ANSWERS TO EXERCISES AND REVISION QUESTIONS(c) Two fluorine atoms accept electrons:
2F(g)2e–2F–(g) H^ — 4 (2339)678 kJ mol–^1
Substituting into equation (A),
H^ — 5 1121 148 2200 158 678
2949 kJ mol–^1
or,H^ —LH^ — 5 2949 kJ mol–^1.
13O
Average(5311075)/2803 kJ mol–^1.
13P
(i)The standard enthalpy change for the process
CH 4 (g)CH 3 (g)H(g) will approximately equal H^ —C–H.
Therefore the enthalpy change for the reverse
process:
CH 3 (g)H(g)CH 4 (g)
will be H^ —C–H, i.e. 412 kJ mol–^1.
(ii)Comparing the bond lengths with the bond enthalpy
data in Table 13.4, the trend supports the proposition
that long bonds are weak bonds (or, conversely, that
short bonds are strong bonds).
(iii)Double and triple bonds are stronger than single
bonds. The general order of bond strength is triple bonds
double bonds single bonds.
13Q
(i)CH 2 CH 2 (g)Cl 2 (g)CH 2 Cl–CH 2 Cl(g)
To completely break up the molecules on the left-hand
side of this equation we need to dissociate: one CC
bond, four C–H bonds and one Cl–Cl bond. From
Table 13.4,
energy absorbed 612 4(412) 242 2502 kJ mol–^1
i.e. the energy change due to bond destruction is
2502 kJ.
To form the reactant molecules, we need to make four
C–H bonds, two C–Cl bonds and one C–C bond. From
Table 13.4,
energy given out 4(412)2(338) 348 2672 kJ
mol–^1
i.e. the energy change due to bond formation 2672
kJ. Therefore, the net change in energy is
H^ —f(298 K) 2502 2672 170 kJ mol–^1
(ii)Bond energies can be used directly only if all the
reactants and products are gases. If some reactants or
products are not gases then the enthalpy change for the
phase change (solid gas or liquid gas) must be
allowed for.
We start by working out the enthalpy change for the
reaction in which all the reactants and products are
gaseous:
2H 2 O(g)2H 2 (g)O 2 (g) (A)
We now apply the average bond energies in Table 13.4
for the heat absorbed in breaking bonds in reactant
molecules, this involves breaking four O–H bonds:
4 463 1852 kJ mol–^1
From Table 13.4 the heat released when bonds in
product molecules are assembled, which involves
making two H–H bonds and one OO bond is
(2 436) 497 1369 kJ mol–^1
The overall change in enthalpy is therefore
H^ —A 1852 1369 483 kJ mol–^1
However, we want the enthalpy change for the reaction
2H 2 O(l)2H 2 (g)O 2 (g) (B)
and we are given
H 2 O(l)H 2 O(g) H^ —f44 kJ mol–^1or
2H 2 O(l)2H 2 O(g) H—^ f88 kJ mol–^1
H 2 O(g) possesses a higher enthalpy than H 2 O(l): this
means that the enthalpy change for reaction (B) will be
larger than that of reaction (A) by 88 kJ. (If you can’t see
this, draw an energy diagram for the processes involved):
H—^ B(298 K) 483 88 571 kJ mol–^1
Revision questions
13.1Following are false: (i),(ii),(iv)(2 mol SO 3 (g)
made) and (vi).
13.2 (i)See text.
(ii)Combustion of methyl benzene:
C 7 H 8 (l)9O 2 (g)7CO 2 (g)4H 2 O(l)
H—c^ [4H—^ f(H 2 O(l)) 7 H^ —f(CO 2 (g))][H^ —f(C 7 H 8 (l))
9 H^ —f(O 2 (g))]
From Table 13.2:
[(285.83 4)(393.51 7)][12.00]
(3897.89)(12.0)3909.9 kJ per mole of
methyl benzene
(iii)1 mol of methyl benzene has a molar mass of
approximately 92 g mol–^1.
Heat given out in the complete combustion of 1 g of
methyl benzene is 3909.9/92 42 kJ g–^1.
1 cm^3 of methyl benzene has a mass of 0.86 g.
The heat given out in the complete combustion of
1 cm^3 of methyl benzene is approximately
42 0.8636 kJ cm–^3.
13.3 (i)H^ —f[Hf—^ (P(s) (red))] [H^ —f(P(s) (white))]
18 0 18 kJ mol–^1
i.e. red phosphorus is the more energetically stable form
of phosphorus.
(ii)
H—^ f[Hf^ —(H 2 O(l))H^ —f(CH 3 CO 2 C 2 H 5 (l))]
[H—f^ (C 2 H 5 OH(l))H—f^ (CH 3 CO 2 H)(l)]
(285.83481)(277.69485)
(766.83)(762.69)
4 kJ mol–^1
(iii)
H—f^ [3H^ —f(N 2 (g)) 9 H^ —f(H 2 O(g))
5 H^ —f(Al 2 O 3 (s)) 6 H^ —f(HCl(g))]
[6H^ —f(NH 4 ClO 4 (s)) 10 H^ —f(Al(s))]
TheH^ —fof elements in their reference states is zero.
