Chemistry Notes - Equilibrium

7/27/2019 Chemistry Notes - Equilibrium

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4/26

SCH4U: Chemical Equilibrium (Text Ref. pg. 429-433)

1a) What is a quantitative reaction?

1b) What must be true of a chemical system such as the decomposition of CaC03(s) if it is to beconsidered a Quantitative reaction?

f \ \\C o . c o , --'7 C o , 0 (S) -; C o, ' '3 'r

(c.ltl. (!!,") N\ ~\ "rQc (\ If',1c) What happens to the CaC03(s) reaction if the system is closed and the CO2 is not allowed to

escape?Tk (C tI.,\ i"t r , , - o s . e . 5 . o..\\~ Yj l h a . ~ f'CL ') , '

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2) Consider the reaction: N204(g) ~ 2N02(g) and the data in Table 2 (pg, 431)a) -Sketch a graph of Concentration of Reactants and Products vs Time for both experiments,

Exper iment 1 Exper iment 22~?2 ~ N2~'4- ~ )( -4 'X .

N204). 2N02 20, 1~ 0

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5/26

THE EQUILIBRIUM CONSTANT: K c vs K pSo far we have been considering the equilibrium constant (K.q) in terms of the equilibriumconcentrations of the reactants and products. This form of K e q is often referred to as K c

However, the concentration of a gas is proportional to the pressure of that gas. So in equil ibriainvolving gases, the concentration of each gas in the system is proportional to the partialpressure of each gas in the system. ie: [N2l a pN2' [H2l a pH2and [NH3l a pNH3Therefore, for systems involving gases, K e q can also be expressed in terms of the partialpressure of each gas at equilibrium, or K p .

Generally, K , ' * K p , but they both have all the properties of K e q and are constant at constanttemperature.Problems:1) Write the K e q expression for each of the following equilibria as K, and K p :

'l..U .\ - \ L qa) H2(g) + CI2(g) ..2HCI(g) = It 1[ uJb) 2NO(g) + 02(g) ..2N02(g) Ill. f > -c) 2N20S(g) ..4N02(g) + 02(g)2) Consider the following equilibrium system: 2NO(g) + CI2(g) ..2NOCI(g) \ : : : 1 - ' . ~ ~ " \ ) 0 " \ 1a) Write the K p expression for this system r 'l.. rb) At a given temperature, the partial pressures of the gases in an equilibrium mixture of l V \ - - l c : i - S \....

this system are determined to be: pNO = 66kPa, pCI2 = 18kPa and pNOCI = 15kPa.Determine the value of K p for this system. l) .1 0 0 C }C Y~3) At 25C, K p = 7.04x10-2 for the following system: 2N02(g) ...N204(g).a) If pN02 at equilibrium is 15kPa, determine the pN204 . \ . \ ': l . ~ k i l o . " )b) If total pressure is equal to the sum of the partial pressures, determine the total pressureof this system at equilibrium? ( '?l a. ~ \< .~ q.)4) Baking soda when heated decomposes in the following reaction:

2NaHC03(s) . Na2C03(s) + CO2(g) + H20(g)-1'1 t)< +)( +)1.a) Consider the stoichiometry of this system in terms of "x"b) Write, the K, and the K, expressions for this heterogeneous system. k,: [L C \2 ] 0116] k = (~ (Ol') (pc) If at 125C, the K, for this system is 2.4x10"". Determine the equilibriumconcentrations of [C02] and [H20]. O. I . J \ 1 0 ~ ~ 1 ' 1 1 0 \ I Ld) At the same temperature, the K p is 2.6x1 03 Determine the pH20 and the pC02 in thisequilibrium system (in kPa). 'l "\ ~Y o...e) At equilibrium, what is the total pressure of the system at equilibrium?

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6/26

Keq and HETEROGENEOUS EQUILIBRIAReactions in which the chemical components exist in more than one physical state are referredto as heterogeneous reactions. For example the decomposition of solid calcium carbonate:

If we consider the equilibrium constant for this system we find:KeQ [CaOJ[C021[CaC03]

But, as we have discussed before, the concentrations of pure liquids and pure solids atconstant temperature are proportional to their densities, which are constant and independent ofthe amount of the solid or liquid. In these reactions, the "concentration" of the pure solid orliquid is considered as part of the "nature of reactants" and is incorporated into the K constant.:. since [CaC03] = constant and [CaO] = constantIf we re-arrange the KeQexpression to isolate all constants into the K constant:K[CaC031 = [C02][CaO]Problems:1) Write the Keqexpression for each of the following equilibria:a) 2Hg(l) + CI2(g) H92CI2(s) '/tt.h'\b) H20(I) H20(g) \: :1 - l , t> J ?/ .. .r .c) 5CO(g) + 1205(s) 12(g)+ 5C02(g) C \ . ..ll\)11to')-d) Ca(HC03Maq) CaC03(s) + H20(I) + CO2(g)e) CaO(s) + S02(g) CaS03(s)f) PbCI2(s) Pb2+(aq)+ 2CI1(aq)g) CH3COOH(aq) + H20(I) .CH3C001(aq) + H301+(aq)2) The density of water is 1.00 g/mL, determine the "molar concentration" of:a) 18.0 mL of water, 100.0 mL of water.3) The density of solid CaC03 is 2.71 g/cm3 (1 er n" = 1 mL). What is the "molarconcentration" of:a) 10.0 9 of solid CaC03 b) 35.0 9 of solid CaC03


