First-year courses · Sciences
General chemistry II: energy, rates and equilibrium
Thermochemistry, rate laws, equilibrium constants, pH, buffers, Ksp, electrochemical cells, entropy and Gibbs energy: second-term chemistry flashcards.
About this deck
The second term of first-year chemistry shifts from what molecules look like to how and why reactions happen. These cards cover enthalpy, calorimetry and Hess's law; reaction rates, rate laws, integrated rate laws, half-life and the Arrhenius equation; equilibrium constants, reaction quotients, ICE tables and Le Chatelier's principle; acids and bases, pH, Ka and Kb, titrations and buffers; solubility products and the common ion effect; redox, galvanic cells, standard potentials, the Nernst equation and electrolysis; and finally entropy and Gibbs free energy, which tie the course together.
You need first-term stoichiometry, comfort with logarithms (log and ln, and undoing them with 10^x and e^x) and confident use of a scientific calculator. Most examples assume 25 degrees C (298 K).
Many cards here are short calculations, so treat the example as the question: cover it, work it, then compare. For conceptual cards such as Le Chatelier's principle, predict the direction of shift for a new change you invent yourself before moving on. When reviewing, deliberately pair cards across tags: Gibbs energy with equilibrium constant, cell potential with Gibbs energy, Ka with buffers. Those links are the favourite subject of final exam questions.
Classic errors: including pure solids and liquids in equilibrium expressions; thinking a catalyst shifts the position of equilibrium (it only speeds up how fast it is reached); using the Henderson-Hasselbalch equation for a solution that is not actually a buffer; and mixing kJ and J when combining delta H and T delta S. Watch the sign conventions for cell potential as well: the cathode is where reduction happens, in every type of cell.
Before you start: First-term general chemistry (stoichiometry, gases, solutions) and comfort with logarithms.
All 45 cards
Cover the definitions with a sheet of paper and test yourself, or use the study modes above.
Thermochemistry
- Enthalpy change (delta H)
- The heat absorbed or released by a reaction at constant pressure; negative means exothermic (heat released), positive means endothermic.Example: Burning methane: delta H = -890 kJ per mol of CH4.
- Specific heat capacity
- Heat needed to raise 1 g of a substance by 1 degree C (or 1 K); q = mc delta T. Water's value is 4.184 J/(g degrees C).Example: Heating 100 g of water by 10 degrees C: q = 100(4.184)(10) = 4184 J.
- Calorimetry
- Measuring reaction heat from the temperature change of surroundings; heat released by the reaction is absorbed by the solution, so q_rxn = -q_solution.Example: Solution absorbs 4.18 kJ when 0.050 mol reacts: delta H = -4.18/0.050 = -83.6 kJ/mol.
- Hess's law
- Because enthalpy is a state function, the delta H of an overall reaction equals the sum of the delta H values of any steps that add up to it.Example: C to CO (-110.5 kJ) plus CO to CO2 (-283.0 kJ) gives C to CO2: -393.5 kJ.
- Standard enthalpy of formation
- delta H for forming 1 mol of a compound from its elements in their standard states; delta H_rxn = sum over products minus sum over reactants (times coefficients).Example: CH4 + 2 O2 to CO2 + 2 H2O(l): (-393.5 + 2(-285.8)) - (-74.8) = -890.3 kJ.
Kinetics
- Reaction rate
- The change in concentration of a reactant or product per unit time, adjusted by its coefficient so all species give the same value.Example: For 2 N2O5 to 4 NO2 + O2, NO2 forms 4 times as fast as O2.
- Rate law
- An equation rate = k[A]^m[B]^n, where the orders m and n are found from experiment, not from the balanced equation.Example: If rate = k[NO]^2[O2], the reaction is third order overall.
- Finding order from initial rates
- Compare experiments where only one concentration changes: the factor change in rate equals the concentration factor raised to the order.Example: Doubling [A] quadruples the rate: 2^m = 4, so m = 2.
- First-order integrated rate law
- ln[A] = ln[A]0 - kt; the half-life t1/2 = 0.693/k does not depend on starting concentration.Example: k = 0.0231 per min: t1/2 = 30 min, so after 90 min one-eighth remains.
- Second-order integrated rate law
- 1/[A] = 1/[A]0 + kt; the half-life 1/(k[A]0) gets longer as the reaction proceeds.Example: k = 0.50 M^(-1) s^(-1), [A]0 = 1.0 M: first half-life 2.0 s.
