Reactions are Reversible Z Most reactions do NOT proceed to completion and then stop. Z Most reactions proceed in both the forward and reverse directions: A + B C + D ( forward) C + D A + B (reverse)
Reactions are Reversible Z A + B C + D ( forward) Z C + D A + B (reverse) Z Initially only A and B are present, so only the forward reaction is possible. Z As C and D are produced the reverse reaction speeds up, while the forward reaction slows down. Z Eventually the rates become equal.
Z Chemical equilibrium: concentrations of all species remain constant with time.
What is equal at Equilibrium? Z Rates are equal. Z Concentrations are not. Z Rates are determined by concentrations and activation energy. Z The concentrations do not change at equilibrium. Z Recall equilibrium is dynamic! a Reactions keep going at equilibrium! a A + B C + D
Reaction Quotient, Q Z A ratio between the amount of reactants and products can be made at any time during the reaction. This is called the reaction quotient (Q). Z For aA + bB cC + dD the reaction quotient is: Q = [C] c [D] d [A] a [B] b Z Notice the effect of coefficients! Z Q may be written using concentrations (Q c ) or partial pressures (Q p ).
Equilibrium Constant, K Z The ratio between the amount of reactants and products at equilibrium is called the equilibrium constant (K). a K = [C] c [D] d [A] a [B] b a Again, may be K c or K p. a K has no units- it represents a ratio! Z Side note: P = CRT. Derived from ideal gas law since C = n/V (C = concentration). equilibrium expression
Equilibrium Constant, K Z Pure solids and liquids are NOT included in the equilibrium expression; their concentrations are assumed to be one. Why? a Concentrations of pure substances don’t change! Z Also, K is constant at any temperature. * If T changes then K will change!
Example Z Write the equilibrium expression (K p ) for the following reaction: SiH 4 (g) + 2Cl 2 (g) SiCl 4 (g) + 2H 2 (g) Remember to use partial pressures! K p = (P SiCl4 )(P H2 ) 2 (P SiH4 )(P Cl2 ) 2
Altering K Z If we write the reaction in reverse: cC + dD aA + bB Z Then the new equilibrium constant is: K ’ = [A] a [B] b = 1/K [C] c [D] d
Example Z At a given temperature, K = 1.3 x 10 -2 for the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) Z What is the value of K for the following: 2NH 3 (g) N 2 (g) + 3H 2 (g) K = 1/(1.3 x 10 -2 ) = 77
Altering K Z If we multiply the reaction by some number n: (aA + bB cC + dD)n naA + nbB ncC + ndD Z Then the new equilibrium constant is: K ’ = [C] nc [D] n d = ([C] c [D] d ) n = K n [A] na [B] nb ([A] a [B] b ) n
Example Z At a given temperature, K = 1.3 x 10 -2 for the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) Z What is the value of K for the following: 1/2N 2 (g) + 3/2H 2 (g) NH 3 (g) K = (1.3 x 10 -2 ) 1/2 = 0.11
Example- Calculating K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial Equilibrium Z [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Z [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Z [NH 3 ] 0 =0 M [NH 3 ] = 0.157M Z K = _ [NH 3 ] 2 _ = 0.0603 [N 2 ] [H 2 ] 3
Example- Calculating K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial Equilibrium Z [N 2 ] 0 = 0 M [N 2 ] = 0.399 M Z [H 2 ] 0 = 0 M [H 2 ] = 1.197 M Z [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M Z K = 0.0602 Z K is the same no matter what the amount of starting materials.
What does K tell us? Z By looking at the size of K (and also Q), you can determine relative amounts of reactants and products present. Z Large values of K indicate greater amounts of products compared to reactants (forward reaction is favored/reaction lies to the right). Z Small values of K indicate more reactants are present (reverse reaction is favored/reaction lies to the left).
K c and K p Values of K calculated from concentrations or partial pressures can be related: K p = K c (RT) Δn a R = ideal gas constant: 0.0821Latm/molK Make sure to choose the correct R value- look at units of P! Also T must be in K! Δn = change in # moles of gas in the reaction: mol product – mol reactant
Example Z For the reaction C(s) + CO 2 (g) 2CO(g) K p =1.90 at 25°C. What is K c ? 1.90 = K c [(0.0821Latm/molK)(298K)] 1 K c = 0.0776
Lesson Essential Question (sections 5-7) Z How can equilibrium pressures and concentrations be calculated?
APPLICATIONS OF THE EQUILIBRIUM CONSTANT Section 13.5
Using Q Z Recall that it’s calculated the same way as the equilibrium constant, but for a system not at equilibrium (or if you’re not sure if a reaction is at equilibrium or not). Z By calculating Q you can compare the value to K to get an idea of where a reaction is in terms of reaching equilibrium.
