Short answer
As long as external pressure is constant, it does not make sense to discuss $K_b$'s temperature dependence. If external pressure is changed, boiling point of solvent changes and hence also the ebullioscopic constant.
Longer version
Condition for equilibrium (skip if familiar)
Boiling is a reversible equilibrium process. By the second law of thermodynamics,
$$\mathrm{d}S_\text{universe}=0.$$
Assuming constant pressure,
$$\mathrm{d}S_\text{universe}= \mathrm{d}S-\frac{\mathrm{d}H}{T}.$$
Provided additional constant temperature,
$$-T\mathrm{d}S_\text{universe}= \mathrm{d}H - T\mathrm{d}S = \mathrm{d}G_\text{P,T}.$$
We also know that $G=G(P,T,n_1,\ldots, n_k).$ Thus
$$\mathrm{dG}=V\mathrm{d}P-S\mathrm{d}T + \sum_{i=1}^{k}\mu_i\mathrm{d}n_i,$$
and therefore
$$\mathrm{d}G_\text{P,T} = \sum_{i=1}^{k}\mu_i\mathrm{d}n_i=0.$$
Currently, we have two phases: a gas and a liquid. Boiling is a dynamic equilibrium, the differentials are non-zero, $\mathrm{d}n_1 = -\mathrm{d}n_2$, and so
$$\mathrm{d}G_\text{P,T} = 0 \iff G_\text{liquid} = G_\text{gas} \iff \mu_\text{liquid} = \mu_\text{gas}.$$
Clausius$-$Clapeyron equation (skip if familiar)
Suppose a closed system with initial equilibrium, constant temperature and pressure. We saw that
$$G_\text{liquid} = G_\text{gas}.$$
Now change the temperature. To achieve a new equilibrium in a closed system, the pressure would also need to change. At this new equilibrium, $G'_\text{liquid} = G'_\text{gas}$, which is why
$$\mathrm{d}G_\text{liquid} = \mathrm{d}G_\text{gas},$$
or equivalently,
$$V_\text{liquid}\mathrm{d}P - S_\text{liquid}\mathrm{d}T = V_\text{gas}\mathrm{d}P - S_\text{gas}\mathrm{d}T.$$
Now
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{S_\text{gas}-S_\text{liquid}}{V_\text{gas}-V_\text{liquid}}.$$
Note that the RHS depends again only on the initial and final conditions of phase equilibrium. Thus we are allowed to use $\Delta S = \Delta H/T$ to arrive at
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{\Delta H}{T(V_\text{gas}-V_\text{liquid})}.$$
Because $V_\text{gas} >> V_\text{liquid}$, and due to the ideal gas law
$$\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}T}\approx\frac{\Delta H}{RT^2}.$$
Assuming $\Delta H$ is not a function of temperature,
$$\ln \frac{P_2}{P_1} \approx -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\tag{Clausius–Clapeyron}$$
NOTE. There is a small but important detail to keep in mind here. The assumption that $\Delta H$ is independent of temperature will only make our equations more inaccurate but it will not affect our answer. See the definition further below.
Boiling point elevation
We have
$$\ln \frac{P^\text{s}}{P^\circ} \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right)\tag1$$
where
- $P^\text{s}$ – vapor pressure of solution at initial boiling temperature $T^\circ_\text{b}$,
- $P^\circ$ – saturation pressure of solvent, by definition of boiling, $P^\circ = P_\text{ext}$,
- $\Delta H_{vap}(T^\circ_\text{b})$ – entalphy of vaporisation at initial boiling temperature $T^\circ_\text{b}$,
- $T^\text{s}_\text{b}$ – new boiling temperature at some external pressure $P_\text{ext}$,
- $T^\circ_\text{b}$ – boiling temperature of solute at the same external pressure $P_\text{ext}$.
From Raoult's law
$$P^\text{s} \approx P^\circ X_\text{solvent} = P^\circ(1 - X_\text{solute}),$$
and thus from $(1)$
$$\ln (1 - X_\text{solute}) \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right).$$
From first-order Maclaurin series, and since boiling point elevation is usually small for our equations to be a good approximation,
$$\Delta T_\text{b} \approx \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}X_\text{solute}.$$
Granted molar mass $M$ is in units $\mathrm{g/mol}$, and molality $c_m$ has units of $\mathrm{mol\ /\ 1\ kg\ solvent}$, then
$$\Delta T_\text{b} \approx \underbrace{\frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}}_{K_\text{b}}c_m \equiv K_\text{b}c_m$$
or
$$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}. \tag2$$
It does not matter whether our previous equations are mere approximations, this is now a definition of the ebullioscopic constant. It is evident that it is in no manner dependent on external temperature. It is only dependent on the boiling temperature of a pure solvent at some well-defined external pressure $P_\text{ext}$. To change this temperature, we would have to change the external pressure, or pick a different solvent altogether.
If the external pressure is transformed, $T^\circ_\text{b}$ will not remain the same. There is no reason to believe that $(T^\circ_\text{b})^2$ at this new pressure $P'_\text{ext}$ will cancel with the change in vaporisation entalphy. And thus, the ebullioscopic constant is, in fact, not constant.
Conclusion
- If external pressure remains the same, it makes no sense to talk about temperature dependence. This is because the ebullioscopic constant is defined as $$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}$$
which is dependent only on the original boiling temperature of the pure solvent at some well-defined external pressure.
- If external pressure is changed, the pure solvent's boiling point $T^\circ_\text{b}$ will change to a new value. There is no cause to believe that the square of the boiling temperature will cancel with the altered entalphy of vaporisation. Hence, ebullioscopic constant will change, and is, in this sense, dependent on temperature. The exact function will vary with the solvent at hand but is most likely approximated by polynomials with empirical coefficients if need be.
