Volumes of Gases in Chemical Reactions

Volumes of Gases in Chemical Reactions

Gasses are often reactants or products in chemical reactions
Balanced chemical equations deal with the number of moles of reactants consumed or products formed
For a gas, the number of moles is related to pressure (P), volume (V) and temperature (T)

Example

The synthesis of nitric acid involves the reaction of nitrogen dioxide gas with water:

3NO₂(g) + H₂O(l) -> 2HNO₃(aq) + NO(g)

How many moles of nitric acid can be prepared using 450 L of NO₂ at a pressure of 5.0 atm and a temperature of 295 K?

(5.0 atm)(450 L) = n(0.0821 L atm/mol K)(295 K)

= 92.9 mol NO₂

92.9 mol NO₂ (2HNO₃/3NO₂) = 61.9 mol 2HNO₃

Collecting Gases Over Water

Certain experiments involve the determination of the number of moles of a gas produced in a chemical reaction
Sometimes the gas can be collected over water

Potassium chlorate when heated gives off oxygen:

2KClO₃(s) -> 2KCl(s) + 3O₂(g)

The oxygen can be collected in a bottle that is initially filled with water

The volume of gas collected is measured by first adjusting the beaker so that the water level in the beaker is the same as in the pan.

When the levels are the same, the pressure inside the beaker is the same as on the water in the pan (i.e. 1 atm of pressure)

The total pressure inside the beaker is equal to the sum of the pressure of gas collected and the pressure of water vapor in equilibrium with liquid water

P_t = P_O2 + P_H2O

The pressure exerted by water vapor at various temperatures is usually available in tables:

Temperature (°C)	Pressure (torr)
0	4.58
25	23.76
35	42.2
65	187.5
100	760.0

Example

A sample of KClO₃ is partially decomposed, producing O₂ gas that is collected over water. The volume of gas collected is 0.25 L at 25 °C and 765 torr total pressure.

a) How many moles of O₂ are collected?

P_t = 765 torr = P_O2 + P_H2O = P_O2 + 23.76 torr

P_O2 = 765 - 23.76 = 741.2 torr

P_O2 = 741.2 torr (1 atm/760 torr) = 0.975 atm

PV = nRT

(0.975 atm)(0.25 L) = n(0.0821 L atm/mol K)(273 + 25K)

n = 9.96 x 10^-3 mol O₂

b) How many grams of KClO₃ were decomposed?

9.96 x 10^-3 mol O₂ (2KC lO₃/3 O₂) = 6.64 x 10^-3 mol KClO₃

6.64 x 10^-3 mol KClO₃ (122.6 g/mol) = 0.814 g KClO₃

c) If the O₂ were dry, what volume would it occupy at the same T and P?

P_O2 = (P_t)(X_O2) = 765 torr (1.0) = 765 torr (1 atm/760 torr) = 1.007 atm

(1.007 atm)(V) = (9.96 x 10^-3 mol)(0.0821 L atm/mol K)(273 + 25 K)

V = 0.242 L

Alternatively...

If the number of moles, n, and the temperature, T, are held constant then we can use Boyle's Law:

P₁V₁ = P₂V₂

V₂ = (P₁V₁)/ P₂

V₂ = (741.2 torr * 0.25 L)/(765 torr)

V₂ = 0.242 L

1996 Michael Blaber