What Q tells us: Z If Q<K a Not enough products (has not yet reached equilibrium). a Shift to the right to reach equilibrium. Z If Q>K a Too many products (has gone past equilibrium). a Shift to the left to reach equilibrium. Z If Q=K system is at equilibrium
Example Z For the reaction 2NO(g) N 2 (g) + O 2 (g) K = 2.4 x 10 3 at a certain temperature. If there are 0.024mol NO, 2.0mol N 2, and 2.6mol O 2 in a 1.0L flask, is the reaction at equilibrium? If not, in which direction will the reaction shift to reach equilibrium? Z Q = (2.0)(2.6) = 9.0 x 10 3 Q > K (0.024) 2 rxn. shifts left
Procedure 1. Write balanced equation. 2. Write equilibrium expression. 3. Calculate Q, determine shift direction. 4. Start ICE table– determine I (initial concentrations). 5. Define change, C – add to I to find E (equilibrium concentrations). 6. Substitute E into expression and solve 7. Check values in expression, see if they equal K.
Example Z If K = 5.10 and there is 1.000mol of each substance present in a 1.000L flask, what are the equilibrium concentrations? CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) Z Initial concentrations: all 1.000M Z Determine Q: Q = (1.000M)(1.000M) = 1.000 Q<K, so (1.000M)(1.000M) rxn. shifts right
Example Cont. Z Now make your ICE table! Write ICE in a column below and to the left of the eqn. Z Use balanced equation coefficients and the direction the reaction shifts to find change. Z Add I and C to get E. Z ICE table helps you keep track of everything: CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) I 1.000 1.000 1.000 1.000 C -x -x +x +x E 1.000-x 1.000-x 1.000+x 1.000+x
Example Cont. Z Now plug the E values into the equilibrium expression and solve for x: 5.10 = [CO 2 ][H 2 ]_ = (1.000+x)(1.000+x) [CO] [H 2 O] (1.000-x)(1.000-x) 5.10 = (1.000 + x) 2 solve for x: x = 0.387 mol/L (1.000 – x) 2
Example Cont. Z [CO] & [H 2 O] = 1.000 – 0.387 = 0.613M Z [CO 2 ] & [H 2 ] = 1.000 + 0.387 = 1.387M Z Verify: K = (1.387)(1.387) = 5.12 (0.613)(0.613)
Practice Z K = 115 for a reaction where 3.000mol of all species were added to a 1.500L flask. What are the equilibrium concentrations? H 2 (g) + F 2 (g) 2HF(g) Q = 1.000, so rxn. will shift right. 115 = (2.000 + 2x) 2 => x = 1.528 (2.000 – x) 2 Z [H 2 ] = [F 2 ] = 0.472M ; [HF] = 5.056M
Using the Quadratic Formula Z You may have wondered when you would use this! Z Need it to solve some equilibrium problems (form: ax 2 + bx + c). Z Previous problems worked out nicely- just had to take the square root of both sides. Z Can use graphing calculator program (if you have it) to solve. Otherwise know it:
Example #1 – Quadratic Formula Z K = 115 for a reaction where 3.000mol of H 2 and 6.000mol F 2 were added to a 3.000L flask. What are the equilibrium concentrations? H 2 (g) + F 2 (g) 2HF(g) Z Q = _[HF] 2 _ [H 2 ][F 2 ] Z Reaction will shift right.
Example #1 Cont. H 2 (g) + F 2 (g) 2HF(g) I 1.000 2.000 0 C -x -x +2x E 1.000-x 2.000-x 2x K = 115 = _____(2x) 2 _____ (1.000-x)(2.000-x) (1.000-x)(2.000-x)(115) = 4x 2 (2.000 -3.000x+x 2 )(115) = 4x 2 230. – 345x + 115x 2 = 4x 2 111x 2 – 345x + 230. = 0 a = 111 b = -345 c = 230.
Example #1 Cont. Use quadratic formula: x = 345 ±√(-345) 2 – 4(111)(230.) 2(111) x = 2.14M and x = 0.968M Only one value of x is correct: [H 2 ] = 1.000M – x [H 2 ] = 1.000M – 2.14M = -1.14M [H 2 ] = 1.000M – 0.968M = 0.032M [F 2 ] = 2.000M – 0.968M = 1.032M [HF] = 2(0.968) = 1.936M a = 111 b = -345 c = 230. Can’t have a negative concentration!
What if K is very small? Z If K is < 10 -3 we can consider it to be negligible with respect to concentrations. a Can make calculations simpler. a Can ignore x only where it would make little change in the concentration. a MUST check to see if the calculated equilibrium concentration is less than 5% different from the initial – if it’s more than 5%, then you can’t disregard x.
