Chemistry Key Equations Reference

$$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $$

$$ E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K $$

$$ E^\circ_{\text{cell}} = \frac{0.0257 \text{ V}}{n} \ln K = \frac{0.0592 \text{ V}}{n} \log K \quad (\text{at } 298.15 \text{ K}) $$

$$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q $$

$$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592 \text{ V}}{n} \log Q \quad (\text{at } 298.15 \text{ K}) $$

$$ \Delta G = -nFE_{\text{cell}} $$

$$ \Delta G^\circ = -nFE^\circ_{\text{cell}} $$

$$ w_{\text{elc}} = w_{\text{max}} = -nFE_{\text{cell}} $$

$$ Q = I \times t = n \times F $$

$$ \Delta S = \frac{q_{\text{rev}}}{T} $$

$$ S = k \ln W $$

$$ \Delta S = k \ln \frac{W_f}{W_i} $$

$$ \Delta S^\circ = \sum \nu S^\circ(\text{products}) - \sum \nu S^\circ(\text{reactants}) $$

$$ \Delta S = \frac{q}{T} $$

$$ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} $$

$$ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \Delta S_{\text{sys}} + \frac{q_{\text{surr}}}{T} $$

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

$$ q = c \times m \times \Delta T = c \times m \times (T_{\text{final}} - T_{\text{initial}}) $$

$$ \Delta U = q + w $$

$$ \Delta H^\circ_{\text{reaction}} = \sum n \times \Delta H^\circ_f(\text{products}) - \sum n \times \Delta H^\circ_f(\text{reactants}) $$

$$ \text{For } mA + nB \rightleftharpoons xC + yD\text{:} $$

$$ Q_c = \frac{[C]^x[D]^y}{[A]^m[B]^n} $$

$$ Q_P = \frac{P_C^{\,x}\,P_D^{\,y}}{P_A^{\,m}\,P_B^{\,n}} $$

$$ P = MRT $$

$$ K_c = Q_c \text{ at equilibrium} $$

$$ K_p = Q_p \text{ at equilibrium} $$

$$ K_P = K_c(RT)^{\Delta n} $$

$$ M_pX_q(s) \rightleftharpoons pM^{m+}(aq) + qX^{n-}(aq) $$

$$ K_{sp} = [M^{m+}]^p[X^{n-}]^q $$

$$ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad (\text{at } 25°\text{C}) $$

$$ \text{pH} = -\log[\text{H}_3\text{O}^+] $$

$$ \text{pOH} = -\log[\text{OH}^-] $$

$$ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} $$

$$ [\text{OH}^-] = 10^{-\text{pOH}} $$

$$ \text{pH} + \text{pOH} = pK_w = 14.00 \quad (\text{at } 25°\text{C}) $$

$$ K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} $$

$$ K_b = \frac{[\text{HB}^+][\text{OH}^-]}{[\text{B}]} $$

$$ K_a \times K_b = 1.0 \times 10^{-14} = K_w $$

$$ pK_a = -\log K_a $$

$$ pK_b = -\log K_b $$

$$ \text{pH} = pK_a + \log \frac{[\text{A}^-]}{[\text{HA}]} $$

$$ \text{Percent ionization} = \frac{[\text{H}_3\text{O}^+]_{\text{eq}}}{[\text{HA}]_0} \times 100 $$

$$ \text{For } aA \to bB: \quad -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = \frac{1}{b}\frac{\Delta[B]}{\Delta t} $$

$$ [A]_t = -kt + [A]_0 \quad \text{(integrated rate law)} $$

$$ t_{1/2} = \frac{[A]_0}{2k} \quad \text{(half-life)} $$

$$ \ln[A]_t = -kt + \ln[A]_0 \quad \text{(integrated rate law)} $$

$$ t_{1/2} = \frac{0.693}{k} \quad \text{(half-life)} $$

$$ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \quad \text{(integrated rate law)} $$

$$ t_{1/2} = \frac{1}{[A]_0 k} \quad \text{(half-life)} $$

$$ k = Ae^{-E_a / RT} $$

$$ \ln k = \frac{-E_a}{R}\left(\frac{1}{T}\right) + \ln A $$

$$ \ln \frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$

$$ PV = nRT $$

$$ P = \frac{F}{A} $$

$$ P = h\rho g $$

$$ P_{\text{Total}} = P_A + P_B + P_C + \cdots = \sum P_i $$

$$ P_A = X_A P_{\text{Total}} $$

$$ X_A = \frac{n_A}{n_{\text{Total}}} $$

$$ \text{Rate of diffusion} = \frac{\text{amount of gas passing through an area}}{\text{unit of time}} $$