- Zero-order rate law
- Rate stays constant regardless of concentration: [A] = [A]0 - kt.Example: Drug release from some slow-release patches is roughly zero order.
- Arrhenius equation
- k = A e^(-Ea/RT): the rate constant grows with temperature and falls with activation energy Ea.Example: Ea = 50 kJ/mol: going from 300 K to 310 K makes k about 1.9 times larger.
- Catalyst
- A substance that speeds up a reaction by providing a lower-activation-energy pathway; it is not consumed and does not change the equilibrium constant.Example: Platinum in a car's catalytic converter speeds oxidation of CO to CO2.
- Rate-determining step
- The slowest elementary step in a reaction mechanism; it controls the overall rate, and its molecularity gives the rate law (if it is the first step).Example: For NO2 + CO to NO + CO2, the slow first step is NO2 + NO2 to NO3 + NO, so rate = k[NO2]^2.
Equilibrium
- Dynamic equilibrium
- The state where forward and reverse reactions proceed at equal rates, so concentrations stay constant although reactions continue.Example: In a sealed bottle of soda, CO2 keeps dissolving and escaping at equal rates.
- Equilibrium constant Kc
- Products over reactants at equilibrium, each concentration raised to its coefficient; pure solids and liquids are left out.Example: N2 + 3 H2 to 2 NH3: Kc = [NH3]^2/([N2][H2]^3).
- Reaction quotient Q
- Same expression as K but using current, not equilibrium, concentrations. Q < K: shifts toward products; Q > K: toward reactants; Q = K: at equilibrium.Example: K = 10, Q = 2: the reaction proceeds forward.
- Le Chatelier's principle
- A system at equilibrium that is disturbed shifts in the direction that partly offsets the change in concentration, pressure or temperature.Example: Adding N2 to the ammonia equilibrium shifts it toward NH3.
- Kp and Kc relationship
- For gas reactions, Kp = Kc(RT)^(delta n), where delta n is moles of gaseous products minus moles of gaseous reactants.Example: N2 + 3 H2 to 2 NH3 has delta n = -2.
- ICE table
- A table of Initial, Change and Equilibrium concentrations used to solve for unknown equilibrium amounts from K.Example: H2 + I2 to 2 HI, K = 64, start 1.0 M each: (2x)^2/(1 - x)^2 = 64, so x = 0.80 and [HI] = 1.6 M.
Acids and bases
- Bronsted-Lowry acid and base
- An acid donates a proton (H+); a base accepts one.Example: In NH3 + H2O to NH4+ + OH-, water acts as the acid.
- Conjugate acid-base pair
- Two species that differ by exactly one proton; the stronger an acid, the weaker its conjugate base.Example: NH4+ and NH3; H2O and OH-.
- pH
- -log[H3O+]; values below 7 are acidic and above 7 basic at 25 degrees C.Example: [H3O+] = 1.0 x 10^(-3) M gives pH 3.00.
- Ion product of water (Kw)
- [H3O+][OH-] = 1.0 x 10^(-14) at 25 degrees C, so pH + pOH = 14.Example: pOH 4.00 means pH 10.00.
- Strong acid
- One that ionizes essentially completely in water: HCl, HBr, HI, HNO3, HClO4 and H2SO4 (first proton).Example: 0.010 M HCl has [H3O+] = 0.010 M and pH 2.00.
- Weak acid and Ka
- An acid that only partly ionizes; Ka = [H3O+][A-]/[HA]. For a small Ka, [H3O+] is about sqrt(Ka C).Example: 0.10 M acetic acid, Ka = 1.8 x 10^(-5): [H3O+] = 1.34 x 10^(-3) M, pH 2.87.
- Ka times Kb equals Kw
- For a conjugate pair, Ka Kb = Kw, so pKa + pKb = 14 at 25 degrees C.Example: Acetic acid pKa 4.74, so acetate has pKb 9.26.
- Acidic and basic salts
- A salt's solution is basic if its anion is the conjugate base of a weak acid, and acidic if its cation is the conjugate acid of a weak base.Example: NaF solution is basic; NH4Cl solution is acidic; NaCl is neutral.
