Chapter 16 - Principles of Chemical Equilibrium

Transcription

1 Chapter 16 - Principles of Chemical Equilibrium -allreactions are "reversible" - principle of micro-reversibility - the "committed step" - much theory - not always obvious - for some the reverse reaction is very difficult - a measure of reversibility is the "equilibrium constant"

2 16.1 Dynamic Equilibrium the book provides a number of examples: - vapour over a liquid - substance in a saturated solution - iodine partitioning (solid/vapour) º all of these are physical equilibria what about chemical equilibria? - protonation/deprotonation of an acid - special type of equilibrium - Chapter 17 -CaO + CO 2 º CaCO 3-3O 2 º 2O 3

3 all equilibria are dynamic that is, A º B all equilibria are simply a competition between a forward and reverse reaction equilibrium is achieved when the rate of the forward reaction is exactly equal to the rate of the backwards (or reverse) reaction - the net change is zero but this is not the same thing as saying there is no reaction - the reactions occur but balance each other out for no net change

4 note that the individual reactions (the forward and reverse reactions) for an equilibrium are elementary steps so the coefficients are the stoichiometric factors hence, if: 2A º B then: k forward [A] 2 = k reverse [B]

5 16.2 The Equilibrium Constant Expression there are many ways to write the equilibrium constant expression but they all equate to: K eq = [products] / [reactants] that is, a numerical constant gives the proportion of products relative to the proportion of reactants recall: by definition, products = right hand side reactants = left had side

6 the equilibrium constant expresses a ratio so, what happens if part of the equilibrium is missing? for example, CO (g) + 2H 2 (g) º CH 3 OH (g) (this is the production of methanol from syn gas ) the equilibrium can be written as: K eq = [CH 3 OH (g) ] / [CO (g) ][H 2(g) ] 2 = 14.5 M -2

7 K eq = [CH 3 OH (g) ] / [CO (g) ][H 2(g) ] 2 = 14.5 M -2 if we start with, say, [CO] = 1.0 M and [H 2 ] = 1.0 M, then to balance this equation, we are going to have to have enough methanol present so that the ratio of methanol to carbon monoxide and hydrogen is 14.5 to 1. Note that since we do not have any methanol to begin with, reducing either the concentration of carbon monoxide or hydrogen will not get us to a balanced equation. The only way to achieve the prescribed ratio is by increasing the amount of methanol present. (In this case, to 14.5 M.)

8 the ratio is called the equilibrium constant expression and K eq is the equilibrium constant normally, when the concentrations are in molarity, then the equilibrium constant can be called K c but if pressure is used then it is also designated as K p We also have the terms K a for acid equilibrium constant and K b for base equilibrium constant and, of course, there is K sp for solubility equilibrium constant which is more frequently referred to as the solubility product º all use a capital K

9 General expression for equilibria: aa + bb º cc + dd K c = [C] c [D] d / [A] a [B] b (wrong in note package) thermodynamics

10 16.3 Relationships involving Equilibrium Constants between K c and the balanced equation: 1) reversing the equation inverts K c i.e. A º B k 1 [A] = k 2 [B] k 1 /k 2 = [B]/[A] B º A k 3 [B] = k 4 [A] k 3 /k 4 = [A]/[B] since, [B]/[A] = ([A]/[B]) -1, k 1 /k 2 = (k 3 /k 4 ) -1 and K 12 = (K 34 ) -1

11 2) multiplying the coefficients or stoichiometric factors by a constant does the same thing to the exponents i.e. 2H 2 + O 2 º 2H 2 O K c = [H 2 O] 2 /[H 2 ] 2 [O 2 ] H 2 + ½O 2 º H 2 O K c = [H 2 O]/[H 2 ][O 2 ] ½

12 3) adding equilibria, we multiply the constants i.e. A º B K 1 = [B]/[A] B º C K 2 = [C]/[B] overall: A º C K = [C]/[A] = ([B]/[A])([C]/[B]) = K 1 K 2

13 Equilibria involving gases, K p for a gas, PV = nrt so, (n/v) = P/RT if we consider an equilibrium involving gases, then we have to account for the RT term in the gas law equations two examples: H 2 + Cl 2 º 2HCl K c = [P HCl /RT] 2 / [P H2 /RT][P O2 /RT] = [P HCl ] 2 / [P H2 ][P O2 ] x (RT) 2 /(RT) 2

14 2H 2 + O 2 º 2H 2 O K C = [P H2O /RT] 2 / [P H2 /RT] 2 [P O2 /RT] = [P H2O ] 2 / [P H2 ] 2 [P O2 ] x (RT) 3 /(RT) 2 = [P H2O ] 2 / [P H2 ] 2 [P O2 ] x (RT) 3 /(RT) 2 K c = K p x RT

15 General expression: K p = K c (RT) )ngas where )n gas is the difference in the number of gas molecules between the products and the reactants i.e. 2H 2 (g) + O 2 (g) º 2H 2 O (g) where )n gas = (2) - (2+1) = -1 2H 2 (g) + O 2 (g) º 2H 2 O (l) where )n gas = (0) - (2+1) = -3

16 Pure substances and Equilibria liquids and solids - when pure - are not included in the Equilibrium Constant Expression Why? Because their concentration is constant and just subsumed by the equilibrium constant i.e. pure water - always 55.5 M - never changes pure gold - density 19.3 g/cm 3 and 1cm 3 = 1 ml therefore, 1 L of gold = 19.3 kg and gold is 98 M - a constant

17 16.4 Magnitude of Equilibrium Constants As it is a ration, it gives the relationship between the products and the reactants. Hence, - large K - more products than reactants - small K - more reactants that products -K. 1 - balance between products and reactants note that a large K tells us that the reaction favours the products but still have the activation barrier to consider - reaction may not reach equilibrium quickly

18 16.5 The Reaction Quotient Concept - allows you to determine the direction of the shift in an equilibrium when it has been disturbed - not really necessary! Le Ch>telier s Principle When a system at equilibrium is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.

