FInals- CHEM 130 summary PDF

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Summary

Acid and Base● What is the ph scale? ○ ph scale help us determine what is going on in the solution ○ how much excess hydronium ions (H3o+) in the solutions ○ related to the concentration of hydronium ions in the solution ● Does the pH stop at zero? ○ No, it goes further to the left and extends beyon...

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Acid and Base ●

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What is the ph scale? ○ ph scale help us determine what is going on in the solution ○ how much excess hydronium ions (H3o+) in the solutions ○ related to the concentration of hydronium ions in the solution Does the pH stop at zero? ○ No, it goes further to the left and extends beyond 0 and beyond 14 What is the range of the ph of a strong acid? ○ 0 to 6.9 is acidic ■ This means that ph that has more H3O+ ions in the solution has more ph ○ 7 is neutral —> strong acid and base ○ 7 to infinity is a strong base ■ More OH- ions in the solution What concentration of hydronium (H3O+) have a ph of negative? ○ concentration should be greater than 10, log 10 results in a negative number What can you say about the reaction between acid and base? ○ they are in thermodynamically controlled reactions. They react quickly and reach equilibrium quickly (reversible reaction!) ○ rate of forwarding = rate of backward = get equilibrium quickly ○ acid + base = salt + water ○ strong acids/base dissociate completely = G˚ cC + dD ○ K= [C]^c [D]^d over [A]^a [B]^b (ka and Kb are derived from here) ○ K= [prod]/[reac] ○ Concentrations of solids and liquids are ignored (~1)

Greater the value of K

product-favored equilibria

Smaller the value of K

Reactant-favoured equilibria

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What are strong acids? ○ Strong acids = dissociate essentially to completion ○ List of strong acids include: ■ HCl, HBr, HI, HCIO4, HNO3, H2SO4 ■ Large Ka values but low pka values ■ NOT HF ○ DISSOCIATE COMPLETELY = PRODUCTS WILL THEN BE FORM AND K>1 and G naught any hydroxide in the first two columns of the periodic table have large Kb values ○ 1 Ca(OH)2 → Ca2+ + 2OH○ If Ca(OH)2 is 0.059 moles then it has 2( 0.059) moles OHWhat are weak acids? ○ Weak acids partially dissociate (ionize) in solution. ○ Solutions of weak acids contain a small fraction of dissociated ions and many intact weak acid molecules What are weak bases? ○ weak bases partially dissociate (ionize) in solution Ka expressions ○ The reaction of an acid and water to produce its conjugate base and H3O+ is called ACID DISSOCIATION reaction ○ weak acids have small Ka values (ka7) ○ weak base + strong acid will lead to acidic solution (ph [H3O+] / [A-] [H3O+] < [OH-] [A-]=1.0 M [A-] > [H3O+]

K= Products/reactants = [H3O+]^[A-]^/[HA]^

Q= Products/reactants (INSTANTANEOUS) = [H3O+]^[A-]^/[HA]^ Ph = -log[H3O+]

MEMORIZE THIS TABLE STRENG TH of acid/bas e

Ka value

Pka value

K value

Equilibriu m lie

G naught

First added

STRONG

HIGH

LOW

K>1

Products

NEGATIVE (Gnaught 3C ○ molecular complexity increases (minor factor) say for example: SO2 is less complex than SO3 more different arrangement of atoms/more area covered ■ change in the number of molecules of gas is going to have a more significant effect on the entropy from molecular complexity ■ gases are very disordered What is spontaneity? ○ spontaneous process will continue once started without any external action (ex: combustion = exothermic process) ○ non-spontaneous process will not occur unless external action is applied ○ Remember: spontaneity is not the rate of the reaction! It is whether the reaction will happen ○ Exothermic/Endothermic is NOT EQUAL to Spontaneous/not spontaneous ■ ice melting at room temperature ■ process involve solid to liquid and it absorbs heat

