Chemical Equilibrium

What happens to an equilibrium when: - You reverse an equilibrium? - You multiply coefficients in a balanced equilibrium by a constant - You divide coefficients in a balanced equilibrium by a constant - Add individual equations

- Invert the value of K - Raise the value of K to the factor - Take the corresponding root of K - Multiply K values -- Math is in Camscan

3 H2SnO3(s) + 4 NO(g) ⇌3 Sn(s) + 4 H+(aq) + 4 NO3-(aq) + H2O(l) What is the expression of Kc? [Press i ]

A

Describe the effect of a catalyst on equilibrium

A catalyst changes the mechanism of a reaction by lowering the Ea (increasing rate) *Does not effect equilibrium*. Only rate - (Hreactants and Hproducts remain the same)

What is Le Chatelier's Principle?

A system will always move to oppose a stress in a way that minimizes the effect of the stress

What does K mean?

An equilibrium constant that tells us how far a reaction will go

How to determine limiting reagent? What subtracts from what?

Do the given *moles over the stoich* in balanced reaction. Lower value is limiting. All of the limiting is used up and goes to 0 Other reactants should decrease *proportionately*. So if 3mol was lost in one 2B species, then the A species should decrease by 1.5.

The following reaction reaches equilibrium in a closed reaction vessel 3Fe(s) + 4H2O(g) >< Fe3O4(s) + 4H2(g) H = -150.8kJ Which of the following actions increases the mass of Fe3O4 in the equilibrium mixtures i) Reducing the volume of the reaction vessel ii) Increasing the temperature iii) Adding some Fe(s) iv) Removing some H2 (g) A ii and iv. B. ii, iii and iv C. iii and iv D. iv E. i, ii, iii and iv

Exo = release heat Want net rxn to right Vol wont affect since same moles gas both sides Adding heat goes left Iron not part of expression so irrelevant Removng H2 makes go right!!! *D*

The following reaction reaches equilibrium in a closed reaction vessel 3Fe(s) + 4H2O(g) >< Fe3O4(s) + 4H2(g) H = -150.8kJ per mol Fe3O4 Which of the following maximizes the mass of Fe3O4 in the equilibrium mixture? A. High temperature B. Low temperature C. High pressure D. Low pressure E. High temperature and low pressure

Exothermic so low temperature good Pressure/Volume changes wont affect since gases are same both sides stoich-wise Kc= [H2]^4 / [H2O]^4

2 SO2 (g) + O2 (g) >< 2 SO3 Kc = 2.8e2 (at 1000K) In a 10L flask, the moles of species are as follows: 0.68 mol SO3, 0.32 mol SO2, 0.16 mol O2. What would happen if 1.00mol SO3 is added?

First, is the initial flask in equilibrium? Q = (0.68)^2 / (0.032)^2 (0.016) = 2.8E2 This is the same as Kc, so it is at equilibrium. Once add 1.0mol SO3, the numberator will be greater so Q > Kc so reaction will go left to balance this, so the concentrations of the reactants would increase.

For aA + bB >< cC + dD Where Ho > 0 Impact of increasing temp

Forward endothermi = reverse exothermic So increasing temp will favour the forward (think of adding heat as a reactant for endothermic) Cooling down is the same as removing heat and favouring exothermic

Kc for the vaporization of H2O?

H2O(l) >< H2O(g) r = k1 x [H2O(l)] r-1 = k-1 x [H2O(g)] Kc = [H2O(g)]/[H2O(l)] Kc = [H2O(g)] since ignore liquids

k changes with temp?

Increases with temp

v't hoff tells us what about enthalpy?