19 much easier to say if an equilibrium is disturbed, it will respond in a way so as to minimize the disturbance but which way will that be? that is what Q, the reaction quotient, is supposed to tell us Q is the instantaneous equilibrium quotient that is, it is the value of the equilibrium constant expression calculated based on the concentration of the species present at any instant in time

20 It is not really necessary to calculate Q. However, consider the following: CO + 2H 2 º CH 3 OH K eq = [CH 3 OH]/[CO][H 2 ] 2 = 14.5 M -2 if [CO] [H 2 ] [CH 3 OH] then 0.0 M 0.0 M 0.1 M Q = M 0.1 M 0.0 M Q = M 0.1 M 0.1 M Q = 100

21 according to the book, if Q = 0 Q < K eq Q = K eq Q > K eq favour left) Q = 4 favour left) then the reaction proceeds to the right and produces more product then the reaction proceeds to the right and produces more product then the reaction is at equilibrium then the reaction reverses and shifts in of the reactants (proceeds to the then the reaction reverses and shifts in of the reactants (proceeds to the

22 16.6 Altering Equilibrium Conditions - applying Le Ch>telier s Principle effect of changing concentration - equilibrium shifts so as to remove the effect - finds a new equilibrium effect of changing pressure - two ways to do this - adding an inert gas and adding one of the reactants - both change the concentration terms, P gas

23 - change in partial pressure is a change in concentration and the system responds accordingly note: a decrease in overall pressure - equilibrium shifts in favour of the side with the most number of gas molecules an increase in overall pressure - equilibrium shifts to the side with the fewest gas molecules

24 effect of temperature book says raising the temperature or an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction. Lowering the temperature causes a shift in the direction of the exothermic reaction. - of course, this means that you have to know the thermicity of the reaction!! more important is the Arrhenius equation as a change in temperature is the only thing that can change the value of the rate constant and the equilibrium constant is a function of both the forward and reverse rate constants! changes in temperature are the only thing that can change the value of K eq

25 effect of catalyst - speeds up both the forward and reverse reaction (by changing the activation energy of both) - changes how fast equilibrium is obtained - doesn t change the position of the equilibrium

26 16.7 Equilibrium Calculations eight simple steps: 1) write out the equilibrium (most likely already done for you) 2) write out the equilibrium constant expression 3) write out and fill in an ICE table ICE stands for: Initial Change Equilibrium this table should contain values for each reactant and product in the equilibrium expressed mathematically

27 4) substitute the equilibrium expressions from the ICE table into the equilibrium constant expression 5) multiply out and manipulate the resulting algebraic expression to get a quadratic (or simpler) equation 6) substitute the coefficients into the quadratic formula and solve 7) substitute the resulting answer into the equilibrium expressions from the ICE table to get the equilibrium concentrations of all of the reaction species 8) CHECK YOUR ANSWER!

28 step 1) A + B º C step 2) K eq = [C]/[A][B] step 3) [A] [B] [C] initial change - x - x x equilibrium x x x step 4) K eq = (x)/( x)( x) = 10.3

29 step 5) (x) / ( x)( x) = 10.3 (x)/ ( x 2 ) = 10.3 x = 10.3 ( x 2 ) x = 10.3x x x x = 0 (quadratic: ax 2 + bx + c = 0 )

30 step 6) x = [ -b ± (b 2-4ac) ½ ] / 2a = [-(-3.06) ± ( (10.3)(0.103)) ½ ] / 2(10.3) = [3.06 ± ( ) ½ ] / 20.6 = [3.06 ± (5.12) ½ ] / 20.6 = [3.06 ± 2.263] / 20.6 x = or x =

31 step 7) if x = [A] e = = not possible!! a concentration can not be negative! hence, x = [A] e = = [B] e = = [C] e =

32 step 8) K eq = ( ) / ( ) 2 = 10.3 Note: What if I had chosen wrong? That is, if I had set up the ICE table with [A] = x, [B] = x, and [C] = -x? The answer would have given me a negative value for x and step 7 would still have given me the correct values!! You don t have to work out the reaction quotient to determine the equilibrium values. If you correctly follow all eight steps, every time, then you will minimize the chance of getting it wrong. And you can check your answer!

33 Examples A flask initially contains only M H 2 and M F 2. These two compounds react to product HF. The concentration of HF at equilibrium is M. What is the value of K eq? step 1)

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35 more commonly, questions are like: At 25 C, the equilibrium constant for the reaction: Ca 2+ (aq) + SO 4 2- (aq) º CaSO 4 (aq) is 156 M -1. If the initial concentrations are [Ca 2+ ] = 0.26 M, [SO 4 2- ] = 0.21 M, and [CaSO 4 ] = 0.83 M, calculate the concentration of the species at equilibrium. step 2)

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38 What if we set up the ICE table the other way around? step 3) [Ca 2+ ] [SO 2-4 ] [CaSO 4 ] initial change + x + x - x equilibrium x x x

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40 Another example, Br 2 + I 2 º 2IBr if K eq = 1.36, [Br 2 ] o = 0.146M, [I 2 ] o = M, and [IBr] o = 0.0M, then what are the concentrations at equilibrium? step 2)

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43 if [IBr] o = M, then what would the concentrations be at equilibrium? step 1)

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46 An example of the most complicated type of question: for the equilibrium, HI + Br 2 º HBr + IBr if K eq = 10.4 and [HI] o = M, [Br 2 ] o = M, [HBr] = M, and [IBr] = 0.0 M, then what are the concentrations at equilibrium? step 2)

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