Chemistry 2720 Fall 2000 Test 1 Solutions

1.

(a)

The mass of water is

$$m_\mathrm{H_2O} = 0.14(15\,\mathrm{g}) = 2.1\,\mathrm{g}.$$

The heat required to warm the water to $100^\circ\mathrm{C}$ is

$$q_1 = m_\mathrm{H_2O}\tilde{C}_P\Delta T = (2.1\,\mathrm{g})... ...\mathrm{J\,K^{-1}g^{-1}})(80\,\mathrm{K}) = 0.70\,\mathrm{kJ}.$$

The heat required to vaporize the water once it reaches the boiling point is

$$q_2 = m_\mathrm{H_2O}\Delta\tilde{H}_\mathrm{vap} = (2.1\,\mathrm{g})(2257\,\mathrm{J/g}) = 4.74\,\mathrm{kJ}.$$

The total heat required is therefore

$$q = 0.70+4.74\,\mathrm{kJ} = 5.4\,\mathrm{kJ}.$$

(b)

The process under consideration is

$$\mathrm{H_2O_{(l)}} \rightarrow\mathrm{H_2O_{(g)}}.$$

The equilibrium constant for this process is $K = a_\mathrm{g}/a_\mathrm{l} = (P/P^\circ)/X$. For pure water, X=1 so that $K=(P/P^\circ)$. Therefore

$$\ln\left(\frac{P_1}{P_2}\right) = \frac{\Delta\bar{H}^\circ}{R} \left(\frac{1}{T_2}-\frac{1}{T_1}\right).$$

In our case, take $T_1=373.15\,\mathrm{K}$ and $P_1 = 101\,325\,\mathrm{Pa}$. We want to find the boiling temperature corresponding to the pressure $P_2 = 1.5\times 10^5\,\mathrm{Pa}$. The enthalpy of vaporization of water at $100^\circ\mathrm{C}$ is 40.66kJ/mol. Therefore

$$\begin{eqnarray*}% latex2html id marker 43 \frac{1}{T_2} & = & \frac{R}{\Delta\... ...erefore T_2 & = & 384.66\,\mathrm{K}\equiv 112^\circ\mathrm{C} \end{eqnarray*}$$

(c)

The pressure is constant, so the work done is

$$w = -P_\mathrm{ext}\Delta V.$$

The initial volume is

$$V_1 = \frac{20\times 10^{-3}\,\mathrm{L}}{1000\,\mathrm{L/m^3}} = 2\times 10^{-5}\,\mathrm{m^3}.$$

The final volume is 40 times larger:

$$V_2 = 40V_1 = 8\times 10^{-4}\,\mathrm{m^3}.$$

The pressure, in Pascals, is

$$P = (0.9\,\mathrm{bar})(10^5\,\mathrm{Pa/bar}) = 9\times 10^4\,\mathrm{Pa}.$$

The work is therefore

$$w = -(9\times 10^4\,\mathrm{Pa})(8\times 10^{-4} - 2\times 10^{-5}\,\mathrm{m}^3) = -70\,\mathrm{J}.$$

2.

(a)

The maximum work is $-\Delta\bar{G}$. In this case, the reaction only involves solids, so $\Delta\bar{G}=\Delta\bar{G}^\circ$.

$$\begin{eqnarray*}% latex2html id marker 85 \Delta\bar{G}^\circ & = & \Delta\bar... ...imes 10^{-3}\,\mathrm{kg/mol}}\\ & = & -4.15\,\mathrm{MJ/kg} \end{eqnarray*}$$

The maximum specific electrical work is therefore 4.15MJ/kg.

(b)

We need to find the limiting reagent:

$$\begin{eqnarray*}n_\mathrm{Zn} & = & \frac{10\,\mathrm{g}}{65.38\,\mathrm{g/mol}... ...c{15\,\mathrm{g}}{86.936\,\mathrm{g/mol}} = 0.17\,\mathrm{mol} \end{eqnarray*}$$

Since we need two stoichiometric equivalents of MnO₂ per stoichiometric equivalent of zinc, the former is clearly limiting. Accordingly, the amount of zinc that reacts is

$$n_\mathrm{Zn,used} = n_\mathrm{MnO_2}/2 = 0.086\,\mathrm{mol}.$$

The maximum total work is therefore

$$w'_\mathrm{max} = (271.1\,\mathrm{kJ/mol})(0.086\,\mathrm{mol}) = 23\,\mathrm{kJ}$$

3.

(a)

At equilibrium, $\Delta\bar{G} = 0$ so

$$\Delta\bar{G}^\circ = -RT\ln K.$$

We therefore need K for the reaction

$$\mathrm{TlF_{4(s)}} \rightarrow \mathrm{Tl^{4+}} + 4\mathrm{F^-_{(aq)}}$$

The equilibrium activity of Tl⁴⁺ is $4.5\times 10^{-4}$. We get four equivalents of fluorine for every thallium ion, so $a_\mathrm{F^-} = 4(4.5\times 10^{-4}) = 1.8\times 10^{-3}$. Thus

$$\begin{eqnarray*}% latex2html id marker 134 K & = & (a_\mathrm{Tl^{4+}})(a_\math... ...\mathrm{K})\ln(4.7\times10^{-15})\\ & = & 82\,\mathrm{kJ/mol} \end{eqnarray*}$$

(b)

The free energy of reaction can be written

$$\Delta\bar{G}^\circ = \Delta\bar{G}^\circ_{f\mathrm{(Tl^{4+}... ...irc_{f\mathrm{(F^-)}} - \Delta\bar{G}^\circ_{f\mathrm{(TlF_4)}}$$

Rearranging, we get

$$\Delta\bar{G}^\circ_{f\mathrm{(TlF_4)}} = \Delta\bar{G}^\ci... ...+ 4\Delta\bar{G}^\circ_{f\mathrm{(F^-)}} - \Delta\bar{G}^\circ.$$

Accordingly, if we can obtain values for the standard free energies of formation of thallium (IV) ions and of fluoride ions, we can calculate the standard free energy of formation of TlF₄.