Increasing temp = shift in direction of endothermic Decreasing temp = shift in the direction fo exothermic If reaction is thermoneutral, then wont be affected by temperature

For H2O(l) >< H2O(g) K? van't hoff

K = Ph2o / nothing since ignore H2O(l) But pH2O = vapor pressure So C-C is a unique derivative of this equation

In an experiment, 1.0mol each of H2(g)and I2(g) are placed in an empty 5L flask and the flask is then sealed. These gases react as follows. H2(g) + I2(g) >< 2HI(g) Which of the following statements is/are true? i) The value of K increases as the system approaches equilibrium ii) The value of K decreases as the system approaches equilibrium iii) The total pressure increases as the system approaches equilibrium A. i and iii are true B. ii and iii are true C. Only iii is true D. None of these are true E. All of these are true

K doesnt increase/decrease the Q ratio does Q = [HI]^2 / [H2][I2] Q = Kc at eq Net rxn right No change in pressure cause stoich is same: If A + 2B >< AB2 Total pressure decreases at eq (3mol --> 1mol) *D*

Suppose that A and B react to form C according to the equation below. A + 2B → C What are the equilibrium concentrations of A, B and C if 2.0 mol A and 2.0 mol B are added to a 1.0 L flask? Assume that the equilibrium constant for this reaction is Kc = 8.3x1014.

K largeso favours right so reverse is small so x by SSA Kc = 8.3e14 = [C] / [A][B]^2 Qc makes go right i) 2 | 2 | n/a i') 1 | 0 | 1M << 1/2 of 2mol from limiting 2B >> 1 mol of A left over from 1:2-->1 reaction >> assuming 100% product w/ high Kc c) +x | +2x | -x >> go back e) 1+x | 0+2x | 1-x >>e here is "i' + c" 8.3e14 = 1-x / [(1+x)(2x)^2] >> SSA for x (only if +/- x tho) 8.3e14 = 1 / (2x)^2 8.3e14 = 1 / (4x^2) 4x^2 = 1 / 8.3e14 4x^2 = 1.20e-15 x^2 = 3.012e-16 x=sqr(3.012e-16) x= 1.7e-8 Yes looks much smaller so can be negligible [A]eq = 1+x = 1.000000025 = *1.0* [B]eq = 2x = *3.5e-8* [C]eq = 1-x = *1.0*

At 2,200 K, the equilibrium constant for the reaction below is Kc=11.3. Br2(g) --> 2 Br(g) What is the percent dissociation of Br2 at 2,200 K, if the initial concentration of Br2 is 1.75 mol/L?

Kc = 11.3, cant make SSA Qc = [Br]^2 / [Br2] Qc = [0] / [1.75] = 0 Qc < Kc = rxn goes right i) 1.75 | na c) -x | +2x e) 1.75-x | 2x Kc = [Br]^2 / [Br2] 11.3 = [2x]^2 / [1.75-x] 11.3[1.75-x] = 4x^2 19.775-11.3x = 4x^2 4x^2 + 11.3x - 19.775 = 0 a = 4 b = 11.3 c = -19.775 2a = 8 -b = -11.3 sqr[b^2-4ac] = 21.07 (+) x = 1.22125 (-) x = -4 << cant be negative x = 1.22125 (x/Co) x 100 (1.22125/1.75) = *69.8%*

What are the equilibrium concentrations when 0.1mol HCN is dissolved in water to make 1L of solution? At the temperature of the experiment, Kc = 6.2e-10 HCN (aq) + H2O(l) >< CN- (aq) + H3O+(aq)

Kc = [CN-][H3O+] / [HCN] ^^ Water ignored because solvent Small Kc = goes left = little forward reaction progression Q = [CN-][H3O+] / [HCN] = 0 < Kc I: 0.1 | n/a | na C: -x | +x | +x E: 0.1-x | x | x Sub x's into Kc x^2 / (0.1-x) Use the 5% esimation to say that x is so small that minusing it from anything leaves the remainder value essentially the same, so it can be ignored. x^2 / (0.1-x) = x^2 / 0.1 x = sqr(Kc x 0.1) = 7.9e-6M Double check 5% assumption was correct 7.9e-6 / 0.1 = 0.008% < 5% Assumption valid

Kc for concerted and multi-step reactions? What does sub-eq mean?

Kc = k1/k-1 Kc = (k1/k-1)(k2/k-2)...(kn/k-n) Eq implies rate of forward and reverse are equal

Kp = Kc nuance? Whats delta n?

Kp = Kc x (RT)^(Δn) Where Δn = #stoichproducts - #stoichreactants R = 0.082058 atm L / k mol

Kp hint?

