Chemistry 2720 Fall 2000 Assignment 4 Solutions

1.

The second law says that the entropy of the Universe always increases. The argument given only considers the entropy change of the substance being frozen and not of its surroundings. A substance can be frozen if the sum of these two contributions to the entropy change of the Universe is positive.

2.

(a)

The reaction is

$$\mathrm{CH_3CH_2OH_{(l)}} + 3\mathrm{O_{2(g)}} \rightarrow 2\mathrm{CO_{2(g)}} + 3\mathrm{H_2O_{(l)}}.$$

There is a decrease in the number of moles of gas. Moreover, ethanol can form only one hydrogen bond while three water molecules can be involved in multiple hydrogen bonding interactions. Both of these factors indicate an increase in the organization of the products relative to the reactants and so a decrease in entropy.

(b)

$$\begin{eqnarray*}\Delta\bar{S}^\circ & = & 2(213.74) + 3(69.91) - \left[160.67 +... ...\,K^{-1}mol^{-1}}\\ & = & -138.87\,\mathrm{J\,K^{-1}mol^{-1}} \end{eqnarray*}$$

 

(c)

We would have to adjust the entropies of the reagents and products to the new temperature. This is done by adding the entropy at $25^\circ\mathrm{C}$ to the change in entropy associated with heating each substance to $500^\circ\mathrm{C}$. For water and ethanol, we would need to calculate the change in entropy on warming the liquid to the boiling point, the entropy of vaporization at the boiling point and the change in entropy on warming the vapor to the final temperature. For oxygen and carbon dioxide, we would only need to calculate the entropy change on warming these gases from $25^\circ\mathrm{C}$ to $500^\circ\mathrm{C}$. Accordingly, we would need the following data:

• the entropies at $25^\circ\mathrm{C}$ (from the table in question 2b)
• the boiling temperatures of water and of ethanol
• the heats of vaporization of water and of ethanol at their respective boiling points
• the specific heat capacities of ethanol and water in both the liquid and vapor phases, of oxygen and of carbon dioxide

3.

The easiest way to answer this question is to calculate the standard free energy change:

$$\begin{eqnarray*}\Delta\bar{G}^\circ & = & 2\Delta\bar{G}^\circ_{f(\mathrm{CO_2}... ... - (-174.8)\,\mathrm{kJ/mol}\\ & = & -1325.5\,\mathrm{kJ/mol} \end{eqnarray*}$$

Since $\Delta\bar{G}^\circ<0$, the reaction is spontaneous under standard conditions.