The Equilibrium Constant, K, and The Reaction Quotient, Q SCH 4U

Chemical Systems at Equilibrium For reversible reactions in closed systems, the reaction mixture always contains both reactants and products. At equilibrium, the forward and reverse reaction rates are ____________ and the reactant and product concentrations are ____________. EQUAL CONSTANT

Just like we learned yesterday that the percentage reaction is always constant (under what conditions?)......These two smart Norwegian dudes noticed, for a system at equilibrium, the following relationship exists...

Equilibrium Law For the general rxn: aA + bB cC + dD (a-d are coefficients; A-D are chemical species) The following Equilibrium Law expression can be written: K = [C] c [D] d [A] a [B] b

K is a constant called... The Equilibrium Constant Note: This is for a system AT EQUILIBRIUM, therefore conc’s are equilibrium conc’s! Also, species are in aqueous or gas phase......if liquids or solids, they do NOT appear in the expression (b/c conc. unchanging)

No matter what reactant/product concentrations we start at, once equilibrium is established, the value of K is always the same!* Ahem.... Equilibrium Constant... (see p.439 if you don’t believe me) *It is still temperature-dependent though!

Writing an Expression for K Ex. 1 N 2(g) + 3 H 2(g) 2 NH 3(g) K = __________ [NH 3 ] [N 2 ][H 2 ] 2 3

Type 1: Calculating K For the same system, given the following equilibrium concentrations, find K: [N 2(g) ] = 1.50 x mol/L [H 2(g) ] = 3.45 x mol/L [NH 3(g) ] = 2.00 x mol/L K = (2.00 x ) 2. (1.5 x ) (3.45 x ) 3 K = 6.49 x at 500°C

Type 2: Calculating Equil. Conc. H 2(g) + I 2(g) 2 HI (g) K= at 458°C Given:[H 2 ] equil = 1.07 mol/L [I 2 ] equil = 1.07 mol/L Calculate [HI] at equilibrium K = [HI] 2 [H 2 ] [I 2 ] [HI] = √ K[H 2 ][I 2 ] Rearranging: = 7.54 mol/L

The Magnitude of K K = [C] c [D] d [A] a [B] b reaction proceeds toward completion (much more products) concentrations of reactants and products are approx. equal at equilibrium very small amounts of products formed (much more reactants) products reactants If K >> 1 If K ≈ 1

Comparing K’s H 2(g) + I 2(g) 2HI (g) K = 50 (at 450°C) CO 2(g) + H 2(g) CO (g) K = 1.1 (at 900°C) How do these systems at equilibrium compare?

Type 3: Using K and ICE Table (knowing initial conc’s and K) H 2(g) + I 2(g) 2 HI (g) K= 64 (at partic. temp) Given:[H 2 ] initial = [I 2 ] initial = 3.0 M Find all equilibrium concentrations. H 2(g) + I 2(g) = 2 HI (g) Initial3.0 M 0 Change Equilibrium -x +2x 3.0 M - x 2x K = [HI] 2 [H 2 ] [I 2 ] = [2x] 2 [3.0M – x][3.0M – x] = [2x] 2 [3.0M – x] 2

Therefore at equilibrium: [H 2 ] = 3.0M – 2.4 = 0.6 M [I 2 ] = 3.0M – 2.4 = 0.6M [HI] = 2(2.4M) = 4.8 M K = [2x] 2 [3.0M – x] 2 64 = [2x] 2 [3.0M – x] 2 8 = 2x 3.0 M – x x = 2.4

Type 4: Using K and ICE Table (knowing initial and equil concs) Given above concentrations, find K H 2(g) + I 2(g) = 2 HI (g) Initial2.0 M1.0 M0 Change Equilibrium1.80 M First need to find equilibrium concentrations! K = [HI] 2 [H 2 ] [I 2 ] -x +2x 2.0M -x1.0M -x Need to solve for x! 1.80M = 2x 0.9M = x = 1.1M= 0.1M

Now can plug in equilibrium conc’s to find value of K: K = (1.80) 2 (1.10)(0.1) K = 29

