CHEM1101
2014-J-11
June 2014
• Calculate the volume change when 20.0 g of solid trinitrotoluene C7H5N3O6(s)
explosively decomposes via the following process at 2000. °C and 2.0 atm.
2C7H5N3O6(s) → 12CO(g) + 5H2(g) + 3N2(g) + 2C(s)
Assume all gases behave as ideal gases and neglect the volume of any solid phases.
Show all working.
The molar mass of C7H5N3O6 is:
molar mass = (7 × 12.01 (C) + 5 × 1.008 (H)
+ 3 × 14.01 (N) + 6 × 16.00 (O)) g mol-1
= 227.14 g mol-1
The number of moles in 20.0 g is therefore:
number of moles = mass / molar mass = 20.0 g / 227.14 g mol-1 = 0.0880 mol
The chemical reaction, when 2 mol of C7H5N3O6 decomposes, 12 mol of CO(g), 5
mol of H2(g) and 3 mol of N2(g) are produced. When 0.0880 mol decomposes,
moles of gas = 0.0880 × (12 / 2 (CO) + 5/2 (H2) + 3/2 (N2)) mol = 0.88 mol
Using PV = nRT, the volume of this amount at 2000. oC (2273 K) and 2.0 atm is:
V = nRT / P
= (0.88 mol × 0.08206 L atm K-1 mol-1 × 2273 K) / (2.0 atm) = 82 L
Answer: 82 L
Marks
3
CHEM1101
2012-J-8
June 2012
• The equation for the detonation of nitroglycerine, C3H5N3O9(l), is given below.
4C3H5N3O9(l) → 6N2(g) + 12CO2(g) + 10H2O(g) + O2(g)
What mass of nitroglycerine is required to produce 1000 L of product gases at
2000 °C and 1.00 atm? Assume all gases behave as ideal gases. Show all working.
Using the ideal gas equation, PV = nRT, the total number of moles of gas
produced is:
n = PV / RT
= (1.00 atm × 1000 L) / (0.08206 atm L K-1 mol-1 × (2000 + 273) K)
= 5.36 mol
From the chemical equation, detonation of 4 mol of nitroglycerine gives (6 + 12 +
10 + 1) mol = 29 mol of gases. Therefore:
number of moles of nitroglycerine required = (4/29) × 5.36 mol = 0.739 mol
The molar mass of nitroglycerine is:
molar mass = [3 × 12.01 + 5 × 1.008 (H) + 3 × 14.01 (N) + 9 × 16.00 (O)] g mol-1
= 227.1 g mol-1
The mass of nitroglycerine is therefore:
mass = number of moles × molar mass = (0.739 mol) × (227.1 g mol-1) = 168 g
= 200 g (1 s.f.)
Answer: 200 g
3
CHEM1101
2012-N-3
November 2012
c) The Australian air quality guidelines stipulate a concentration of less than
5.0 × 10–9 mol L–1.
Does your answer in part b) exceed
Australian air quality guidelines?
YES / NO
What is the partial pressure of NO2 (in Pa) that corresponds to the Australian
guidelines at 25 °C and 100 kPa?
From the ideal gas law:
PV = nRT
or
P = (n / V) × RT = c × RT
When c = 5.0 × 10–9 mol L–1, at 25 °C (equivalent to 298 K):
P = (5.0 × 10–9 mol L–1) × (0.08206 atm L mol-1 K-1) × (298 K)
= 1.2 × 10–7 atm
As 1 atm = 101.3 kPa, this corresponds to:
P = (1.2 × 10–7 × 101.3) kPa = 1.2 × 10–5 kPa = 0.012 Pa
Answer: 0.012 Pa
Marks
4
CHEM1101
2010-J-15
June 2010
• You would like to make a gas thermometer using a mole of N2 at 1 atm. Assuming
that you can treat the gas as ideal, determine how much the volume increases (in mL)
per degree °C.
The ideal gas law is PV = nRT. At a constant pressure of 1 atm and with n = 1
mol, the change in temperature and volume are thus related by:
ΔV = (nR/P) × ΔT
= (0.08206 L atm K-1 mol-1)(1 mol) / (1 atm) × ΔT
= 0.080 L = (80 mL) × ΔT
The volume increases by 80 mL for every Kelvin (and therefore for every °C).
Answer: 80 mL
Marks
2
CHEM1101
2010-N-8
November 2010
• Automobile airbags are inflated by the decomposition of sodium azide according to
the following equation.
6NaN3(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) + 9N2(g)
What mass of NaN3 is required to produce 63 L of nitrogen gas at 25 °C and
1.76 atm?
Using the ideal gas law, PV = nRT, 63 L of N2 at 25 °C at a pressure of 1.76 atm
corresponds to:
n = PV / RT
= (1.76 atm) × (63 L) / ((0.08206 L atm mol-1 K-1) × (25 + 273) K))
= 4.5 mol
From the chemical equation, 6 mol of NaN3 leads to 9 mol of N2. The amount of
NaN3 required is therefore:
moles of NaN3 = (6 / 9) × 4.5 mol = 3.0 mol
The molar mass of NaN3 is (22.99 (Na) + 3 × 14.01 (N)) g mol-1 = 65.02 g mol-1.
This amount therefore corresponds to:
mass of NaN3 = molar mass × number of moles
= (65.02 g mol-1) × (3.0 mol) = 2.0 × 102 g
Answer: 2.0 × 102 g
3
CHEM1101
2009-N-11
November 2009
 A helium balloon is filled on the ground, where the atmospheric pressure is
768 mmHg. The volume of the balloon is 8.00 m3. When the balloon reaches an
altitude of 4200 m, its volume is found to be 16.8 m3. Assuming that the temperature
remains constant, what is the air pressure at 4200 m in mmHg?
Marks
2
From the ideal gas law, PV = nRT. As the number of moles and the temperature
is constant, the pressure and volume at the two heights are related by:
P1V1 = P2V2
Hence:
(768 mmHg) × (8.0 m3 = P2 × (16.8 m3)
P2 = 366 mmHg
Answer: 366 mmHg
 The volume of a gas is 40.0 mL at –15 °C and 1.30 atm. At what temperature (°C)
will the gas have a pressure of 1.00 atm and a volume of 65.0 mL?
As the number of moles is constant, the pressure, volume and temperature of the
gas are related by;
𝑷𝟏 𝑽 𝟏
𝑻𝟏
=
𝑷𝟐 𝑽 𝟐
𝑻𝟐
Hence:
𝟏.𝟑𝟎 𝐚𝐭𝐦 (𝟒𝟎,𝟎 𝐦𝐋)
−𝟏𝟓+𝟐𝟕𝟑 𝐊
=
𝟏.𝟎𝟎 𝐚𝐭𝐦 (𝟔𝟓.𝟎 𝐦𝐋)
𝑻𝟐
T2 = 322.5 K = 49.5 °C
Answer: 49.5 °C
2