Kp = P(products)^p / P(reactants)^r

Kp? Conversion w/ Kc? R?

Kp = PH2O Kp = Kc(RT) The R here is different than in the past

Implications of: A large Kc A small Kc 5% rule?

Large Kc = high numerator (lots of product) with few reactants = reaction goes to completion Small Kc = high denominator (lots of reactant) with few products = reaction does not go forward much 5% rule is that the one concentartion will be <5% different than another, and is neglibly different so can consider it the same

At 1100K, Kc = 4.2e-6 for the gas phrase reaction below. 2H2S (g) >< 2H2 (g) + S2(g) In an experiment carried out at 1100K, the initial concentrations are: [H2S]o = [H2]o = [S2]o = 1.0M Set up an ICE table and then decide: Which of the following statements must be true when the system reaches a state of dynamic equilibrium A. [H2S] = [H2] B. [H2S] = [H2] = [S2] C. [H2S] < [H2] D. [H2] < [S2] E. [H2] = [S2]

Low Kc so not likely to occur left to right Q = [H2]^2[S2] / [H2S]^2 = 1 (if 1mol each) Q > Kc then since 1 > 4.2e-6 Reaction goes left: i) 1 | 1 | 1 c) +2x | -2x | -x e)1+2x | 1-2x | 1-x So reverse direction is favoured (incerse of Kc left right is actually big) 1-2x is a smaller number than 1-x, so *D*

N2O4(g) ><2NO2(g) In a particular experiment N2O4 (0.0250M) is placed in an empty flask. At equilibrium [N2O4]eq = 0.0202M. a) What is the value of Kc? b) What are the equilibrium concentration if the initial concentration are [N2O4]o = [NO2]o = 0.0125M

N2O4(g) ><2NO2(g) ( Q = p / r = 2NO2 / 0.0250 = 0 / 0.0250 = 0 So goes right I: 00250M | n/a C: -x | +2x E: 0.0202M | 2x (0.0250 - x) = 0.0202 x = 0.0048M a) at equilibrium Kc = p / r = [NO2]^2 / [N2O4] = 2x^2 / 0.0202 = 4.6e-3 b) Equal concs w/ different stoichs arent at equilibrium Q = [NO2]^2 / [N2O4] = (0.0125)^2 / (0.0125) = 0.0125 0.0125 > 4.6e-3 = goes left Q > Kc I: 0.0125 | 0.0125 C: +x | -2x E: 0.0125 + x | 0.0125 - x Kc = [NO2]^2 / [N2O4] = [0.0125-2x)^2 / [0.0125+x] = Math for x = 2.15e-3M Dont accept negative x's or x's > initial concentrations [N2O4] = (0.0125+x)M = 0.0146M [NO2]^2 = (0.0125-2x)M = 0.0082M Verify by subbing into Kc to get same Kc

For the reaction below, Kc = 1e10 at 500K 2X(g) + Y(g) >< 2Z(g) In an experiment, 2M each of X and Z are sealed in a 1L vessel and allowed to reach equilibrium at 500K. Which one of the following statements concerning the equilibrium mixture is true? A. The equilibrium concentration of X will be much less than 2M B. The equilibrium concentration of Z will be much less than 2M C. The equilibrium concentration of Z will be much greater than 2M D. The equilibrium concentration of Y will be approx 1M E. The equilibrium concentration of Y will be much less than 1M

Net rxn left to generate Y K largeso favours right so reverse is small so x by SSA I) 2 | na | 2 C) +2x | +x | -2x E) 2 | x | 2 *E*

What species in reactions do you ignore?

