Here we want to calculate the heat of adsorption from a couple measurements of physical adsorption
of nitrogen on charcoal.
These measurements were made at low coverages of nitrogen, so we make some assumptions to
calculate the heat of adsorption.
So we have two measurements at different temperatures and pressures.
I should point out that we're probably not going to get a very accurate heat of adsorption
measurement from just two measurements at different temperatures.
Accuracy of the measurements has to be really quite good, and to get a good measure we should
take measurements at a number of temperatures, but to demonstrate the idea we'll use these
two temperatures.
And what we're going to do is assume Langmuir-type behavior for adsorption, so this is the fractional
coverage, theta is fractional coverage, versus pressure.
High enough pressure, this is saturation, namely we've covered the surface with nitrogen,
and at lower pressures, some regime we have a linear behavior, and so the Langmuir isotherm
says the fractional coverage is an adsorption equilibrium constant times the pressure over
1 plus that same adsorption equilibrium constant and pressure.
We're going to look at the case of low coverage, so we're saying this is approximately equal
to k times the pressure.
if we calculate k at the two conditions, at 190 Kelvin, we don't know the actual fractional
coverage, we just know how much is adsorbed.
So how much is adsorbed of course divided by the saturation coverage will give us the
fractional coverage, so we're saying it's proportional to the fractional coverage, so
I'm just going to use alpha as the proportionality constant, times 0.92, this is going to be
equal to our adsorption equilibrium constant times the pressure, and of course the same
thing at 250 Kelvin.
So we can solve for K1, which corresponds to the adsorption equilibrium constant at
190 in terms of alpha, and that's just 0.1917 times alpha.
K2 is 0.02875 times alpha.
Now, K is related to the heat of adsorption by a pre-exponential factor, e to the lambda
over RT, where lambda is the heat of adsorption, absolute value of the enthalpy change, because
adsorption is typically exothermic, so we're going to write K1 divided by K2 in terms of
lambda and temperatures.
K0 doesn't change, lambda doesn't change, gas constant doesn't change, temperature is
different.
So K0's cancel, this is lambda over R, 1/T1 minus 1/T2.
Now I can substitute in values.
Of course these are absolute temperatures.
I can simplify this, this is the exponential, let's write it out this way to make it a little
clearer to see, lambda over R, 1.263 times 10 to the minus 3.
So alphas of course cancel, this ratio on the left side is 6.667, and what I'm going
to do, of course at T equal to the right side, I'm going to take the log of the left side,
6.667, going to be equal to the log of the right side, which then becomes lambda over
R times this value, so I can now solve for lambda over R. So I've done some of the calculations,
I've put in here units, Kelvin, to make sure where things come from, so when I solve for
this, 12.5 kJ/mol, that's the heat of adsorption.
Realistically, 13 kJ/mol, but again, I want to mention it's very sensitive to these measurements,
so to get a very accurate heat of adsorption, you'd really need more measurements, but the
ideas would still apply.