The Reaction Quotient, Q Q = [C] c [D] d [A] a [B] b K is calculated using conc’s at equilibrium Q is calculated using conc’s that may or may not be at equilibrium

... Tells us if system is at equilibrium or, if not, which way system must shift to reach equilibrium! If Q = K If Q < K If Q > K System at equilibrium Have more products than system at equilibrium, must shift to left (toward reactants) Have less products than system at equilibrium, must shift to right (toward products)

Example N 2(g) + 3H 2(g) 2NH 3(g) K= at 450°C Given:[N 2 ] = 4.0 mol/L [H 2 ] = 2.0 x mol/L [NH 3 ] = 2.2 x mol/L Is system at equilibrium? Q = [NH 3 ] 2 [N 2 ] [H 2 ] 3 Q < K, not at equilibrium, needs to shift right, toward products =

Type 5: Calculating Q First Calculate the equilibrium concs of H 2, I 2, and HI if initial concs are 0.5M, 0.5M, and 4.0M respectively and K is 50. Set up ICE Table... H 2(g) + I 2(g) = 2 HI (g) Initial0.5 M 4.0 M Change Equilibrium -x +2x WATCH OUT!! None of the reactant conc’s are zero – have to first det. in which direction system is shifting!

How do we do that? → Find Q! Q = [HI] 2 = (4.0) 2 = 64 [H 2 ][I 2 ] (0.5)(0.5) Now compare Q to K... Q > K therefore system shifts to left (toward reactants) H 2(g) + I 2(g) = 2 HI (g) Initial0.5 M 4.0 M Change Equilibrium + x - 2x 0.5M + x 4.0M - 2x

How do we proceed? (Don’t know x yet) Use K to find value of x = (4.0M – 2x) 2 (0.5M + x)(0.5M + x) 7.07 = (4.0M – 2x)x = 0.05 (0.5M + x) Then can calculate equil. conc’s... (Do it!)

Note: If an equil. problem cannot be solved by taking the square root of both sides of the equil. equation, the quadratic equation must be used. x = – b + √b 2 – 4ac 2a

Using the 100 Rule 2CO 2(g) 2CO (g) + O 2(g) K = 6.40 x Calc. equil. concentrations if mol of CO 2 is placed in a 1L closed container and heated to 2000°C Since there are initially no products, Q=0 and reaction will proceed to right... at 2000°C

Use K to find x... K = [CO] 2 [O 2 ]= 6.40 x [CO 2 ] 2 = (2x) 2 (x) (0.25 – 2x) 2 = 4x 3 (0.25 – 2x) 2 2 CO 2(g) = 2 CO (g) + O 2(g) Initial0.250 M00 Change- 2x+2x+x Equilibrium0.250 – 2x2xx This is a cubic (x 3 ) function, hard to solve directly Can we simplify it?

Yes! K is very small (10 -7 ) in comparison to initial conc of CO 2(g)...which means very little CO 2(g) decomposes at this temp Therefore we can assume x to be very small (2x as well) and negligible...

i.e – 2x ≈ So our expression becomes... 4x 3 ≈ 6.40 x (0.25) 2 x ≈ 2.15 x (can check validity of assumption)

100 Rule If concentration to which x is added or subtracted from is 100x or more greater than the value of K, then we can ignore x e.g. [CO 2 ] initial = = 3.91 x 10 5 K 6.40 x See Example p. 472 Which is >> 100!

No Class Tomorrow In total 2 lessons behind the other class... It is your responsibility to make sure you are okay with ALL the homework questions that follow... you have 4 nights/ 3 days! Come see me before school or at lunch on Monday if there are any problems. We will be having an assessment on Monday

Practice Problems *See (do!) sample problem on p. 472 (100 rule) p. 444 # 2, 3 p. 445 # 5, 6 p. 447 # 7 p. 449 # 9 p.472 # 5, 6 p. 476 # 8 p. 480 # 9, 10 p. 481 # 2, 3a, 3c, 4-8