Pure liquids, pure solids, or liquid solvents (H2O)

The reaction below reaches equilibrium in a closed reaction vessel CaCO3 (s) >< CaO(s) + CO2(g) What is the effect, if any, on the equilibrium partial pressure of CO2 if some CO2 is added to the equilibrium mixture A. The equilibrium partial pressure of CO2 increases B. The equilibrium partial pressure of CO2 decreases C.The equilibrium partial pressure of CO2 reimains the same D. Not enough info

Q = [CO2] Qp = Pco2 Keq = Pco2, eq May move to oppose, but then reach eq, and since = Pco2, stays same *C*

Some CH4(g), intially at 1.000 atm, is placed in an empty reaction vessel. At equilibrium, the total pressure is 1.981 atm. Assuming the partial pressures are in atmospheres, what is Kp for the reaction below? 2 CH4(g) → C2H2(g) + 3 H2(g)

Q = p / r = [Pc2h2] x [Ph2]^3 / [ch4]^2 Qp = 0 / [1]^2 = 0 Goes right I: 1 | na | na C: -2x | +x | +3x E: 1-2x | x | 3x At equilibrium: - Ptot = 1.981atm = Pch4 + Pc2h2 + Ph2 1.981 = (1-2x) + x + 3x 0.981 = -2x+1+3x 0.981 = 2x 0.4905atm = x - Pch4 = 1-2x = 0.019atm - Pc2h2 = x = 0.4905 atm - Ph2 = 3x = 1.4715 atm Kp = p / r = ([0.4905][1.4715]^3) / [0.019]^2 = 4239 = *4.33e3*

Qc? Eq? Kc Direction proportional to Kc?

Q is a function with p/r and has properties in equilibrium. Qc = Kc at equilibrium Qc > Kc = higher numerator = excess product = moves left Qc < Kc = higher denominator = excess reactant = moves right

A solution is prepared by adding some solid CaF2 to water and allowing the solid to come to equilibrium with the solution. The equilibrium constant for the reaction below is Kc = 3.9x10-11. CaF2(s) → Ca2+(aq) + 2 F -(aq) What is the equilibrium concentration of Ca2+ in the solution? Assume that some of the solid remains undissolved.

Q= Ksp = [Ca2+][F-]^2 Kc = 3.9e-11 >> Q = Ksp at equilibrium. >> Stoich allows the equilibrium concentrations to be determined from Ksp. CaF2(s) → Ca2+(aq) + 2 F -(aq) i) -- | 0 | 0 c) -- | +3.9e-11 | +7.8e-11 e) -- | *3.9e-11* | 7.8e-11

At 40oC, Kc=0.0303 for the reaction below. H2 + Cl2 --> 2 HCl In an experiment carried out at 40oC, the initial concentrations were [H2]=[Cl2]=[HCl]=2.00 mol/L. What is the equilibrium concentration of HCl?

Qc = [products] / [reactants] Qc = [HCl]^2 / [Cl2][H2] Qc = [2]^2 / [2][2] Qc = 1 Qc > Kc so reaction goes left to reactants H2 + Cl2 --> 2 HCl i) 2.00 | 2.00 | 2.00 c) +x | +x | -2x e) 2+x | 2+x | 2-2x Kc = [HCl]^2 / [Cl2][H2] Kc = (2-2x)2 / [2+x][2+x] Kc =(4x^2-8x+4) / (x^2 + 4x + 4) 3.03e-2(x^2 + 4x + 4) = (4x^2-8x+4) 0.0303x^2 + 0.1212x + 0.1212 = 4x^2-8x+4 Subtract all of left side off of right 0 = 3.9697x^2 -8.1212x + 3.8788 [] sub into quadratic equation a= 3.9697 b = -8.1212 c = 3.8788 >> -b = 8.1212 >> sqr(b^2-4ac) = 2.088827422 >> 2a = 7.9394 (+) (8.1212+2.088827422) / 7.9394 = 1.285994839 (-)(8.1212-2.088827422) / 7.9394 = 0.759802073 [HCl]eq = 2-2x (+) = 2-2(1.285994839) = -0.57 << Cant be negative (-) = 2-2(0.759802073) = *0.480* So x must be the *0.480M*

What is an equilibrium? Kc?Hint?

