Formula Sheet and Data Tables

The exam will have the printed version of the formula sheet.

These data tables can be used on assignments but will not be provided on the exam.  Relevant data will be given in the exam questions.

• Ionization constants of weak acids
• Ionization constants of weak bases
• Solubility product constants for sparingly soluble ionic compounds
• Standard thermodynamic properties of substances
• Standard reduction potentials

Constants and conversion factors

[latex]\begin{align*} R &=62.36 \text{ L}\cdot\text{torr }\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \\ &= 0.08206 \text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \\ &= 0.08314 \text{ L}\cdot\text{bar}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \\ &= 8.314 \text{ L}\cdot\text{kPa}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \\ &= 8.314 \text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \\ &= 8.314 \text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \end{align*}[/latex]

[latex]\begin{align*} F &= 96485 \text{ C} \cdot \text{mol}^{-1} \\ &= 96485 \text{ J} \cdot \text{mol}^{-1}\cdot \text{V}^{-1} \end{align*}[/latex]

[latex]1 \text{ atm }=101.325 \text{ kPa } = 760 \text{ torr} = \text{exactly } 760 \text{ mmHg}[/latex]

[latex]T(K) = T(^\circ \text{C})+273.15[/latex]

Unit 1 Gases

[latex]\begin{array}{ccc} PV=nRT &&& \dfrac{P_{\text{f}}V_{\text{f}}}{n_{\text{f}}T_{\text{f}}} = \dfrac{P_{\text{i}}V_{\text{i}}}{n_{\text{i}}T_{\text{i}}} &&&E_{\text{K, ave}}= (3/2)RT/N_{\text{A}} \\\\ u_{\text{rms}}= \sqrt{3RT/M} &&&M=\frac{mRT}{PV} &&&d=\frac{PM}{RT} \\\end{array}[/latex]

[latex]\begin{array}{cc} \frac{\text{rate (diffusion or effusion) of gas A}}{\text{rate (diffusion or effusion) of gas B}} = \sqrt{\frac{\text{molar mass of gas B}}{\text{molar mass of gas A}}} \\\\ \frac{\text{rate (diffusion or effusion) of gas A}}{\text{rate (diffusion or effusion) of gas B}} = \sqrt{\frac{\text{density of gas B}}{\text{density of gas A}}}\end{array}[/latex]

[latex]\begin{array}{ll}P_{\text{tot}}= P_{\text{A}}+P_{\text{B}}+\dots &&& x_\text{A}= \frac{n_{\text{A}}}{n_{\text{tot}}}=\frac{P_{\text{A}}}{P_{\text{tot}}}=\frac{V_{\text{A}}}{V_{\text{tot}}}\end{array}[/latex]

Unit 2 Kinetics

For a reaction a A + b B ⟶ c C + d D,

[latex]\text{reaction rate }=k[\text{A}]^x[\text{B}]^y=-\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t}=-\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta t}=\frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t}=\frac{1}{d}\frac{\Delta[\text{D}]}{\Delta t}[/latex]

[latex]k=Ae^{\frac{-E_{\text{a}}}{RT}}[/latex] [latex]\ln k=\frac{-E_{\text{a}}}{R}\frac{1}{T}+\ln A[/latex] [latex]\ln\left(\frac{k_2}{k_1}\right) = \frac{−E_\text{a}}{R}\left(\frac{1}{T_2}−\frac{1}{T_1}\right)[/latex]

For first-order and pseudo first-order reactions,

[latex]\ln \frac{[\text{A}]}{[\text{A}]_0}=-kt[/latex] [latex]\ln [\text{A}]=-kt+\ln [\text{A}]_0[/latex] [latex]t_{1/2}=\frac{0.693}{k}[/latex]

Unit 4 Acid-Base and Solubility Equilibria

Acid ionization, HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

[latex]K_{\text{a}}=\frac{[\text{H}_3\text{O}^+][\text{A}^−]}{[\text{HA}]}[/latex] [latex]\text{p}K_{\text{a}}=-\log K_{\text{a}}[/latex] [latex]\text{% ionization}=\frac{[\text{H}_3\text{O}^+]_{\text{eq}}}{[\text{HA}]_\text{initial}}×100\%[/latex]