H—^ f(NH 4 ClO 4 (s)) is obtained from Table 13.2.
H—^ f[(241.82 9)(1675.7 5)
(92.31 6)][295.31 6]
[2176.388378.5553.86][1771.86]
9337 kJ mol–^1
(Notice that the effect of producing a highly exothermic
compound (Al 2 O 3 (s)), is to make the overall reaction
highly exothermic.)
13.4Ignoring the heat capacity of the calorimeter and of
the reactants and products,
q mmixtureCmixtureT
q 100 4.18 0.411.7 102 J
...H1.7 102 J
Neither reactant is in excess. The number of moles of
Ag+(aq) (which equals the number of moles of Cl–(aq)) is(^0500) 0.0502.5 10 – 3
1000
The enthalpy change when 1 mol of either reactant is
consumed is
H^1
2.5 10 –^3 1.7^10
2
H 6.8 104 68 kJ mol–1
13.5HCOOH(l)^1 ⁄ 2 O 2 (g)CO 2 (g)H 2 O(l)
H—^ c277.2 kJ mol–^1
H—^ c[H^ —f(COO(g))H—^ f(H 2 O(l))]
[H—^ f(HCOOH(l))^1 ⁄ 2 H^ —f(O 2 (g))]
whereH—f^ (HCOOH(l)) is the only unknown,
H—^ f(CO 2 (g))393.51 kJ mol–^1 andH^ —f(H 2 O(l))
285.83 kJ mol–^1 (from Table 13.2) and H—^ f(O 2 (g))
0. Substitution and rearrangement gives:
H—^ f(HCOOH(l)(393.51285.83)(277.2)
402.1 kJ mol–^1
13.6 (i)C 6 H 12 O 6 (s)6O 2 (g)6CO 2 (g)6H 2 O(l)
(ii)M(C 6 H 12 O 6 )180 g mol–^1. The complete
combustion of 180 g of glucose releases 2816 kJ of
energy. The complete combustion of 10 g of glucose
would release (10/180) 2816 160 kJ (2 significant
figures). If the combustion of glucose within the body
were only 40% efficient, only 64 kJ of energy would be
available for work for every 10 g of glucose consumed.
13.7 (i)The third standard ionization energy of
aluminium is the standard enthalpy change for the
process
Al^2 +(g)Al^3 +(g)e–
H—^ ion(3)2751 kJ mol–^1
(ii)The equation
Al(s)Al^3 +(g)3e– (A)
is the sum of the equations:
Al(s)Al(g) (B)
Al(g)Al+(g)e– (C)
Al+(g)Al^2 +(g)e– (D)
Al^2 +(g)Al^3 +(g)3e– (E)
The enthalpy change for reaction (A) is the sum of the
enthalpy changes for reactions (B)–(E) (If you are unsure
about this, sketch an energy diagram):
H—A^ H—B^ H^ —CH^ —DH^ —E
326 584 1823 2751 5484 kJ mol–^1
13.8
(a) Ca(s) Ca(g) H—^1
(b) Ca(g) Ca+(g)e– H—^ 2A
(c) Ca+(g)Ca^2 +(g)e– H—^ 2B
(d) O 2 (g)O(g)O(g) H—^3
(e) O(g) 2e–O^2 – (g) H—^4
(f) Ca^2 +,O–(s)Ca^2 +(g)O^2 – (g) H—^ L
The formation of calcium oxide,
Ca(s)^1 ⁄ 2 O 2 (g)Ca^2 +,O–(s) H—^ f
may also be split up into steps, as for NaCl. We follow
the pattern for NaCl in the text, with
H—^2 H2A—^ H^ —2B. Therefore,
H—^ fH^ — 1 H^ — 2 ^1 ⁄ 2 H—^3 H— 4 (H^ —L)
Notice that only half of H^ — 3 is required because only
one atom of O is required to combine with one atom
of Ca.
H—^ f 193 1740 248.5 703 3513
629 kJ mol–^1
13.9
2F 2 (g)S(g)SF 4 (g) H—^ f994 kJ mol–^1
In breaking bonds in the reactant molecules, we break
two F–F bonds:
heat absorbed 2 158 316 kJ
The heat released when bonds in product molecules are
assembled involves making four S–F bonds, each of
bond enthalpy x. The total enthalpy change is 4x (which
is negative because heat is given out when bonds are
formed). Therefore, the overall change in enthalpy is
994 4 x 316
which gives
xH—^ S–F(994316)/4
x328 kJ per mole of S–F bonds.
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