Question

Consider the process,

$$\begin{gathered} \mathbf{A}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{A}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{75}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{}\mathbf{}\mathbf{155}\mathbf{°}\mathbf{C} \end{gathered}$$

which is carried out at constant pressure. The total $\mathbf{\Delta S}$for this process is known to be $\mathbf{75}\mathbf{.}\mathbf{0}{\mathbf{JK}}^{\mathbf{-}}{\mathbf{mol}}^{\mathbf{-}}$. For$\mathbf{A}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{andA}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{,}$the ${\mathbf{C}}_{\mathbf{p}}$values are $\mathbf{75}\mathbf{.}\mathbf{0}{\mathbf{JK}}^{\mathbf{-}}{\mathbf{mol}}^{\mathbf{-}}$and$\mathbf{29}\mathbf{.}\mathbf{0}{\mathbf{JK}}^{\mathbf{-}}{\mathbf{mol}}^{\mathbf{-}}$respectively, and are not dependent on temperature. Calculate ${\mathbf{\Delta H}}_{\mathbf{vaporization}}$for$\mathbf{A}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$at 125°C (its boiling point).

Short answer

${\mathbf{\Delta H}}_{\mathbf{vaporization}}$for$\mathrm{A}\left(\mathrm{l}\right)$at 125°C (its boiling point) is $\mathbf{24}\mathbf{.}\mathbf{9}\mathbf{KJ}\mathbf{/}\mathbf{mol}$.

Step-by-step solution

Introduction to the Concept

The entropy change, $\mathbf{\Delta S}$at temperature, T at constant pressure is determined by the given formula,

$\mathbf{\Delta S}\mathbf{=}\frac{\mathbf{\Delta H}}{\mathbf{T}}$

Here, $\mathbf{\Delta H}$is change in Enthalpy.

Step 2: Introduction to the calculation of ΔHvaporization

The given process is,

$\mathrm{A}\left(\mathrm{l}\right)\to \mathrm{A}\left(\mathrm{g}\right)$

Under the constant pressure

$∆{\mathrm{S}}_{\mathrm{total}}=75.0\mathrm{J}/\mathrm{Kmol}$

${\mathbf{C}}_{\mathbf{p}}$Of localid="1658384779782" $\mathrm{A}\left(\mathrm{l}\right)\mathrm{andA}\left(\mathrm{g}\right),$the values are localid="1658384786116" $75.0{\mathrm{JK}}^{-}{\mathrm{mol}}^{-}$and localid="1658384793544" $29.0{\mathrm{JK}}^{-}{\mathrm{mol}}^{-}$respectively.

And the conversion boiling point islocalid="1658384799659" $398\text{K}$

The change from localid="1658384808894" $\text{A}\left(\text{l}\right)\to \text{A}\left(\text{g}\right)$is divided into three steps.

localid="1658384817048" $\mathrm{A}(\mathrm{l},348\text{K})\to \mathrm{A}(\mathrm{l},398\text{K}) \left(1\right)$

localid="1658384824705" $\mathrm{A}(\mathrm{l},398\text{K})\to \mathrm{A}(\mathrm{g},398\text{K}) \left(2\right)$

localid="1658384832219" $\mathrm{A}(\mathrm{g},398\text{K})\to \mathrm{A}(\mathrm{g},428\text{K}) \left(3\right)$

Step 3: Determination of the calculation of ΔHvaporization

The change in entropy of step (1) and (3),

Step 1: Heating the liquid A to reach its boiling point

$\mathrm{A}(\mathrm{l},348\text{K})\to \mathrm{A}(\mathrm{l},398\text{K})$

The change in entropy expression as follows,

${\mathrm{\Delta S}}_1={\mathrm{C}}_{{\mathrm{P}}_{\mathrm{A}}\left(1\right)}\ln \left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$

Let’s put the values,

${\mathrm{\Delta S}}_1=(75.0\text{J/Kmol})\times \ln \left(\frac{398\text{K}}{348\text{K}}\right)$

${\mathrm{\Delta S}}_1=10.1\text{J/Kmol}$${\mathbf{\Delta S}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{1}\mathbf{\text{J/Kmol}}$

Step 3: Heating of gas A from 398K to 428K

localid="1658384907489" $\mathrm{A}(\mathrm{g},398\text{K})\to \mathrm{A}(\mathrm{g},428\text{K})$

The change in entropy expression as follows,

localid="1658384917060" ${\mathrm{\Delta S}}_3={\mathrm{C}}_{\mathrm{PA}\left(\mathrm{g}\right)}\ln \left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$

Let’s put the values,

localid="1658384924258" ${\mathrm{\Delta S}}_3=(29.0\text{J/Kmol})\times \ln \left(\frac{428\text{K}}{398\text{K}}\right)$

localid="1658384962841" ${\mathrm{\Delta S}}_3=2.1 \mathrm{J}/\mathrm{mol}$

Total entropy,

localid="1658384938751" $$\begin{gathered} {\mathrm{\Delta S}}_{\mathrm{total}}={\mathrm{\Delta S}}_1+{\mathrm{\Delta S}}_2+{\mathrm{\Delta S}}_3 \\ {\mathrm{\Delta S}}_2={\mathrm{\Delta S}}_{\mathrm{total}}-({\mathrm{\Delta S}}_1+{\mathrm{\Delta S}}_3) \\ {\mathrm{\Delta S}}_2=75\mathrm{J}/\mathrm{Kmol}-(10.1+2.1)\mathrm{J}/\mathrm{Kmol} \\ {\mathrm{\Delta S}}_2=62.8\mathrm{J}/\mathrm{Kmol} \end{gathered}$$

localid="1658384954722" $$\begin{gathered} {\mathrm{\Delta H}}_{\mathrm{vaporization}}={\mathrm{\Delta S}}_{\mathrm{Vap}}\times {\mathrm{T}}_{\mathrm{bp}} \\ =62.8\mathrm{J}/\mathrm{Kmol}\times 398\mathrm{K} \\ =24994.4\mathrm{J}/\mathrm{mol} \\ =24.9\mathrm{KJ}/\mathrm{mol} \end{gathered}$$

localid="1658384979221" ${\mathrm{\Delta H}}_{\mathrm{vaporization}}=24.9\mathrm{KJ}/\mathrm{mol}$

Therefore the localid="1658382461339" ${\mathbf{\Delta H}}_{\mathbf{vaporization}}$is localid="1658382489109" $\mathbf{24}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol}$.