7/26

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11/26

SCH4U - Chemical Equilibrium and LeChatelier's PrincioleLeChatelier's Principle states:

A stress is defined as any introduced change that will effect the rate of the chemical reactionsinvolved. As such, the followinq factors mav be used to stress an eauilibrium system:1) Consider the following equilibrium system:

(Pink) (Blue\Applied Stress System Response Observed Change Changes in [components]to offset stress in System in new Equilibrium1ll\-] F O . ~ I H j r ~ F w D r~rI \\J'r\\ rt'\af C L h \ \ ) Q . . L?ll [J''Jl'J Lq-j hV\)\Jf.:tSo R'1 'f"> -n \\..lrry rr vor e , \ ') i f ' l k ~ R l1 ' [PJJ-t ~"2-()] FI\-\J'DvR5 F~I'\ f " , Y \ lurn b \UlLf LR1~ [rl +1 [H2O] ff\lfo\)a~ ~t:\{ R'i(\ I\J(II fnOf"1l. p I '" k LRji (J'1 \.~~ r''I"tll\~ ["''' ~"1\ \f ""~ 'vrl) i'Y)t\\'Q. b.\"u [1


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13/26

SCH4U: LeChateli er and the Ef fec t o f TemperatureChanging the temperature will affect the rate of both the forward and the reverse reactions,changing the value of the rate constant "k".. The system will re-adjust the concentrations of reactants and products to settle into anewequilibrium with a new Ke for the new temperature.Consider the following system:

Ep(k .l}

Forward Rxn: endothermic with Ea = 125 kJ/molReverse Rxn: exothermic with Ea = 75 kJ/molAt equilibrium rate forward = rate reverseWhen concentration of reactants/products are constant,rate a "k" and "k" is dependent on temperature.Rate a "k" = Ae-Ea/RT"k" a e-Ea/RT

Rxn

If this system is at equilibrium at 300 K, what will happen to the rate of each reaction if the. d 1 K?emperature is Increase to 3 5k (Forward Rxn) at 300 K k (Forward Rxn) at 315 K

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14/26

-2-Increasing temperature will increase the rate of an endothermic rxn more than it increases therate of the exothermic rxn.

...Higher .T~mp favours ENDOTHERMIC RXNA similar analysis will show that decreasing temperature will slow down both reaction rates, butthe endothermic reaction will be affected to a greater degree.

Lower Temp favours the EXOTHERMIC RXN

SCH4U: LeChatel ier and the Effect of CatalystCatalysts work by decreasing the activation energy of both forward and reverse rxns, Letsassume that a catalyst for the previously studied rxn lowers the Ea by 25 kJ/mol.(ie: Ea forward (w/o cat)= 125 kJ/mol Ea forward (withat)= 100 kJ/molEa reverse(W/Ocat)= 75 kJ/mol Ea reverse(Withcat)= 50 kJ/mol)k (Forward Rxn) at 300 K (without catalyst) k (Forward Rxn) at 300 K (with catalyst)

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Forward rxn: kf(Withat/kf(W/Oat)= \O.~~k (Reverse Rxn) at 300 K (without catalyst) k (Reverse Rxn) at 300 K (with catalyst)

~ -bo ./g ., k~ - C o . ! Q TrJ . . 1 2 - Q

Reverse rxn: kr(Withall1kr(w/oall= ~.I~The catalyst increases the rate of both forward and reverse reaction equally, as a result, it hasno effect on the equilibrium system.


15/26

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16/26

SCH4U: Equilibrium and LeChatelier's PrincipleAlthough equilibrium prevents chemical reactions from reaching 100% yield, thereaction still follows the stoichiometric mole ratio when it reacts. We can, therefore,consider the amounts that react in terms of a quantity called "X".1) Consider the following chemical system:

N2(g) + 3H2(g) l'" 2NH3(g)If we begin by adding 1 mole of N 2 and 3 moles H 2 to a closed 1 L reaction vessel, acertain amount of each will react and a certain amount of NH 3 will be produced beforethe system reaches equilibrium.If we call the amount of N? that reacts "X" , then according to the mole ratio:In terms of "X", what is the amount of H 2 that must also react? s " .In terms of "X", what is the amount of N H 3 that must be produced? Q x \'Illustrate these changes by completing the following Concentration vs Time graph:

rI~T 3[mol/Ll

Concetration vs Time

2_ _ _ 1 '21


18/26

SCH4U: Application of LeChateli er 's Pr inci pl e1) For each of the given equilibria, state the system response to the applied stresses, and

state how the concentration/amount of each component will change as the newequilibrium is reached.

A)i)

ii)

iii)

B)

i)

ii)

iii)

- ' f . , iv)

C)i)

ii)

' ' ' f .: . iii)

D)i)ii)iii)iv)8

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Adding more CaO(s)rn~~CJ 1 ~ If'(\ l c . . . " , ltlJ 't ~\:J2S02(g) + 02(g) '" 2503(g) .6H = -198kJ

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Chemistry Notes - Equilibrium

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