Rate of forward rxn = rate of reverse rxn - No net change in [species] aA+bB><cC+dD Kc = [products]^p / [reactants]^r - Concentration raised to power of stoich value ^^ make sure *equation is balanced*

An empty flask is filled initially with S8(g), at a pressure of 1.00atm. At equilibrium, the total pressure in the flask is 3.25atm What are Kp and Kc for the reaction shown below, assuming T = 1225K? S8(g) >< 4 S2 (g)

S8(g) >< 4 S2 (g) Lots of S8 w/ no S2 so rxn goes right ( Q = p / r = Ps2 / Ps8 = 0 / 1 = 0 I: 1.00atm | n/a C: -x | +4x ^^ Comes from stoich + direction of equilibrium E: 1-x | 4x At equilibrium: - Ptot = 3.24atm = Ps8 + Ps2 = (1-x) + (4x) x = 0.75atm - Ps8 = 1-x = 0.25 atm - Ps2 = 4x = 3.00 atm Kp = p / r = [3.00]^4 / [0.25] = 324 Concert to Kc Kp = Kc(RT)^(Δn) Kc = Kp / (RT)^3 = 2.53e-4

For the reaction below, Kc = 1e20 A + 2B >< C In an experiment, 1M each of A and B are placed in a 1L container and then the container is quickly sealed. Set p an ICE table and then decide: When the system reaches equilibrium, which of the following statements is true. A. [A] << 1M B. [B] << 1M C. [C] << 1M D. Both [A] and [B] are very small (<<1M) E. None of the above

So tendency to go left to right i) 1 | 1 | na c) -x | -2x | +x e) x determined by limiting reagent. Which is B since for every mole of A, need 2 moles of B,but only have 1 mole of B. Lets see what happens when it goes all the way to completion I) 1 | 1 | na I') 0.5 | 0 | 0.5 C)+x | +2x | -x << opposite direction but weak so SSA E) 0.5 | 2x | 0.5 *B* since x is super small 2 by x is super small

The reaction below reaches equilibrium in a closed reaction vessel. The enthalpy change for the forward reaction is +135 kJ/mol. Does the mass of Na2CO3 increase, decrease or stay the same if a catalyst is added to the equilibrium mixture? 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) The mass of Na2CO3 : [] increases [] decreases [] stays the same

Stays the same. Cat affects rate, not position of equilibrium

2 SO2 (g) + O2 (g) >< 2 SO3 In a 10L flask, the moles of species are as follows: 0.68 mol SO3, 0.32 mol SO2, 0.16 mol O2. After increasing pressure, the new moles are 0.83 SO3, 0.17 SO2, 0.085 O2 What happened to the equation?

The volume would decrease More products and less reactants. Reaction went to the right.

For A>< B Assume concerted Difference between Eaf-Ear? If endo, which reaction increases more with higher temp?

Using arhenius k = Ae^-Ea/RT ^^ Where k is for any rate constant K = kf/kr = [Af x e^-Eaf / RT] / [Ar x e^-Ear / RT] = A e^-(Eaf-Ear)/RT But difference between Eaf and Ear is Hreactants-Hproducts

Consider the following reaction. PbCO3(s) → Pb2+(aq) + CO32−(aq), Kc = 7.40x10-14 A solution is prepared by adding some solid PbCO3 to water and allowing the solid to come to equilibrium with the solution. What are the concentrations of Pb2+ and CO32− in the solution?

Vanhoff Kc = [Pb2+] [CO32− ] = 7.40x10-14 Kc = [C]^c [D]^d / [A]^a x [B]^b 7.4e-14 = [x]^2 x = sqr(7.4e-14) x = 2.7e-7

Note for v't hoff equation

Very similar to the C-C equation use *thermodynaic equilibrium constant* where P given in atm and concentrations in M So they give up Kp if pressures or Kc if dissolved species, but if a mix of both just K aA(g) + bB(aq) >< cC(s) Ignore pure solid K = 1 / PA^a x [B]^b

The following reaction reaches equilibrium in a closed reaction vessel 3Fe(s) + 4H2O(g) >< Fe3O4(s) + 4H2(g) Which of the following actions changes the value of Kc? i) Reducing the volume of the reaction vessel ii) Adding some Fe(s) iii) Adding a catalyst iv) Removing some H2(g) A i ad iv. B. i, ii and iv C. iv only D. all of i, ii, iii and iv E. none of i, ii, iii, and iv

Volume changes it (same moles gas both sides no effect) Catalyst affects rate not Kc Fe not in Kc equation Removing H2 would change equilibrium concentrations, but the only thing that changes the Kc value itself is temperatuer *E*

When will you have sensitive system?

