Question

Calculate the values of $\mathit{\Delta }\mathit{S}$and $\mathit{\Delta }\mathit{G}$ for each of the following processes at 298 K:
${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{(}\mathbf{l}\mathbf{,}\mathbf{298}\mathbf{K}\mathbf{)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{(}\mathbf{g}\mathbf{,}\mathbf{V}\mathbf{=}\mathbf{1000}\mathbf{L}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{)}$

${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{(}\mathbf{l}\mathbf{,}\mathbf{298}\mathbf{K}\mathbf{)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{(}\mathbf{g}\mathbf{,}\mathbf{V}\mathbf{=}\mathbf{100}\mathbf{L}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{)}$
The standard enthalpy of vaporization for water at298 K is $\mathbf{44}\mathbf{.}\mathbf{02}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$. Does either of these processesoccur spontaneously?

Short answer

Here, the first reaction process is spontaneous and the second is non-spontaneous.

Step-by-step solution

Step 1: Introduction to the Concept

If the value of change in standard Gibbs free energy is negative, any reaction is spontaneous in nature. If the value of change in standard Gibbs free energy is positive, the reaction is non-spontaneous.

Step 2: Determination of volume

Let's break down the processes into two parts:$H_{2}O$vaporization and volume expansion.

The volume that$H_{2}O$ expands to as it vaporizes is computed as follows:

$V=\frac{nRT}{P}$

$=\frac{\left(1.00\text{mol}\right)\times (0.08206\times \frac{L\cdot atm}{K\cdot mol})\times \left(298\text{K}\right)}{\left(1.00\text{atm}\right)}$

$=24.5\text{L}$

Step 3: Determination of Entropy change  ΔSo

1. Using Hess's law and standard entropies, where n is the stochiometric coefficient.

$H_{2}O(l,298K)\to H_{2}O(g,298K,24.5L)$

Here, the Entropy change is:

$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}\mathbf{\sum }\mathit{n}{\mathit{S}}^{\mathbf{o}}\mathbf{\left(}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{d}\mathbf{u}\mathbf{c}\mathbf{t}\mathbf{s}\mathbf{\right)}\mathbf{-}\mathbf{\sum }\mathit{n}{\mathit{S}}^{\mathbf{o}}\mathbf{\left(}\mathbf{r}\mathbf{e}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{t}\mathbf{s}\mathbf{\right)}$

$=S^{\circ }\left[H_{2}O(g,298\text{K,24.5L})\right]-S^{\circ }\left[H_{2}O(l,298\text{K})\right]$

$=(189-70)J/K$

$=119J/K$

Step 4: Determination of change in free energy

2.To increase the volume of a molecule from $\mathbf{24}.\mathbf{5}\mathbf{L}/\mathbf{mol}\mathbf{to}\mathbf{1000}\mathbf{L}/\mathbf{mol}$

$\mathit{\Delta }{\mathit{S}}_{\mathbf{3}}\mathbf{=}\mathit{R}\mathbf{\times }\mathit{l}\mathit{n}\mathbf{\times }\left(\frac{V_2}{V_1}\right)$

$=\left(8.314{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\times \ln \times \left(\frac{1000}{24.5}\right)$

${\text{=30.8JK}}^{-1}{\text{mol}}^{-1}$

As a result,the total entropy change of the initial process is equal to the sum of entropy change in two steps:

$\Delta S=\Delta S_1+\Delta S_{vap}+\Delta S_3$

$\text{=}\left(\text{119+30.8}\right){\text{JK}}^{-1}{\text{mol}}^{-1}$

${\text{=150JK}}^{-1}{\text{mol}}^{-1}$

The change in free energy at 298 K is calculated using the formula below.

$\Delta G=\Delta H-T\Delta S$

$\text{=44.02KJ-}\left(298\text{K}\right)\times \left({\text{150JK}}^{-1}\right)\times \left(\frac{1\text{kJ}}{{10}^3\text{J}}\right)$

$\text{=44.02-44.7KJ}$

$\text{=-0.7KJ}$

This process is spontaneous because the change in free energy is negative.

For the other process, use the same two-step strategy.

Step 5: Determination of Entropy change for the vaporization step 

3.Use standard entropies with Hess's law for the vaporization stage, where n is the stochiometric coefficient.

$H_{2}O(l,298K)\to H_{2}O(g,298K,24.5L)$

$\Delta S^o=\sum n{S}^o\left(products\right)-\sum n{S}^o\left(reactants\right)$

$$\begin{gathered} =S^{\circ }\left[H_{2}O(g,298\text{K,24.5L})\right]-S^{\circ }\left[H_{2}O(l,298\text{K})\right] \\ =(189-70)J/K \\ =119J/K \end{gathered}$$

Step 6: Determination of change in free energy

4.To increase the volume of a molecule from

$\Delta S_3=R\times ln\times \left(\frac{V_2}{V_1}\right)$

$=\left(8.314{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\times \ln \times \left(\frac{100}{24.5}\right)$

${\text{=11.7JK}}^{-1}{\text{mol}}^{-1}$

As a result,the total entropy change of the initial process is equal to the sum of entropy changes in two steps:

$$\begin{gathered} \Delta S=\Delta S_1+\Delta S_{vap}+\Delta S_3 \\ \text{=}\left(\text{119+11.7}\right){\text{JK}}^{-1}{\text{mol}}^{-1} \\ {\text{=131JK}}^{-1}{\text{mol}}^{-1} \end{gathered}$$

The change in free energy at 298 K is calculated using the formula below.

$$\begin{gathered} \Delta G=\Delta H-T\Delta S \\ \text{=44.02KJ-}\left(298\text{K}\right)\times \left({\text{131JK}}^{-1}\right)\times \left(\frac{1\text{kJ}}{{10}^3\text{J}}\right) \end{gathered}$$

$$\begin{gathered} \text{=44.02-39.0KJ} \\ \text{=5.0KJ} \end{gathered}$$

Thisprocess is non-spontaneous because the change in free energy is positive.