Chemistry Notes- Chemical Equilibrium PDF

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Reversibility: The ability for a reaction to run in both the forward and the reverse directions Equilibrium: the condition in which the forward and reverse reaction rates are equal Concentration of all reactants and products remain the same; constant, not necessarily the same RateM =RateB kM* cM = kB*cB kM/kB = cB/cM Ratio of products and reactants is the equilibrium constant= K, big K Catalysts don’t change equilibrium constant Law of Mass Action ←−¿ aA+bB −→ cC+dD ¿ Rate = k1 [A]a[B]b Reverse rxn rate =k-1[C]c[D]d k1 [A]a[B]b= k-1[C]c[D]d k1 [ C]c [ D]d = = K equilibrium constant k−1 [ A ]a [ B ]b K expressed in terms of activity of products and reactants, so it’s therefore unitless AHCl=[HCl]/1M Pure solids and liquids have an activity of 1, so dont include them “right”, product heavy “left”, reactant heavy If original reaction is multiplied by a factor n, new equilibrium constant raised to power n For gases, PV=nRT can be rearranged n P = V RT n = molar concentration; R is constant, T constant, therefore molar concentration V directly proportional to pressure c d [ C] [ D ] Kc = a [ A ] [ B ]b P c C Pd D K p= Pa A Pb B Kp= Kc(RT) ∆n ∆ n =(c+d)-(a+b)

Catalyzed decomposition of hydrogen peroxide Rate of change of a gaseous product Molarity from volume using ideal gas law VO2/delta t * P/RT/Vsoln = moles/ volume E1: method of initial rates Rate2/Rate1 = Using ratios Use trials 1&6 where initial molarities are equal and use Arrhenius equation Same molarities, different T J/molK for R 8.3145 Rate6/Rate1= k6/k1 Reaction rate from graph Slope= rate Ln(k6/k1)= Ea/R(1/T1 – 1/T6) Validity of mechanism based on rate law Does experimental rate law match any conditions Treatment of Error Vh2O2; Avg and standard d And [KI] Error propagation for results & Ea Error propagation with log ln If multiple trials, standard deviation Using V vs relative volume Volume proportional to moles Relative volume proportional to molarity Blue dye and bleach Step 1: determine the rate law Using a spectrometer to measure change in concentration Beers law To determine order with respect to [blue], make sure hypochlorite molarity is constant, effectively constant Take data, convrert to concentration, ln concentration, inverse concentration to find order k’ = k {OCl-]^m log k’ = m*log[OCl] + log k LINEST Reporting/ using error from graphed resukts Get sigfigs from linest

Ka is a magnitude measuring how likely the acid is to dissociate in water Ka = [H3O+][A-]/[HA]

Small, highly charged metal ions: Al3+ and Fe3+ in water generates extra H3O+ in water, making it more acidic Cations and Anion in salts of acid/base affect pH when weak Us Ka Kb to calculate pH of salt solution NaCN solution: Since HCN is a weak acid, the cyanide ion must have a significant affinity for protons The value of Kb can be calculated from Kw and Ka Trends in acidity  For binary acid, two trends are responsible for acid strength  Polarity o Important as you move across the periodic table not acidic, C-H, N-H, O-H, F-H weakly acidic o Increasing electronegativity predicts the greater polarization  Bond strength o Important as you move down a column in the periodic table weak acid H-F, H-Cl, H-Br, H-I strong acid Solving acid-base equilibria problems  List the major species in the solution  Look for reactions that can be assumed to go to completion  ... Common ion effect When you add a conjugate base to a weak acid (or a conjugate acid to a weak base) you create a buffer A buffer is a solution that resists changes in pH when small amounts of strong acids/bases are added HA + H2O arrows H3O+ + AThe ability of a buffer to resist a change of pH is due to the buffer capacity (based on magnitude of concentrations of HA and A-) pH=pKa when molar concentrations of HA and A- are equal buffers work because common ion effect A−¿ pH= pKa - log HA ¿ pKa of acid is based on chosen species if [HA] and [A-] are known, pH can be calculated if pH is known, the ratio of [HA] and [A-} can be calculated pH is independent of [HA] and [A-] only when concentrations are equal Original buffered solution pH Step 1: Do stoichiometric calculations to fetermine new concentrations. Assume reaction with H+ and OH- goes to completion Do equilibrium calculations

Thermodynamics First law: Energy is neither created nor destroyed, constant amount of energy in the universe Second Law: The entropy of the universe is always increasing: the energy of the universe is continuously evolving towards a more probable distribution Any process that leads to an increase in the entropy of the universe is spontaneous Configuration: a type of energy distribution Microstate (permutation): A specific arrangement of energy corresponding to a configuration Which microstate will you see? The one with the largest number of microstates. This is called the dominant configuration The connection between weight (w) and entropy (s) is given by Boltzmann’s Formula S= kB*lnW Weight/ Omega (w): the number of microstates associated with a configuration Change in entropy ( ∆ S )for the expansion of a monatomic ideal gas from V1 to V2. S2-S1. Original weight W new weight 2W. After expansion, this particle will have twice the number of positions available. Vf ∆ S = R ln2= N*kB*ln Vi ) N particles ¿ How to increase entropy:  Increase the number of KE (kinetic energy) modes  Translational motion through space  Vibrational motion directed through chemical bonds  Rotational motion about the center of mass o Phase change o Increase the volume of the system  Increase energy in available in KE modes ∆ S System ∆ G Universe = ( ∆ S System + (

∆ H System T

∆ G Universe =-T ∆ S System + ∆ H System G 0 not spontaneous in forwards At standard Pressure Standard free energy of formation

)(

∆T )

Chemical bonds are most efficient form of energy E°cell and work The flow of electrons can be used to do work Galvanic cell: Volt: Work (J)/charge (c) Potential of cell E°cell: -w/q w= work done be the system, q= charge Electrons flow in a galvanic cell because reaction is spontaneous Electrons will stop flowing at equilibrium  Cell potential decreases over time

∆ G approaches 0

q E°cell = -w 1 F = the amount of charge on 1 mol of e- Faraday 1F= 96,485 C/mol eq= nF ; n=mol ew= -nF*E°cell upper limit on the work we can obtain from a galvanic cell at standard state measured under conditions of near-zero current flow electrical work is performed when a charge is moved through a potential ∆ G and work ∆ G=∆ H -T ∆ S Enthalpy can be transferred as heat or “useful” work qp = ∆ H if wuseful = 0

T ∆ S...