1
(ii)The process Na+,Cl–(s)Na(g)^1 ⁄ 2 Cl 2 (g) involves
the sum of (H^ — 5 )(H^ — 4 )(H— 3 )(H^ — 2 )
 771  364  121  494 520 kJ mol–^1
13N
The formation of magnesium fluoride is
Mg(s)F 2 (g)Mg^2 +,F 2 – (s) H^ —f1121 kJ mol–^1
This may also be split up into steps, as with NaCl in the
text:
H— 5 H—^ fH^ — 1 H 2 —H^ — 3 H—^4 (A)
Note that:
(a)H^ — 2 involves two ionization steps:
Mg(g)Mg+(g)e– H^ —f744 kJ mol–^1
Mg+(g)Mg^2 +(g)e– H^ —f1456 kJ mol–^1
so that H^ — 2  744  1456 2200 kJ mol–^1.
(b) Step 3 involves the production of two fluorine
atoms:
F 2 (g)2F(g) H—^3 158 kJ mol–^1
H
/ kJ mol

-^1

Reactants

Progress of reaction

ΔH = O Products

NH 3 3 +

H 3 N

H 3 N

NH 3

NH 3

NH 3

Co

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