Gaseous Equilibria and K_{p}
Mole Fractions
If you have a mixture of gases, A, B, C, etc. then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas.
The mole fraction of gas A is often given the symbol x_{A}. The mole fraction of gas B would be x_{b} - and so on.
Pretty obvious really!
For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas. The mole fraction of nitrogen is 1/4 or 0.25 and of hydrogen is 3/4 or 0.75.
Partial pressure
The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container. The partial pressure of gas A is often given the symbol P_{A}. The partial pressure of gas B would be P_{B} - and so on.
There are two important relationships involving partial pressures. The first is again fairly obvious.
The total pressure of a mixture of gases is equal to the sum of the partial pressures or:
<Total Pressure P = P_{A} + P_{B} + P_{C} etc..
Gas A is creating a pressure (its partial pressure) when its molecules hit the walls of its container. Gas B does the same. When you mix them up, they just go on doing what they were doing before. The total pressure is due to both molecules hitting the walls - in other words, the sum of the partial pressures.
The more important relationship is the second one where P_{A} is the partial pressure:
\({P_A} = mole\;fraction\;of\;A\; \times Total\;Pressure\)
This means that if you had a mixture made up of 20 moles of nitrogen, 60 moles of hydrogen and 20 moles of ammonia (a total of 100 moles of gases) at 200 atmospheres pressure, the partial pressures would be calculated like this:
Gas |
Mole Fraction |
Partial Pressure (atm) |
Nitrogen |
20/100 = 0.2 |
0.2 x 200 = 40 |
Hydrogen |
60/100 = 0.6 |
0.6 x 200 = 120 |
Ammonia |
20/100 = 0.2 |
0.2 x 200 = 40 |
Partial pressures can be quoted in any normal pressure units.
K_{p} in a homogenous gaseous equilibria
A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use K_{p}, everything must be a gas. A good example of a gaseous homogeneous equilibrium is the conversion of sulphur dioxide to sulphur trioxide at the heart of the Contact Process:
2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g)
Writing an expression for K_{p}
Lets look at a general case with the equation:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, K_{p}.
Just like K_{c}, K_{p} always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with.
K_{p} has exactly the same format as K_{c}, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.
\({K_p} = \;\frac{{P{C^c} \times P{D^d}}}{{P{A^a} \times P{B^b}}}\)
Examples:
Lets look at the contact equation again:
2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g)
K_{p} is given by:
\({K_p} = \;\frac{{PS{O_3}^2}}{{PS{O_2}^2 \times P{O_2}}}\)
K_{p} in heterogeneous equilibria
A typical example of a heterogeneous equilibrium will involve gases in contact with solids.
Writing an expression for K_{p} for a heterogeneous equilibrium is exactly as happens with K_{c}, except you don't include any term for a solid in the equilibrium expression.
Example:
H_{2}O(g) + C(s) ⇌ H_{2}(g) + CO(g)
The K_{p} equation for this is:
\({K_p} = \;\frac{{P{H_2} \times PCO}}{{P{H_2}O}}\)
Notice that I did NOT put the carbon in the K_{p} equation, this is because it is not a gas.