- Titration equivalence point
- The point where moles of added titrant exactly match the moles of analyte according to the reaction; pH there is 7 only for strong acid with strong base.Example: 25.0 mL of 0.100 M HCl needs 25.0 mL of 0.100 M NaOH.
Buffers and solubility
- Buffer
- A solution of a weak acid and its conjugate base (or weak base and conjugate acid) in similar amounts that resists pH change when small amounts of acid or base are added.Example: Acetic acid with sodium acetate; blood uses carbonic acid and bicarbonate.
- Henderson-Hasselbalch equation
- pH = pKa + log([A-]/[HA]) for a buffer; when the two concentrations are equal, pH = pKa.Example: 0.20 M acetate with 0.10 M acetic acid: pH = 4.74 + log 2 = 5.04.
- Solubility product (Ksp)
- The equilibrium constant for a sparingly soluble ionic solid dissolving into its ions; the solid itself is left out.Example: AgCl, Ksp = 1.8 x 10^(-10): molar solubility s = sqrt(Ksp) = 1.3 x 10^(-5) M.
- Common ion effect
- Adding an ion that already takes part in an equilibrium shifts it back, lowering the solubility of a salt or the ionization of a weak acid.Example: AgCl in 0.10 M NaCl: s = 1.8 x 10^(-10)/0.10 = 1.8 x 10^(-9) M.
- Predicting precipitation
- Calculate the ion product Q from mixed concentrations: a precipitate forms if Q > Ksp; none if Q < Ksp.Example: [Ag+] = [Cl-] = 1.0 x 10^(-4) M: Q = 1.0 x 10^(-8) > 1.8 x 10^(-10), so AgCl precipitates.
Electrochemistry
- Oxidation and reduction
- Oxidation is loss of electrons (oxidation number rises); reduction is gain of electrons (oxidation number falls). Memory aid: OIL RIG.Example: In Zn + Cu2+ to Zn2+ + Cu, zinc is oxidized and copper(II) is reduced.
- Oxidation numbers
- Bookkeeping charges assigned by rules (O usually -2, H usually +1, sum equals overall charge) to track electron transfer.Example: In MnO4-: x + 4(-2) = -1, so Mn is +7.
- Galvanic (voltaic) cell
- A cell that turns a spontaneous redox reaction into electrical work; oxidation occurs at the anode and reduction at the cathode, with electrons flowing anode to cathode.Example: Zinc anode and copper cathode joined by a wire and a salt bridge.
- Standard cell potential
- E_cell = E(cathode) - E(anode) using standard reduction potentials; a positive value means the reaction is spontaneous as written.Example: Zn/Cu: 0.34 - (-0.76) = 1.10 V.
- Gibbs energy and cell potential
- delta G = -nFE, where n is moles of electrons transferred and F = 96 485 C/mol.Example: Zn/Cu, n = 2, E = 1.10 V: delta G = -212 kJ.
- Nernst equation
- E = E(standard) - (0.0592/n) log Q at 25 degrees C; it gives the cell potential at nonstandard concentrations.Example: Zn/Cu with Q = [Zn2+]/[Cu2+] = 0.010: E = 1.10 + 0.0592 = 1.16 V.
- Electrolysis and Faraday's law
- Using electric current to drive a nonspontaneous reaction; moles of electrons = charge/F, where charge = current times time.Example: 2.00 A for 965 s is 0.0200 mol e-, depositing 0.0100 mol Cu (0.636 g).
Entropy and Gibbs energy
- Entropy (S)
- A measure of how dispersed energy and matter are, related to the number of microscopic arrangements; it increases from solid to liquid to gas.Example: Melting ice and forming gas from a solid both raise S.
- Second law of thermodynamics
- For any spontaneous process, the total entropy of the universe (system plus surroundings) increases.Example: Heat flows from a hot cup to cooler air, never the reverse on its own.
- Gibbs free energy
- delta G = delta H - T delta S; at constant T and P a process is spontaneous when delta G < 0.Example: delta H = -100 kJ, delta S = -0.200 kJ/K, 298 K: delta G = -40.4 kJ (spontaneous); above 500 K it is not.
- Standard Gibbs energy and K
- delta G(standard) = -RT ln K: a negative value means K > 1 (products favoured).Example: K = 1.0 x 10^3 at 298 K: delta G = -(8.314)(298) ln 1000 = -17.1 kJ/mol.