When higher the Ea, the gradient shows influences of temperature. so small change in temp = large change in k because so steep

What are the equilibrium concentration when 0.1mol each of CO(g) and H2(g) are allowed to come to equilibrium in a 10L vessel. At the temperature of the experiment CO(g) + H2(g) >< C(s) + H2O(g) Kc = 1.5e22

[CO]o = [H2]o = 0.01M ^^ 0.1mol in 10L = 0.01M I: 0.010 | 0.010 | n/a | n/a I': n/a | n/a | solid | 0.010 ^^ Assuming reaction goes entirely to completion ^^ 0.010 H2O made from stoich C: +x | +x | solid | -x ^^ Assuming reaction goes entirely backwards ^^ Since (From Kc) back reaction is small, may be negligible E: Kc = [H2O(g)] / [H2] x [CO] = [0.01-x] / [x][x] 5% assumption to say that 0.01-x is pretty much still 0.01 since X is so small = 0.01 / x^2 [CO] = [H2] = 8.2e-13 [H2O] = 0.01M

What happens when the volume of an equilibrium mixture of gases is reduced.... When the volume is increases... Example for 2SO2 + O2 >< 2SO3?

a net rxn occurs in the direction tha tproduces fewer moles of gas a net rxn occurs in the direction that produces more moles of gas Kc = [SO3]^2 / [SO2]^2[O2] rewrite as #moles/volume since c=m/v - volumes cancel out leaving (nSO3^2 / (nSO2^2 x nO2)) x v So if increase kc....

At 2000 K, Kc = 0.154 for the reaction below. 2 CH4(g) → C2H2(g) + 3 H2(g) If a 1.75-L equilibrium mixture at 2000 K contains 0.80 mol each of CH4(g) and H2(g), answer the following questions. (a) What is the mole fraction of C2H2 present? (Note: The mole fraction of C2H2 is equal to the number of moles of C2H2 divided by the total number of moles.) (b) If the equilibrium mixture at 2000 K is transferred from a 1.75-L flask to 0.250-L flask, will the number of moles of C2H2(g) increase, decrease, or remain unchanged?

a) ?? b) Qc = 1.25 > Kc = moves left Decrease volume w/ stoich ratio of 2:4 moves left

(a) What is the value of the reaction quotient (Qc ) for the reaction below if a 8.00-L flask contains 0.0255 mol NO2(g) and 0.824 mol N2O4(g) at 25°C? N2O4(g) → 2 NO2(g), Kc = 4.61x10-3 at 25°C (b) If the reaction is not at equilibrium, in which direction will the reaction proceed—toward products or reactants?

a) Qc = [products] / [reactants] = [NO2]^2 / [N2O4] [NO2] = 0.0255mol/8L = 0.00319M [N2O4] = 0.824mol/8L = 0.103M Qc = [0.00319M]^2 / [0.103M] = *9.88e-5* b) Qc < Kc So *goes right* to establish equilibrium

H2(g) + S(s) >< H2S(g) where H<0 Which of the following actions increases the mass of H2 in the equilibrium mixture a) Increase vol b) Decrease vol c) Remove a small amount of sulfur d) Adding a small amount of of sulfur e) Increasing temp f) Decreasing temp

add heat H2 + S >< H2S + heat a) No since moles of species are equal on both sides (ignore solid) b) Same as (a) c) Ignore pure solids so nah d) Same as (c) e) Yes this would support endo (reverse) f) No this would support exo (forward)

Given that H2C2O4 (aq) + 2H2O >< C2O4 2- (aq) + 2H3O+ (aq) Kc = 2.9e-6 and 2H2O(;) >< H3O+ (aq) + HO-(aq) Kc = 1e-14 Calculate Kc for the reaction: H2C2O4(aq) + 2 HO- (aq) >< C2O4 2- (aq0 + 2H2O(l)

eq 3 - eq1 + -2eq2 = k3 k1 x 1/k2^2