Chemistry 2720 Fall 2000 Assignment 3 Solutions

1.

(a)

If we add the two reactions given, we almost have the required reaction:

$$\mathrm{C_6H_4(OH)_{2(aq)}} + \mathrm{H_2O_{2(aq)}} \righta... ..._2O_{(l)}} + \mathrm{H_{2(g)}} + \frac{1}{2}\mathrm{O_{2(g)}}$$

The last two terms are just water in its elemental form. If we now add in the formation reaction for water

$$\mathrm{H_{2(g)}} + \frac{1}{2}\mathrm{O_{2(g)}} \rightarrow... ...2O_{(l)}}\qquad\Delta\bar{H}^\circ = -285.830\,\mathrm{kJ/mol}$$

we get the desired reaction. The enthalpy change for the reaction of hydroquinone with hydrogen peroxide is therefore

$$\Delta\bar{H}^\circ = 177 + (-94.6) + (-285.830)\,\mathrm{kJ/mol} = -203\,\mathrm{kJ/mol}.$$

(b)

The density of water at $20^\circ\mathrm{C}$ is 998g/L. (Given the approximations being used in this problem, the density at $20^\circ\mathrm{C}$ is clearly a sufficiently close approximation to the initial density of the solvent.) The heat required to raise the temperature of a liter of water from 18 to $100^\circ\mathrm{C}$ is therefore approximately

$$q = (998\,\mathrm{g/L})(4.184\,\mathrm{J\,K^{-1}g^{-1}})(82\,\mathrm{K}) = 342\,\mathrm{kJ/L}.$$

The stoichiometry of the reaction is 1:1 and the reaction generates heat in the amount of 203kJ/mol. Thus, the required concentrations of hydroquinone and of hydrogen peroxide in the compartment in which the reaction occurs are both

$$c = \frac{342\,\mathrm{kJ/L}}{203\,\mathrm{kJ/mol}} = 1.7\,\mathrm{mol/L}.$$

(c)

The change in the number of equivalents of gas is $\Delta\bar{n}_\mathrm{gas} = 1$. Thus,

$$\begin{eqnarray*}\Delta\bar{E}^\circ & = & \Delta\bar{H}^\circ - RT\Delta\bar{n... ...^{-1}})(298.15\,\mathrm{K})(1)\\ & = & 174.5\,\mathrm{kJ/mol} \end{eqnarray*}$$

2.

First note that at $500^\circ\mathrm{C}$, the combustion of propane will produce water vapor. The enthalpy change at this temperature is calculated from the following cycle:

We calculate the enthalpy changes for each step:

$$\begin{eqnarray*}% latex2html id marker 117 \Delta\bar{H}^\circ_1 & = & (\bar{C... ...3.3 + 116.7\,\mathrm{kJ/mol}\\ & = & -2031.3\,\mathrm{kJ/mol} \end{eqnarray*}$$

3.

For an isothermal process like vaporization the change in entropy is just q_rev/T. In this case, we get

$$\Delta\bar{S} = \frac{84\,\mathrm{J/mol}}{4.22\,\mathrm{K}} = 20\,\mathrm{J\,K^{-1}mol^{-1}}.$$

4.

The entropy change is calculated by

$$\Delta S = \int\frac{dq_\mathrm{rev}}{T}.$$

For a simple heating process, we have $dq_\mathrm{rev} = n\bar{C}_P\,dT$ so that

$$\Delta S = n\bar{C}_P\int_{T_1}^{T_2}\frac{dT}{T} = n\bar{C}_P\ln\left(\frac{T_2}{T_1}\right).$$

In this case,

$$\begin{eqnarray*}% latex2html id marker 167 n & = & \frac{50\,\mathrm{g}}{119.00... ...c{326\,\mathrm{K}}{285\,\mathrm{K}}\right) = 3.0\,\mathrm{J/K} \end{eqnarray*}$$