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K(p) concentrations of products over reactants ONLY GAS!!!!!! Partial pressures- NOT concentration homogeneous equilibrium= everything in the equilibrium mixture is present in the same phase.( to use Kp, everything must be a gas) Just like Kc, Kp always has the same value (provided you don't change the temperature)
EQUILIBRIUM
Le Chatelier's Principle
Equilibrium concentrations are only effected by temperature
***Same number of moles on both sides means no change
***
Using an ICE Table with Concentrations Exampleoften carried out for weak acid titrationsFind the concentration of A- for the generic acid dissociation reaction: HA(aq)+H2O(l)⇌A−(aq)+H3O(aq) with [HA(aq)] initial=0.150M and Ka=1.6×10−21. Fill in given concentrations: Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC???E???2. Calculate the change concentrations by using a variable 'x'Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC-x M+x M+x ME???The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H3O+ are generated when HA dissociates in water 3. Calculate the concentrations at equilibrium Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC-x M+x M+x ME0.150 - x Mx Mx M4. Use the ICE table to calculate concentrations with KaThe expression for Ka is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for Ka gives the following:Ka=x20.150−x=1.6×10−2To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x: (1.6×10−2)(0.150−x)x2+(1.6×10−2)x−(0.150)(1.6×10−2)==x20solve for x and eliminate the negative version of x
K(c)
ACIDS AND BASES
Circle of pH (for K constant)
Calculating Weak Acid and Weak Base pH Find concentrations using what is given Find K(a) or K(b) value if pK(a) or pK(b) is given Make ICE table with concentrations Solve for x using K value Use x to find concentrations Use concentration to find value needed in question (pH, [H+], [OH-], pOH)
Calculate Ksp of Precipitate ExampleCalculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it.First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process. PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)Ksp = [Pb2+][Cl-]2 Second, convert the amount of dissolved lead(II) chloride into moles per liter. (0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2 Third, create an "ICE" table. PbCl2 (s) Pb2+(aq) Cl-(aq) Initial Concentration All solid 0 0 Change in Concentration - 0.0159 M (dissolves) + 0.0159 M + 0.0318 M Equilibrium Concentration Less solid 0.0159 M 0.0318 M Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.Ksp = [0.0159][0.0318]2 = 1.61 x 10-5
Calculating Concentration in Titration using MolarityM(a)V(a) = M(b)V(b)
Normalitycalculate moles of acid or base and convert to moles of OH or Hmoles of OH or H L
ELECTROCHEMISTRY
Assigning Oxidation Numbers: Free elements = 0 Single element ions= change on ion Alkali metals = +1 Alkaline Earth Metals = +2 Fluorine = -1 Hydrogen = +1 (unless in binary with metal, then -1) Oxygen = -2 (unless peroxide O(2) , then its -1 Calculate remaining charges
Balance in Acidic Medium: Write unbalanced equation for the reaction in ionic form Make two half reactions (use oxidation reactions and separate , then balance by adding electrons) Balance the atoms (except for O and H) Add H2O to balance O and H atoms Add electrons to balance Equalize number of electrons in both half reactions by multiplying Add two half reactions together and cancel out things as necessary
Balance in Basic Medium: Follow all steps in acidic medium Add OH- to both sides of equations for every H+ that appears in the final equation For every H and OH, make H20 Reduce the number of H20 if they cancel
Balance by Half Reaction Method: Assign oxidation numbers Write half reactions for elements that changed oxidation numbers Add coefficients and balance elements Add electrons to both reactions to balance charges Multiply electrons to ensure they cancel out when reactions are added Add coefficients back into original equation Check overall balanced equation If elements are present that did not change oxidation numbers, check that they are balanced
Calculate Cell Potential (Voltage): Write two reduction half reactions Look up cell potentials (from chart) The reaction with a more negative value (in the reduction chart) is oxidation, so flip this reaction and write it as oxidation Change the sign of the cell potential for the flipped reaction Add the cell potentials together to get the total voltage When the cell potential is positive it is spontaneous
Electrochemical Cell:
increase in sizeconcentration decreasingpositive electrode
decrease in sizeconcentration increasing negative electrode
KINETICS
Rate Expression
reactants are negative
Reaction Order
units for rate constant, k
M/s
1/s
1/(M*s)
CatalystIncreases reaction rate and rate constant by providing a lower activation energy needed
SOLUTIONS
Types of Acids and BasesBronsted Acid - will donate a H+Bronsted Base- will accept a H+ *** can be both!!!
Equilibrium
Acids and Bases
ElectroChemistry
Kinetics
Solutions
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