Base ionization, B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq)

[latex]K_{\text{b}}=\frac{[\text{HB}^+][\text{OH}^−]}{[\text{B}]}[/latex] [latex]\text{p}K_{\text{b}}=-\log K_{\text{b}}[/latex] [latex]\text{% ionization}=\frac{[\text{OH}^-]_{\text{eq}}}{[\text{B}]_\text{initial}}×100\%[/latex]

[latex]\text{pH}=-\log[\text{H}_3\text{O}^+][/latex]          [latex][\text{H}_3\text{O}^+]=10^{-\text{pH}}[/latex]          [latex]\text{pOH}=-\log[\text{OH}^-][/latex]          [latex][\text{OH}^-]=10^{-\text{pOH}}[/latex]

At 25 ºC,  [latex]K_{\text{w}} = 1.0\times10^{-14}[/latex] and [latex]\text{p}K_{\text{w}}=14.00[/latex]

\begin{array}{cc} K_{\text{w}}=1.0\times10^{-14} = [\text{H}_3\text{O}^+][\text{OH}^-] &&& K_{\text{w}}=1.0\times10^{-14}=K_{\text{a, conj acid}} \times K_{\text{b, conj base}} \\\\
\text{p}K_{\text{w}}=14.00=\text{pH}+\text{pOH} &&& \text{p}K_{\text{w}} = 14.00 = \text{p}K_{\text{a, conj acid}}+\text{p}K_{\text{b, conj base}} \end{array}

[latex]\text{pH}=\text{p}K_{\text{a}}+\log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)[/latex]

Dissolution of a sparingly soluble ionic compound

M_pX_q(s) ⇌ p M^m+(aq) + q Xⁿ⁻(aq) [latex]K_{\text{sp}}=[\text{M}^{m+}]^p\;[\text{X}^{n-}]^q[/latex]

Unit 5 Thermodynamics

[latex]\begin{array}{ccc} \Delta U=q+w && q=mc\Delta T && w=-P\Delta V \end{array}[/latex]

At constant volume, [latex]\Delta U = q[/latex]At constant pressure, [latex]\Delta H = q[/latex]

[latex]\begin{array}{l} \Delta H^\circ = \sum{n} \times \Delta H^\circ_{\text{f}} (\text{products}) - \sum{n} \times \Delta H^\circ_{\text{f}}(\text{reactants}) \\\\ \Delta S^\circ = \sum{n} \times S^\circ (\text{products}) - \sum{n} \times  S^\circ(\text{reactants}) \\\\ \Delta G^\circ = \sum{n} \times \Delta G^\circ_{\text{f}} (\text{products}) - \sum{n} \times \Delta G^\circ_{\text{f}}(\text{reactants})  \end{array}[/latex]

[latex]\Delta G = \Delta H - T \Delta S \Rightarrow[/latex] at standard state, [latex]\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ[/latex]

[latex]\Delta G_{\text{r}}=\Delta G^\circ_{\text{r}}+RT \ln Q[/latex][latex]\Delta G^\circ_{\text{r}}=-RT \ln K[/latex] [latex]K=e^{-\frac{\Delta G^\circ_{\text{r}}}{RT}}[/latex]

Unit 6 Electrochemistry

[latex]E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}[/latex]

[latex]E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{RT}{nF}\ln Q \Rightarrow[/latex]  at 298.15 K, [latex]E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.0592 \text{ V}}{n}\log Q[/latex]

[latex]E^\circ_{\text{cell}}=\frac{RT}{nF}\ln K \Rightarrow[/latex] at 298.15 K, [latex]E^\circ_{\text{cell}}=\frac{0.0592 \text{ V}}{n}\log K[/latex]

[latex]\Delta G=-nFE_{\text{cell}} \Rightarrow[/latex] at standard state, [latex]\Delta G^\circ=-nFE_{\text{cell}}^\circ[/latex]