Reaction rates depend on concentration, temperature, and the presence of a catalyst. The presence of a catalyst is the only thing that affects the Ea. “k” can never be negative (-). Intermediates are never used in rate laws. The rate law only involves reactants and catalyst of slowest step. A catalyst lowers the Ea, and it speeds up the slowest step (allowing it to reach equilibrium faster; it DOES NOT affect the equilibrium at all. The catalyst stabilizes the activated complex. Rate ONLY depends on concentration, so the size of the container DOES NOT matter. [Practice Problem Info Speed of Rate] A higher concentration will take longer. Even though higher concentration means higher rate, the rate is faster initially; however the conc. still has to fall toan intended molarity. The slow step is always the rate-determining step. In equilibrium expression DO NOT count puresolids or liquids. The only thing that affectsKC is temperature. ALWAYS look at “k” values to see what the overall order is in initial rate law if need be!Elementary reactions can ONLY be unimolecular (Aproducts) or bimolecular (A+Bproducts), where two molecules aresimultaneously colliding with one another. If they say a percentage of a substance has degraded subtract that percentage from 100, and use that value for how muchyou have left (e.g., 35% of A hasdegraded, so 100-35=65. We have 65% ofA left). Le Châtelier’s Principle – if a system is already at equilibrium and the conditions are changed so that it’s no longer at equilibrium, the system will react to reach anew equilibrium in a way that partially counteracts the change. Saturated system means that there is an equilibrium betweenthe solid and the ions in aqueous solution (rate of dissociation = rate of the ions crashing back into the solid phase). kobs would be equal to the slope (key word pseudo-first order and flooding). The kobs would also include the reagent with thehighest concentration times the rate constant (k). In the KC the concentrations are proportional to each other. Δ concentration of the products/Δ concentrations of the reactants. Low PressureHigh Volume & High PressureLow Volume. More moles of gasmeans high pressure. The equilibrium will want to shift to side with less number of moles. If there are equal number of moles of gas on both sides, the volume would not affect the equilibrium.  Temperature means that there will be fewer molecules that have enough KE to cross the Ea barrier.Fast RXNsmall Ea; Slow RXNlarge Ea[Substitution]: If a slow step is not the first step, then the concentrations in the rate law for the slow step must be replacedby concentrations of reactants; this can be done by using the earlier reaction steps. Effects of Termperature on the rate constant equation [USE ARRENIHUS EQUATION!]Enzymes: They can speed up rxns by factors of 109 to 1017. They can catalyze specific rxns, because they’re large molecules that bind to the reacting reactions. The active site of an enzyme is the part of the molecule that carries out catalysis. Only certain molecules (substrates) can fit in the active site, and the active site can adjust its structure to accommodate the substrate and can strainbonds or bring atoms into the right positionto react. High temperatures cause the enzyme to become denatured by changingthe enzyme structure and prevent the active site from working. Heterogeneous Catalysis – rnxs that take place on surfaces (nitrogen and oxygen onthe surface of platinum). DEMO: The NO molecules approach the platinum surface. The NO molecules form a bond w/ the platinum atoms in the surface. The molecules dissociate into N and O atoms, w/ each bonded to a platinum atom. The N and O atoms migrate across the surface until they getclose enough to another like atom to form N2 and O2. Finally, the product molecules leave the surface.When reversing a reaction always take the reciprocal of the specific KC value. [KC vs. Kp equation]:HIV1 Protease – The active protease has monomers that are attracted to itself, and the pre-proteins form dimers (by IFs). Scientists want to inhibit the protein production of the virus. The pre-protein must be cut apart to become biologically active, The cleavage enzyme cuts the pre-protein. In theory, the protease can’t cut the pre-protein if there’s anenzyme inhibitor. Thrombin in cows blood did not allow the blood to clot (the thrombin was the inhibitor); the cleavage enzyme formed blots in the cow’s blood by cutting its pre-proteins (forming active clots). DEMO – Iodine mixed w/ hexane and iodine mixed w/ H2O. Essentially the same rxn. As long as you have the same # of moles for the products and/or reactants, you can start w/ products or reactants to get the same rxn. Quantum Dots Experiment #1: Rhodamine 590: Dots flashdark and light, equilibrium being fluorescent and non-fluorescent. They’re excited by a green laser (excite the rhodamine dye to a high energy state by laser, and then the fluorescence goes out).It goes out, because the rxn occurring usesup the dark form.Experiment #2: The quantum dots are more stable than the rhodamine, and they also have a Bright/Dark equilibrium. If excited by a laser, they will fade much lessafter continuous laser excitation. The individual dots appear to be blinking. Theygo through stages where they are bright and where they are absorbing the fluorescence. Here, there is NOT a rxn using up the dark form.If Q = KC, then at equilibriumIf Q < KC, then rxn is product-favored (left to right)If Q > KC, then rxn is reactant-favored (right to left)Equilibrium is affected by: Concentration, Pressure, and TemperatureIf K is large (e.g., 1 x 105) or [K>1], products are favored. If K is small (e.g., 1 x 10-5) or [K<1], reactants are favored. IfK is close to 1 (e.g., 0.1 to 10) or [K≈1], neither the reactants nor products dominate significantly. Abs = εlc ε is the slope of the Beer’s law; “l” is the cell path length (1cm); “c” is the