I am trying to understand what is isosteric heat of adsorption. Based on van't Hoff equation:

$$ \left(\frac{\partial \ln K}{\partial T}\right)_θ = \frac{ΔH^\circ}{RT^2} $$

and $ΔH^\circ$ is defined as the isosteric enthalpy of adsorption. The equilibrium constant for adsorption is:

$$K = \frac{a_{ads}}{a_g \cdot a_s}$$

and it is given by:

$$K = \exp\left(-\frac{ΔG^\circ}{RT}\right)$$

What is the point of differentiating with respect to temperature while keeping surface coverage $θ$ constant. I mean equilibrium constant depends only on $T$, that is $K=f(T)$.

I can't understand even how we are able to force surface coverage to have a constant value.


1 Answer 1


Not saying that's an easy part we can force the surface coverage to have a constant value. This answer may be only partial since I am only showing the result in gas phase. I haven't done the calculation for liquid, but probably we can find relation involving molar concentration, though this does not make it easy to keep coverage invariant.

The isosteric heat of adsorption of compound $c$ is the difference between the partial molar enthalpies of the gas phase and in the adsorbed phase. Thus, we have: $$\Delta H_c^i = H_c^{gas}-H_c^{ad} \tag{1}$$

At equilibrium you have equality of the chemical potential of $c$ in both phases. In the gas phase we have: $$\mu_c^g = \mu_c^{g,*}+RT\ln\left(f_c^g\right)\tag{2}$$

For the adsorption phase, the expression of the enthalpy is quite direct (thank you Gibbs-Helmholtz). We just derive the equality of chemical potential with respect to temperature and at constant amount of $c$ in the adsorbed phase. We note this fixed amount $x_c^{ad}$.

$$H_c^{ad}=\left(\frac{\partial\left(\frac{\mu_c^{ad}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{3}$$

Now comes the "tricky" part (at least at first). For the gas phase phase we can't directly do that because we chose a constant amount of $c$ in the adsorbed phase, thus gas enthalpy can't be in relation with gas chemical potential and the fixed amount $x_c^{ad}$, i.e.: $$H_c^{gas}\neq\left(\frac{\partial\left(\frac{\mu_c^{gas}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{4}$$.

However, we know from thermodynamics how to express the chemical potential of a gas, see $(2)$. f denotes fugacity which is what you would consider for real gas. You would then use an advance EOS to determine it. We can however keep it simple and assume ideal gas for the purpose of this answer. So, we have: $$\left(\frac{\partial\left(\frac{\mu_c^{gas}}{T}\right)}{\partial\left(\frac{1}{T}\right)}\right)_{x_c^{ad}}=H_c^{gas, *}+R\left(\frac{\partial\ln\left(f_c^{gas}\right)}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}} \tag{5}$$

$(1)$ can then be rewritten as: $$\Delta H_c^i = H_c^{gas}-\left(H_c^{gas, *}+R\left(\frac{\partial\ln\left(f_c^{gas}\right)}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}}\right) \tag{6}$$

Considering ideal gas law, fugacity reduces to molar gas fraction $y_c$ times overall pressure (fugacity coefficient is one for ideal gas). I consider here a pure gas so $y_c$ does not even matter. We are left with: $$\Delta H_c^i = H_c^{gas}-\left(H_c^{gas, *}+R\left(\frac{\partial\ln P}{\partial \left(\frac{1}{T}\right)}\right)_{x_c^{ad}}\right) \tag{7}$$

At low pressure, i.e. ideal gas, $$H_c^{gas} \approx H_c^{gas, *} \tag{8}$$

Living us with: $$\Delta H_c^i=RT^2 \left(\frac{\partial \ln P}{\partial T}\right)_{x_c^{ad}} \tag{9}$$

You asked, how do we keep the amount absorbed constant? Well, it's not easy-peasy but here $P$ and $T$ are independent variables so as you change $P$ you can adjust $T$ to keep $x_c^{ad}$ constant and vice-versa.

I hope this makes more sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.