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Wikipedia cites the following formula for the ebullioscopic constant $K_{\mathrm{b}}$ regarding the elevation in boiling point of a volatile solvent upon addition of a non volatile solute (https://en.wikipedia.org/wiki/Ebullioscopic_constant):

(All symbols carry their usual meaning, unless specified otherwise)

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{\Delta H_{\mathrm{vap}}}$$

But my textbook cites another formula :

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{L_{\mathrm{v}}}$$

Where $L_{\mathrm{v}}$ is the latent heat of vaporization. Since there exists the equation $$\Delta H_{\mathrm{vap}} = {mL_{\mathrm{v}}}$$

my textbook further modified its formula to:

$$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}m}{1000\Delta H_{\mathrm{vap}}}$$

Where $m$ is the mass of solvent in grams.

Now my question is, which one's correct? The only difference between the equation on Wikipedia and the one in my textbook is the mass term $m$. I can't seem to find any reliable references on this :/. So a good reference would be appreciated.

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    $\begingroup$ I am familar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term. Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ\ mol^{-1}}$ vs.$\mathrm{kJ \ kg^{-1}}$. The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself, believe. So guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway. $\endgroup$ – getafix Oct 30 '16 at 6:59
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This answer expands upon my first comment:

I am familiar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term.

I have tried to perform the derivation I alluded to, and it is given:

Assuming a binary mixture, A denotes the solvent and B denotes the solute. Superscript (*) implies pure solvent without solute (e.g $T^*$ is boiling point of pure solvent). Also, I assume the enthalpy of vaporisation doesn't change over the temperature range we are interested in.At equilibrium the chemical potential obeys:

$$\mu^*_A(g) = \mu^*_A(l) + RT\ln(\chi_A)$$ $$ \ln(\chi_A) = \frac{\mu^*_A(g)-\mu^*_A(l)}{RT} = \frac{\Delta_{\text{vap}}G}{RT}$$

Differentiating with respect to $T$ and invoking the Gibbs-Helmholtz equation

$$\frac{\mathrm{d}\ln(\chi_A)}{\mathrm{d}T} = \frac{\mathrm{d}(\Delta_{\text{vap}}G/T)}{R\mathrm{d}T} = -\frac{\Delta_{\text{vap}}H}{RT^2}$$

Integrating from $\chi_A = 1 \ \text{to an arbitrary value}$

$$\int_{0}^{\ln(\chi_A} \mathrm{d}\ln(\chi_A) = \frac{-1}{R}\int_{T^*}^{T}\frac{\Delta_{\text{vap}}H}{T^2}\mathrm{d}T$$

$$\ln(\chi_A) = \ln(1-\chi_B) = \frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T}-\frac{1}{T^*}\right)$$

For dilute solutions, $\chi_B < 1$ so the approximate expression

$$\chi_B =\frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T^*}-\frac{1}{T}\right)$$

Since, $T \approx T^*$ $$\left( \frac{1}{T^*}-\frac{1}{T}\right) = \frac{T-T^*}{TT^*} \approx \frac{\Delta T_b}{T^{*2}}$$

Now, we can rearrange to get the desired expression,

$$ \Delta T_b = \overbrace{\left(\frac{RT^{*2}}{\Delta_{\text{vap}}H}\right)}^{K_b} \chi_B$$

Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ \ mol^{−1}}$ vs. $\mathrm{kJ\ kg^{−1}}$ The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself,I believe. So I am guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway.

Note, that the relation given above is in terms of molar fraction however it is easier to work with molalities when you are doing experiments. The same relation can thus be reformulated

$$\Delta T_b = \tilde{K_b}b$$

and this $\tilde{K_b}$ has the definition which involves $L_v$, I believe. (just look at the dimension in both the expressions and you'll see).

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    $\begingroup$ The formatting's beautiful, seriously! Thank you! And yeah, they have the same dimensions ^_^ $\endgroup$ – paracetamol Oct 30 '16 at 10:06
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Wikipedia has the same formula as your textbook .

The formula from Wikipedia $${K_b} = \frac{MRT^2_{\mathrm{b}}}{\Delta H_{\mathrm{vap}}}$$

Simply use the relation $\Delta H_{vap} =ML_{vap}$ and you will arrive at the equation in your textbook.

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