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Kinetics and Rates To find the rate of reaction, find all of the rates of gain or loss of the individual components and then divide each of those by the stoichiometric ratio and take the absolute value

Autoionization of Water Reaction H2O (l) + H2O (l) ↔ H3O (aq) + OH (aq)

Rate Law -

The rate law is the expression of the dependence of rate on the concentration of the reactants The superscripts “x” and “y” are called the orders of the reaction and represents that face

K w = [H3O+] [OH-] = 1.0 x 10^14 x^2 = 1.0 x 10^14 (at 298K) x = 1.0 x 10^7 pH = -log(1.0x10^7)

that the rate does not have to depend linearly on the concentration and they are usually integers or half integers k is the rate constant (constant for a reaction at a given temperature) Rate = k x y [a] [b] To determine the rate law, you need The order of the reaction with respect to each reactant The rate constant

pH = 7 Ka x Kb = Kw The only issue is when you having another acid or base in the reaction. If there was an acid, there will be an initial and final H3O+ concentration. If there was a base, there will be an initial and final OHconcentration. In this case, just use 14 = -log[H3O+] + -log[OH-] because you can plug in the concentration you know from the other acid or the other base. Polyprotic Acid (or Base) Reactions Occurs when an acid donates more than one proton or when a base accepts more than one

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Rate law units are always... (M ) (mol) (atm) (s) (min ) (hr) (yr ) (one of the top over one of the bottom) atm = 12 (atm)(atm)(atm) - for example if you have 3 reactants yr atm Arrhenius Arrhenius Equation stuff in notebook 1. Find the rate law using the data that has the most at one temperature 2. To get the activation energy, you need to find K at 2 different temperatures

a. 3.

ln( K1 )= K2

−E a R

1

( T1 -

1 T2

)

Then use the same equation to figure out K at 400 1 a a. ln( K atK2400 ) = −E ( T12 - 400K ) R

b. When you get to having just the ln in the equation, e both sides to get rid of it Integrated Rate Law See Examples General Equilibrium Equilibrium constant = [products] K eq = [reactants] If K is large, most of the reactants form products If K is small, most of the reactants stay reactants You do the concentration of products when you get to an equilibrium rate divided by the concentration of reactants when you get to an equilibrium rate and then raise them to their stoichiometric coefficient 6 Ex. K eq = [2C ] [B] [A]

2H2 (g) + O2 (g) ↔ 2H2O (l) 2 1 K eq = [H 22 O] = 2 [H2] [O2]

[H2] [O2]

Pure liquids and solids don’t have a concentration so they are just 1 2Na (s) + Cl2 (g) ↔ 2NaCl (s) K eq = [Cl12] Kc = equilibrium constant with concentrations of species Kp = equilibrium constant with partial pressures of species There is always a Kc (units of concentration) There is always a Keq (if the reaction has mixed phases, this will have mixed units) There is only a Kp is that reaction only has units of pressure and it will equal Keq 2 main problems: 1. You know the concentrations and you are calculating the equilibrium constant 2. You know the equilibrium constant and you are trying to figure out the concentrations All equilibrium problems have 3 parts: 1. Balanced Equation 2. Equilibrium constant expression 3. ICE CHART!!! Since K depends on stoichiometry, it doesn’t have consistent units. It is always written as a unitless quantity. Acid-Base Equilibrium An acid is a proton donor A base is a proton acceptor Most common acid? Water!! It has a proton it can donate Most common base? Water!! It can accept a proton H-OH (H can come off and OH can accept a proton) If you have an acid or a base, and you make it an aqueous solution, you automatically have an acid base reaction -

pH is actually a concentration pH is -log[H+] = -log[H3O+] [H+] should be in M pH is the only concentration of H+