$$ \frac{\text{Rate of effusion of gas A}}{\text{Rate of effusion of gas B}} = \sqrt{\frac{m_B}{m_A}} = \sqrt{\frac{M_B}{M_A}} $$

$$ u_{\text{rms}} = \sqrt{\overline{u^2}} = \sqrt{\frac{u_1^2 + u_2^2 + u_3^2 + u_4^2 + \cdots}{n}} $$

$$ KE_{\text{avg}} = \frac{3}{2}RT $$

$$ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$

$$ Z = \frac{\text{molar volume at same } T \text{ and } P}{\text{ideal molar volume at same } T \text{ and } P} = \left(\frac{P \cdot V_m}{RT}\right)_{\text{measured}} $$

$$ \left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT \quad \text{(van der Waals equation)} $$

$$ C_i = kP_i \quad \text{(Henry's Law)} $$

$$ P_A = X_A P_A^* \quad \text{(Raoult's Law)} $$

$$ P_{\text{solution}} = \sum P_i = \sum X_i P_i^* $$

$$ P_{\text{solution}} = X_{\text{solvent}} P^*_{\text{solvent}} $$

$$ \Delta T_b = K_b m $$

$$ \Delta T_f = K_f m $$

$$ \Pi = MRT $$

$$ h = \frac{2T\cos\theta}{r\rho g} $$

$$ P = Ae^{-\Delta H_{\text{vap}} / RT} $$

$$ \ln P = \frac{-\Delta H_{\text{vap}}}{RT} + \ln A $$

$$ \ln\frac{P_1}{P_2} = \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$

$$ n\lambda = 2d\sin\theta $$

$$ c = \lambda\nu $$

$$ E = h\nu = \frac{hc}{\lambda}, \quad \text{where } h = 6.626 \times 10^{-34} \text{ J·s} $$

$$ \frac{1}{\lambda} = R_\infty\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $$

$$ E_n = \frac{-kZ^2}{n^2}, \quad n = 1, 2, 3, \ldots $$

$$ \Delta E = kZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $$

$$ r = \frac{n^2}{Z}a_0 $$

$$ \text{Bond order} = \frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2} $$

$$ \text{Formal charge} = \text{valence electrons (free atom)} - \text{lone pair electrons} - \frac{1}{2}\text{bonding electrons} $$

$$ XY(g) \to X(g) + Y(g), \quad D_{X-Y} = \Delta H^\circ $$

$$ \Delta H = \sum D_{\text{bonds broken}} - \sum D_{\text{bonds formed}} $$

$$ \text{For a solid MX: } MX(s) \to M^{n+}(g) + X^{n-}(g), \quad \Delta H_{\text{lattice}} $$

$$ \text{For an ionic crystal: } \Delta H_{\text{lattice}} = \frac{C(Z^+)(Z^-)}{R_0} $$

$$ \%X = \frac{\text{mass } X}{\text{mass compound}} \times 100\% $$

$$ \frac{\text{molecular or molar mass}}{\text{empirical formula mass}} \text{ (amu or g/mol)} = n \text{ formula units/molecule} $$

$$ (A_xB_y)_n = A_{nx}B_{ny} $$

$$ M = \frac{\text{mol solute}}{\text{L solution}} $$

$$ C_1V_1 = C_2V_2 $$

$$ \text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 $$

$$ \text{ppm} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^6 \text{ ppm} $$

$$ \text{ppb} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^9 \text{ ppb} $$

$$ \text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 $$

$$ \text{Average mass} = \sum(\text{fractional abundance} \times \text{isotopic mass}) $$

$$ \text{Density} = \frac{\text{mass}}{\text{volume}} $$

$$ T_{\text{°C}} = \frac{5}{9}(T_{\text{°F}} - 32) $$

$$ T_{\text{°F}} = \frac{9}{5}T_{\text{°C}} + 32 $$

$$ T_{\text{K}} = \text{°C} + 273.15 $$

$$ T_{\text{°C}} = \text{K} - 273.15 $$