Making calculations easier
When we perform calculations in chemistry we are normally concerned with using the most accurate data and exact equations to obtain an acceptable answer. However, there are times when we are justified in making approximations that make calculations easier or even possible at all. In this article I present three examples showing how approximations can be used. I also highlight the importance of ensuring these approximations can be justified.
Equilibrium problems
Much research has been undertaken on the teaching of equilibrium in chemistry.^{1} Here we will concentrate on one specific mathematical aspect that can make our lives easier while also causing us to consider what is going on beyond the underlying mathematics.
The equilibrium constant K_{c} for the production of ammonia from hydrogen and nitrogen
N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)
is 4.2 x 10^{8} mol^{-2} dm^{6} at 298K. From the equation given K_{c} will be defined as
It follows that for the reverse reaction, the dissociation of ammonia
2NH_{3}(g) ⇌ N_{2}(g) + 3H_{2}(g)
As this is the reciprocal of the expression for K_{c} for the reverse reaction, for the dissociation of ammonia
2NH_{3}(g) | ⇌ | N_{2}(g) | + | 3H_{2}(g) | |
Initially | 0.25 | 0 | 0 | ||
Change | −2x | 3x | |||
At equilibrium | 0.25−2x | 3x |
Let’s now consider what happens when ammonia dissociates. Because K_{c} is very small there will be much less nitrogen and hydrogen (the products) present in the equilibrium mixture than there are of ammonia (the reactant). Let’s assume that the initial concentration of ammonia is 0.25 mol dm^{–3}, and that in concentration terms 2x of this dissociates to give hydrogen and nitrogen. This table shows the concentrations of the three species present initially and at equilibrium, and the changes that occur.
Note how the quantities are related to the stoichiometry expressed in the reaction equation, which is why the decrease in ammonia concentration was chosen to be 2x.
We can now substitute the concentrations at equilibrium into the expression previously given for K_{c}.
We can, in theory, solve this equation for x since we have the value of K_{c}. However, this is complicated by the denominator (0.25 mol dm^{-3} – 2x)^{2}. If we recall the chemistry discussed earlier, we realise the value of x (and consequently of 2x) will be much smaller than 0.25 mol dm^{−3}. If it is sufficiently smaller, we could neglect the −2x term and write an approximate expression for K_{c}.
Note the use of the ≈ symbol that denotes an approximation. For now we will assume this is a valid approximation and so we can write
Rearranging this equation to make x^{4} the subject gives
Taking the fourth root of x^{4} then gives
How do we know whether this is a valid approximation? If we attempt to calculate
0.25 mol dm^{–3} – 2x
We have
(0.25 – 2 x 1.54x10^{–3}) mol dm^{–3}
(0.25 – 3.08x10^{–3}) mol dm^{–3}
(0.25 – 0.00308) mol dm^{–3}
Remembering that if we perform a subtraction we are only justified in giving the answer to the same number of decimal places as in the original data, this shows the term being subtracted will be negligible. This is, therefore, a valid approximation.
Real and ideal gases
A very commonly used equation in chemistry is the gas equation
pV = nRT
that relates the pressure p, temperature T and volume V of an amount n of an ideal gas in terms of the gas constant R. Of course, an ideal gas doesn’t actually exist so attempts have been made to modify the equation to more closely predict the behaviour of real gases.
One such modification is the van der Waals equation,^{2} which is expressed as
This is clearly related to the ideal gas equation, but the pressure has been increased with a term an^{2}/V^{2}, and the volume decreased by nb. The logic behind this is that intermolecular forces reduce the observed pressure so a correction factor needs to be added to give the ideal pressure. This is related to the density of the gas, hence the appearance of n and V. Also, the volume available for the gas molecules to move in is less than that of the container due to the finite size of the molecules themselves. Therefore the volume is reduced by a term which is proportional to the amount n of gas. These terms incorporate constants a and b which are known as the van der Waals constants and are tabulated for different gases.
Let’s consider how both equations predict the behaviour of carbon monoxide gas, for which a = 0.147 Pa m^{6} mol^{-2} and b = 3.9 x 10^{-5} m^{3} mol^{-1}. If we rearrange the ideal gas equation we get
At 1000K, if we have 5.0 mol of CO occupying 2.5 m^{3} then we have
Applying the rules for significant figures and decimal places^{3} gives a more appropriately expressed result of 17 kPa. Note that in the calculation above I have made use of the relationship 1 Pa = 1 J m^{-3}.
Using the van der Waals equation, successive rearranging gives
Calculating the values of the various terms in this equation gives
Again, writing this final expression slightly differently gives
and we see that nb is negligible compared to V.
Putting these together gives
Applying the rules of significant figures^{3} gives 17 kPa and the an^{2}/V^{2} term can again be neglected.
Applying the rules of significant figures gives 17 kPa and the an^{2}/V^{2} term can again be neglected.
It should be apparent that having neglected the -nb and an^{2}/V^{2} terms reduces the van der Waals equation to the ideal gas equation, so the latter is an appropriate approximation in this case.
It is clear that the approximation to an ideal gas will not be valid when V is small. The term an^{2}/V^{2} will then be large, and nb is less likely to be negligible relative to V. Real gases are thus likely to deviate from ideal behaviour at high pressures and low temperatures.
Steady state approximation
The steady state approximation doesn’t require us to choose a value for a quantity, but it is still a powerful example of what can be achieved by approximation. It is most commonly used to analyse multi-step reactions.
For example, consider the gas phase decomposition of ozone.
2O_{3} → 3O_{2}
which is believed to take place via the mechanism
where M can be any molecule.
Before analysing this it is helpful to visualise what is taking place using a graph of concentration c against time t. The concentration of O_{3} falls from its starting value to zero while that of O_{2} increases from zero at a faster rate. The concentration of the intermediate O increases rapidly from zero but only to a value considerably lower than that of O_{2} or O_{3}; its value also remains constant for most of the duration of the reaction.
If r_{1} is the rate at which O_{2} is produced we can write
r_{1} = k_{1}[O_{3}][M] + 2k_{2} [O][O_{3}] – k_{-1}[O_{2}][O][M]
while for the rate r_{2} at which O_{3} is produced
r_{2} = k_{-1}[O_{2}][O][M] – k_{1}[O_{3}][M] – k_{2}[O][O_{3}]
Finally we can write an expression for r_{3}, the rate at which O is produced
r_{3} = k_{1}[O_{3}][M] – k_{-1}[O_{2}][O][M] – k_{2}[O][O_{3}]
The assumption that we make is that r_{3} = 0, which is equivalent to assuming that the line representing the concentration of O in the graph is horizontal. This approximation generally works if the concentration of an intermediate is rather lower than that of the products and reactants.
Once we have set
r_{3} = k_{1}[O_{3}][M] – k_{-1}[O_{2}][O][M] – k_{2}[O][O_{3}] = 0
we can perform various algebraic manipulations to show that r, the overall rate for the reaction, is given by
we can perform various algebraic manipulations^{4} to show that r, the overall rate for the reaction, is given by
The question now arises as to whether the approximation that r_{3} is zero is reasonable. In this type of example the justification is provided by the extent to which the derived rate equation predicts experimental observations.
Paul Yates is deputy head of academic practice at Newman University
References
- A Raviolo and A Garritz, Chem. Educ. Res. Pract., 2009, 10, 5 (DOI: 10.1039/b901455c)
- J O Valderrama, J. Supercrit. Fluids, 2010, 55, 415 (DOI: 10.1016/j.supflu.2010.10.026)
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