Acids are either H with something electronegative (Cl, F, Br) or H with polyatomics Acid Dissociation Reactions - acids with water and water is the base ( K a ) HOAc + H2O ⇔ OAc + H3O pH is -log[H+] = -log[H3O+] -

pH is not the concentration of the acid, it is the concentration of the H+ ,or the H3O+ , that fell of the acid Acids have to have a proton and be attached to something chemically stable as a negative ion

General form of K a reactions HA (aq) + H2O (l) ↔ A- (aq) + H3O + (aq) [H 3 O] K a = [A] [HA] Base Dissociation Reactions - bases with water and water is the acid (K b) N, P, O molecules tend to be bases (but not always) pOH is -log[OH-] General Form of K b reactions B (aq) + H2O (l) ↔ HB+ (aq) + OH- (aq) K b = [HB][B[OH] ] Thermochemistry Continued (Q) Reactant Quotient - equilibrium constant, but not at equilibrium with initial concentrations

proton H3PO4 is triprotic H3PO4 (aq) + H2O (l) - H2PO4- (aq) + H3O+ (aq) H2PO4- (aq) + H2O (l) - HPO4 2- (aq) + H3O+ (aq) HPO4 2- (aq) + H2O (l) - PO4 3- (aq) + H3O+ (aq) They all have different K a ’s You solve 3 equilibrium problems to get the best pH value -When you do the polyprotic acid of sulfuric acid, you will get a very dilute acid. You can’t ignore Kw anymore because the hydronium concentration is less than 10^-7. Therefore, you have do find Kw using the autoionization of water plugging in the concentration you found from the dissociation of sulfuric acid. You will also know you have to do that because you get a basic pH for your acid. You always solve these problems from the highest K to the lowest K. That is why you can’t just do the autoionization of water first. Buffers A buffer is special because it resists changes in pH if you add something else A buffer is just an acid (or base) and its conjugate base (or acid) in the same equation There is a short cut for buffers, but only for buffers HENDERSON-HASSELBALCH [base] pH = pKa + log ( [acid] ) [acid]

pOH = pKb + log ( [base]) Besides having an acid (HAOc) and a base (NaOAc), which is a salt, making a buffer, there is also a buffer when there is a acid (HCl for example) and a base (NH3 for example). However, for the one between HCl and NH3, you also have water so you have 2 acids and 2 bases. 1. Make a list of acids and bases 2. Write out all equations and K’s 3. Do the ice charts from biggest K to smallest K until you find a buffer or get to the end 4. Find pH Buffer Capacity The amount of acid and base that the solution can absorb without changing the pH more than 1 pH unit +/- 1 pH unit (Rule of Thumb) pH = 4.00 buffer pH stays between 3 and 5, it is still considered “buffered” 1. 2. 3. 4.

Solve for the buffer’s pH Add 1 and subtract 1 from that pH Solve for x if you add an acid in H-H Solve for x if you add a base in H-H

Salts For All Salts In Water 1. Separate into ions 2. Cation is the conjugate acid and anion is the conjugate base 3. For the cation - either add an OH (for metal ions with no H) or, if there is one, remove the H to see what base it “came from” 4. For the anion - add an H to see what acid it “came from” 5. Look up the Ka and Kb of the “parent” acid/base 6. Divide the Kw by those to get the K’s you need 7. Ignore all strong acids and bases (FB or FS) TITRATION SEE EXAMPLES Thermochemistry ΣΔ H f Products - (ΣΔ H f Reactants) = Δ H rxn (measured in J or kJ) You have to add the products and subtract that from the sum of the reactants as well You need to take into account the coefficients of the balanced equation as well Entropy - is a measure of the distribution of states; sometimes defined as “disorder” or “randomness”, but it is really more complicated than that and represents the total number of different micro-states available to the system Entropy is a state function ΔS (Measured in J/K or kJ/K) Solid - liquid - gas (entropy increases as you go to the right) Mixtures have more entropy than pure substances Laws of Thermodynamics 1. Conservation of energy 2. 3. -