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■ endothermic ■ spontaneous = that's why we can't hold ice at room temperature ■ dissolved salt precipitating out of water as NaCl (s) ■ liquid to solid (release heat to fridge) ■ exothermic ■ not spontaneous at room temperature but should be spontaneous below 0 degrees ■ spontaneous asking will it happen w/o the presence of some kind of force, NOT how fast it will happen, has nothing to do with the rate of reaction What is the difference between the first and second law of thermodynamics? ○ first law of thermodynamics states = energy transfer ○ second law of thermodynamics states = the entropy (disorder) of the universe is always increasing What is the second law of thermodynamics? ○ entropy of the universe is always increasing ○ no process is possible in which the sole result is complete (100%) conversion of heat into work ○ entropy of an isolated system increases in the course of spontaneous change ○ the sign of ∆Suni = states that the reaction is.. ■ ∆Suni > 0 » spontaneous ■ ∆Suni = 0 » reversible ■ ∆Suni < 0 » impossible Molecular interpretation of entropy ○ bolzmann developed a relationship between entropy and microstates ○ increase in microstates, more probable states more ways to shuffle, increase in entropy ○ S = klnW ■ k= Boltzmann constant, 1.38x10-23 What is the third law of thermodynamics? ○ a reference point is needed in order to determine numerical values of entropy ○ At T= 0K, all motion stops, in pure perfect crystal there is perfect order, entropy is zero Determine the entropy of a sample ○ the entropy of a sample is reported as the increase in entropy from the perfect crystalline form at T= 0K to the defined conditions ○ Standard molar entropy (S˚)= absolute entropy of 1 mole of a substance in its standard state (units J K-1 mol-1) (derived from third law of thermodynamics, where pure crystalline has an entropy of 0) Evaluate entropy change ○ definition of entropy system= can be defined as heat flow of reversible system under the condition of constant temperature ■ We can bring it back to equilibrium because it is reversible at constant temp ○ ∆Ssys = q rev / Tsys ○ ∆Ssurr = qrev/ Tsurr ○ ∆Ssys = -∆Ssurr ○ However, we cannot apply the heat flow of reversible system under the condition of constant temperature Determine the change of entropy of rust ○ we know that rust happens spontaneously so Suniv should be greater than zero ○ however, based on the equation, 2 Fe (s) + 3/2 O2 (g)—> 1 Fe2O3 (g), the number of moles in the reactant side is greater than the product side, so we know that the

entropy of the system decreases, the entropy of the surroundings increases

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The heat of the system is negative (expect the reaction to be exothermic, since heat is released out) - Phase changes that occur in exothermic: gas-liquid-solid - What does this mean for: - The entropy of the system decreases because it is more ordered - The entropy of the surroundings, increases because it is less ordered, (you can also think about it as heat out of the system, heat form of energy, adds to the energy in surroundings, more energy dispersal) As a general rule: - entropy increases with increased temperature because more molecules have energy to move around so more entropy or possible microstates/disorder - Increasing the pressure of a gas at a constant volume = increases the temperature of the gas, which yields an increase in entropy. - The pressure is proportional to temperature - The volume is inversely proportional to the pressure - The volume is proportional to the temperature - The volume is proportional to the number of moles FOR ANY CHANGE TO OCCUR, THE CHANGE IN ENTROPY FOR THE UNIVERSE MUST BE POSITIVE → THIS IS THE CONDITION OF SPONTANEOUS ∆Suniv (+)

∆G (-)

SPONTANEOUS

∆Suniv 0

∆G = 0

EQUILIBRIUM

∆Suniv (-)

∆G (+)

NON-SPONTANEOUS

Gibbs free energy/Van hoff ●

Gibbs free energy (J or kJ) ○ Recap: ∆S uni = determine spontaneity of process, however, we can't measure that because the uni is too big so we define another state variable » Gibbs free energy so we can determine whether the spontaneity of reaction by only looking at the system ○ Gibbs free energy is affected by BOTH enthalpy and entropy ■ G = H + TS (for equilibrium state) ■ ∆Gsys = ∆H - T∆S (for constant pressure and constant temperature process) ■ Check units!! Usually delta H is in kJ and delta S is in J ■ We can determine the ∆G of system given ∆H and ∆S ■ ∆H = (-) (exothermic) → ∆S = (+) » spontaneous at all times ■ combustion ■ ∆ H= (+) (endothermic) → ∆S = (-) » non-spontaneous at all times ■ ∆ H = (-), ∆S = (-) » spontaneous at low T ■ ∆ H = (+), ∆S = (+) » spontaneous at high T

∆G sys = -T∆Suniv (∆Suniv is positive because of second law) (applies in constant T and pressure) (No enthalpy present ∆H) ■ We can determine ∆G of system given ∆Suniv ■ Spontaneous » ∆S uni (+), ∆G =(-) ■ Equilibrium » ∆S uni (0), ∆G =(0) ■ Non-spontaneous » ∆S uni (-), ∆G =(+) ○ amount of free energy a reaction makes available to do work depends on two factors enthalpy (amount of heat reaction transfer) and entropy (amount disorder it creates at a given temperature) ■ ∆G 0, More A than B will be formed, K0 ■ ∆H and ∆S is constant over temperature range ■ the slope is a decline (-) because the equation starts with [-∆H/R ] (negative)(positive)= neg number ○ exothermic ■ ∆H...