CHEM1101 Answers to Problem Sheet 9
1.
Oxygen has an oxidation number of -2 so the oxidation number of N can be found so
that the molecules are neutral. Molecules with odd numbers of electrons must have
unpaired electrons and so are paramagnetic.
(b) NO
N II (+2)
Odd number of electrons à paramagnetic
(d) NO2
N IV (+4)
Odd number of electrons à paramagnetic
2.
(c) N2O
N I (+1)
Even number of electrons.
(e) N2O4
N IV (+4)
Even number of electrons.
RDX (C3H6N6O6) is a powerful explosive.
(a)
2C3H6N6O6(s) + 3O2(g) à 6CO2(g) + 6H2O(g) + 6N2(g)
(b)
In the explosive decomposition, it is unlikely that sufficient O2(g) will be available
in the timescale of the reaction so that all of the oxygen available is that which is
present in RDX. With less oxygen available, it is likely that CO rather than CO2
will be produced:
C3H6N6O6(s) à 3CO(g) + 3H2O(g) + 3N2(g)
(c)
A solid reacts to form lots of gaseous molecules resulting in a large volume
change and hence PΔV work. The reaction is highly exothermic as the products
are very stable so considerable heat is generated. This heat also ensures that the
volume of the gas is large (Charles’ Law)
(b)
Using ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)
For the complete combustion reaction as written:
ΔH o = [6 × Δ f H o (CO2 ) + 6 × Δ f H o (H2O)] − [2 × Δ f H o (RDX)]
= [6 × −394 + 6 × −242]kJ mol −1 − [2 × +65]kJ mol −1 = −3946kJ mol −1
As this is for the combustion of two moles of RDX,
Δ comb H o = −1970 kJ mol −1
Molar mass = (3×12.01 (C) + 6×1.008 (H) + 6×14.01 (N) + 6×16.00 (O)) g mol-1
= 222.138 g mol-1
So,
Δ combH = -1970 kJ mol-1 or
−1970kJ mol −1
222.138g mol −1
Hence 100 g will generate 100 × 8.88 = 888 kJ.
= −8.88 kJ g-1
3.
(a)
∆rG° = ∆rH° - T∆rS° = (-55.3 × 103 kJ mol-1) – (298 K × -175.8 J K-1 mol-1)
= -2912 J mol-1 = -2.91 kJ mol-1
As ∆rG° = -RTln Kp,
ln Kp =
−Δ r G°
2912 J mol −1
so Kp = 3.24
=
RT
(8.314 J K −1mol −1 ) × (298K)
Note that as SI units are used for energy, the gas constant is also used in SI units.
(b)
∆rG° = ∆rH° - T∆rS° and ∆rH° and ∆rS° are both negative. As the temperature
increases, the - T∆rS° term will grow more and more positive and so ∆rG° will
increase: become less negative and eventually positive. Hence Kp will decrease as
the temperature increase.
The equilibrium constant varies with temperature but not with pressure.
(c)
(i)
As the reaction is exothermic, Le Chatelier’s principle predicts that
inreasing the temperature will reduce the amount of product (N2O4) and
increase the amount of reactant (NO2) present at equilibrium.
(ii)
As the reaction leads to a decrease in the number of moles of gas, it will
also lead to a decrease in the volume. Le Chatelier’s principle predicts that
increasing the pressure will favour product formation.
[C]3
4.
(a)
2
Kc = [A] [B]
(d)
Kc =
1
2
[A] [B]
[C]3
(b)
2
Kc = [A] [B]
(e)
Kc =
[C]3
(c)
2
Kc = [A] [B]
1
[A]2
5. The equilibrium is:
PCl5(g) D PCl3(g) + Cl2(g)
(i)
The initial concentration of PCl5(g) is
(ii)
The reaction table is:
initial
change
equilibrium
(iii)
K = 1.00 × 10-3 =
PCl5(g)
1.00
-x
1.00-x
2.00mol
moles
=
= 1.00 M
volume
2.00 L
PCl3(g)
0
+x
x
x2
[PCl 3 (g)][Cl 2 (g)]
=
(1.00-x )
[PCl 5 (g)]
+
Cl2(g)
0
+x
x
As K is small, the amount of PCl5 that dissociates is also small. Hence, 1.00 –x ~
1.00. This approximation is useful as it means that solving the quadratic equation
is unnecessary.
Using this approximation,
x2 = (1.00 × 10-3) × 1.00
x = 0.032.
Hence,
[PCl5(g)] = (1.00 – 0.032) M = 0.97 M,
[PCl3(g)] = [Cl2(g)] = 0.032 M
The assumption that x is negligible compared to the amount of PCl5 can be
checked using the “5% rule”: if the assumption results in a change in
concentration that is less than 5%, it is justified. The change in concentration is
found to be 0.032 M so the percentage change compared to the initial
concentration of 1.00 M is:
percent change =
0.032
×100 = 3.2%
1.00
(The answer obtained using the quadratic formula is x = 0.032).
6.
(a)
C(graphite) + ½ O2(g) → CO (g)
Sn(s) + O2(g) → SnO2(s)
(b)
The Ellingham diagram is shown below. The cross-over point is at about 600 K.
This is the point where the reaction starts to favour products over reactants.
100
ln(Kp)
80
60
40
20
0
0
7.
200
400 600 800 1000
Temperature (K)
If the equilibrium partial pressure of N2 is x, then the equilibrium partial pressure of H2
is 3x, as the stoichiometric ratio of N2 : H2 is employed. With the equilibrium partial
pressure of NH3 being 50 atm. the equilibrium constant is:
Kp =
(pNH )2
3
(pN )(pH )
2
2
3
=
(50)2
(x )(3x )
3
=
(50)2
27x
4
=1.00×10-4 so x = 31 atm.
The equilibrium partial pressures are therefore:
pN 2 = 31 atm, pH2 = 3 × 31 = 93 atm and pNH3 = 50 atm.
The total pressure is (31 + 93 + 50) atm = 174 atm.