The entropy of the universe is always increasing for spontaneous changes A perfect crystal at 0K has no entropy ΔH0, then it will happen (PREFERED) ΔH>0 and ΔS 0 then the reaction is not spontaneous (spontaneous in the opposite direction, to the left) If ΔG < 0 then the reaction is spontaneous to the right If ΔG = 0 then the reaction is at equilibrium ΔG decides whether the bad ΔH wins or the good ΔS wins or vise versa if they are the same

Q=

[products not at eq uilibrium] [reactants not at equilibrium]

Compare Q to K Q is bigger than K - I either have too many products or not enough reactants (Reaction has to go to the left) Q is smaller than K - I either have too many reactants or not enough products (Reaction has to go to the right) Q is equal to K - you are at equilibrium ΔG = ΔG(formation) + RT ln(Q) This also gives us a way to calculate K When we reach equilibrium Q = K When we reach equilibrium ΔG = 0 ΔG forward = ΔG(formation) + RT ln(Q) ΔG reverse = -ΔG forward = -ΔG(formation) + RT ln(1/Q) ΔG reverse = -ΔG forward = 0 0 = ΔG(formation) + RT ln(K) ΔG(formation) = ΔH - TΔS = -RT ln(K) KSP Solubility = MAXIMUM amount of compound that can dissolve in water (this is actually an equilibrium) Ksp is the maximum amount of that solvent I can dissolve in water Ksp doesn’t get affected by the solvent -

Ksp reactions don’t get to equilibrium The “K” is the PRODUCT of the SOLUBLE ions. Hence, this reaction is called a “solubility product” Return to stoichiometry If you have +3x and +2x, you have to do [3x]^3[2x]^2 for the Ksp equation Solubility product - when you dissolve a solid in water Precipitation reaction - when the ions turn into the solid In a precipitation reaction, the K is just “upside down” AB (s) → A+ (aq) + B- (aq) Ksp = small (REGULAR KSP) A+ (aq) + B- (aq) → AB (s) Ksp = big (1/KSP) Precipitation Some of these reactions might not get to equilibrium so you have to “test” to see if we even have a problem. Q is just the concentrations of products and reactants when you are NOT at equilibrium. Ksp = [Ba2+]^3 [PO43-]^2 = 6x10^-39 Q = [Ba2+]^3 [PO43-]^2 = any other concentrations If Q is less than K… 1. You are not at equilibrium 2. You could dissolve more solid: the products (dissolved ions) are too small If Q is greater than K… 1. You are not equilibrium 2. You gave too many products (dissolved ions). They can’t stay dissolved, they need to precipitate To do these problems 1. Dissociate ions and determine the ions that “could” form precipitates 2. 3. 4.

Look up Ksp’s for both and go from biggest to smallest (or smallest to biggest for dissolution reaction) Determine which runs out first Use this as solid concentration to determine x

5. SEE NOTEBOOK FOR THE EXAMPLE If you add an acid to a strong base, it takes away some of the hydroxide. This means that the solubility goes up because you have to subtract how much hydroxide taken by the acid from the concentration of the hydroxide One Beaker - one water source Two Beakers - 2 water sources One beaker problems (What is the number? problems)It is a solubility issue -

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Ksp = 32 32 = (A2+) (B-)^2 32 = (x) (2x)^2 32 = 4x^3 8 = x^3 x = 2M The solubility is 2M for AB, It is always 2M for AB This means that is you are less than 2M, everything is dissolved but you are not at equilibrium

If you are at 2M, everything is dissolved and you are at equilibrium You are never over 2M. At least not for too long Two beaker problems It is really a mixing problem This is a PRECIPITATION problem BIGGEST KSP = MOST SOLUBLE SMALLEST KSP = LEAST SOLUBLE OTHER - Kw = 1 x 10^-14 -Pka = -log(ka) -pH = Pka + log(base/acid) -pOH = Pkb + log (acid/base) -Solve KSP problems, smallest ksp to biggest ksp -Kw = KaKb...