Example #2- Small K Values Exactly 1.0mol NOC l is placed in a 2.0L flask. The value of K = 1.6x10 -5. What are the equilibrium concentrations? 2NOC l (g) 2NO(g) + C l 2 (g) *Rxn. will shift right b/c there are no products initially. *The equilibrium expression will be: K = [NO] 2 [C l ] = 1.6x10 -5 [NOC l ] 2
Example #2- Small K Values 2NOC l (g) 2NO(g) + C l 2 (g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *This would result in a complicated math equation to solve. *Simplification: K is small (<10 -3 ), so x is negligible only with respect to NOC l. -Small K means reaction barely proceeds right, so 0.50M NOCl won’t change noticeably.
Example #2- Small K Values 2NOC l (g) 2NO(g) + C l 2 (g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *Simplification: 0.50 – 2x ≈ 0.50M 1.6x10 -5 = (2x) 2 (x) = _4x 3 _ (0.50) 2 (0.50) 2 x 3 = (1.6x10 -5 )(0.50) 2 => x = 0.010 4
Example #2- Small K Values Verify assumption that x is negligible with respect to NOC l : [NOC l ] = 0.50 – 2(0.010) = 0.48 *This must be less than 5% different from the initial concentration: 0.50-0.48 = 0.02 0.02/0.50 x 100 = 4% Assumption was ok!
Example #2- Small K Values Calculate the rest of the equilibrium concentrations: [NOC l ] = 0.50M (this was the simplification!) [NO] = 2x = 2(0.010) = 0.020M [C l ] = x = 0.010M
Le Chatelier’s Principle Z If a change is applied to a system at equilibrium, the reactions will shift to reduce the change. Equilibrium is reestablished. Z Three types of change: a Concentration a Pressure a Temperature
Changes in Concentration Z Adding more of a substance will cause the reaction to shift in the direction that consumes it. Z Removing a substance will cause the reaction to shift in the direction that produces it. Z Example: N 2 O 4 (g) 2NO 2 (g) Z Adding NO 2 shifts the reaction to the left. Z How can you get the reaction to shift right?
Changes in Concentration Z What does this do to the amounts of reactants and products? Z The amounts of reactants and products will change in order to reestablish equilibrium. Z The ratio of reactants to products will be the same once equilibrium has been reestablished.
Changes in Pressure Z P can be changed by altering volume. a Note this is only applicable to gases! Z If V is decreased, P increases, and the reaction shifts in the direction that produces the least moles of gas. a Why? Z If V is increased, P decreases, and the reaction shifts in the direction that produces the most moles of gas. a Why? Z Equilibrium is reestablished in both cases. Fewer moles of gas = lower P. More moles of gas = higher P.
Example N 2 O 4 (g) 2NO 2 (g) Z What will happen if the volume of the container is decreased? Z If the volume of the container increases? Z Note: if there are equal # moles on both sides, changing P has no effect on equilibrium. Rxn. shifts left. Rxn. shifts right.
Change in Temperature Z Different from previous two changes! Z Doesn’t just change the equilibrium position, also changes the equilibrium constant K. Z The direction of the shift depends on whether it is exo- or endothermic. a It helps to think of E (heat) as a reactant or product.
Change in Temperature Consider the same reaction previously examined: 58kJ + N 2 O 4 (g) 2NO 2 (g) Z Endothermic: think of heat as a reactant. Z Increasing T means increasing heat; reaction shifts right to consume additional heat. Z What happens if T decreases? a Reaction shifts left to generate more heat.
Practice N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Consider the above reaction. What happens if: Z Additional H 2 is added? Z T increases? Z V decreases? Z How could the reaction shift left other than increasing T? Rxn. shifts right Rxn. shifts left Rxn. shifts right
AP Practice Question At constant T, a change in V will NOT affect the moles of substances present in which of the following? a) H 2 (g) + I 2 (g) 2HI (g) b) CO(g) + Cl 2 (g) COCl 2 (g) c) PCl 5 (g) PCl 3 (g) + Cl 2 (g) d) N 2 (g) + 3H 2 (g) 2NH 3 (g)
AP Practice Question C(s) + H 2 O(g) CO(g) + H 2 O(g) endo. An equilibrium mixture of the reactants is placed in a sealed container at 150°C. The amount of the products may be increased by which of the following changes? a) Decreasing the V of the container. b) Raising T & increasing the V. c) Decreasing T of the container. d) Increasing V and adding 1mol of CO(g).
AP Practice Question 2CH 4 (g) + O 2 (g) 2CO(g) + 4H 2 (g) ΔH<0 In order to increase the value of the equilibrium constant, K, which of the following changes must occur? a) Increase T. b) Increase V. c) Decrease T. d) Add CO(g).