# Equilibrium II 15.6 – Using Keq 15.7 – Le Chậtelier’s Principle 19.7 20.5.

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Equilibrium II 15.6 – Using Keq 15.7 – Le Chậtelier’s Principle 19.7 20.5

Calculating Equilibrium Concentrations The same steps used to calculate equilibrium constants are used to calculate equilibrium concentrations. Generally, we do not have a number for the change in concentration. We need to assume that x mol/L of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions.

#1At 2000 o C, the equilibrium constant for the reaction: 2 NO (g)  N 2 (g) + O 2 (g) is K c = 2.40  10 3. If the initial concentration of NO is 0.200 M, what are the equilibrium concentrations of NO, N 2 and O 2 ? 2 NO(g)  N 2 (g) + O 2 (g) init 0.200M 0 0 change - 2x x x equil. 0.200 – 2x x x

#2At 100 o C, Kc = 0.078 for the reaction: SO 2 Cl 2  SO 2 + Cl 2. In an equilibrium mixture of the three gases, the concentrations of SO 2 Cl 2 and SO 2 are 0.108 M and 0.052 M, respectively. What is the partial pressure of Cl 2 in the equilibrium mixture?

#3At 373 K, K p = 0.416 for the equilibrium: 2 NOBr  2 NO + Br 2. If the pressures of NOBr and NO are equal, what is the equilibrium pressure of Br 2 ?

#4For the reaction: H 2 + I 2  2 HI K c = 55.3 at 700 K. In a 2.00 L flask containing an equilibrium mixture of the three gases, there are 0.056 g H 2 and 4.36 g I 2. What is the mass of HI in the flask? (b)

Applications of Equilibrium Constants - Predicting the Direction of Reaction For a general reaction: aA + bB cC + dD We define Q, the reaction quotient, as: Where [A], [B], [C], and [D] are molarities (for substances in solution) or partial pressures (for gases) at any given time.

Q is a K expression with non-equilibrium concentration values.

Q = K eq only at equilibrium. If Q < K eq then the forward reaction must occur to reach equilibrium. –Reactants are consumed, products are formed. –Q increases until it equals K eq. If Q > K eq then the reverse reaction must occur to reach equilibrium. –Products are consumed, reactants are formed. –Q decreases until it equals K eq.

#5At 100 o C, the equilibrium constant for the reaction: COCl 2  CO + Cl 2 has the value K c = 2.19  10 -10. Are the following mixtures of COCl 2, CO, and Cl 2 at 100 o C at equilibrium? (a) [COCl 2 ] = 2.00  10 -3 M [CO] = 3.3  10 -6 M [Cl 2 ] = 6.62  10 -6 M (b) [COCl 2 ] = 4.50  10 -2 M [CO] = 1.1  10 -7 M [Cl 2 ] = 2.25  10 -6 M (c) [COCl 2 ] = 0.0100 M [CO] = [Cl 2 ] = 1.48  10 -6 M

15.7 Le Châtelier’s Principle Le Châtelier’s principle: If a system at equilibrium is disturbed by a change in temperature, a change in pressure, or a change in the concentration of one or more components, the system will shift its equilibrium position in such a way as to counteract the effects of the disturbance.

Change in Reactant or Product Concentration If a chemical system is at equilibrium and we add or remove a product or reactant, the reaction will shift so as to reestablish equilibrium. For example, consider the Haber process: N 2 (g) + 3H 2 (g) 2NH 3 (g) If H 2 is added while the system is at equilibrium, Q < K eq The system must respond to counteract the added H 2 (by Le Châtelier’s principle). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increase until Q = K eq.

Effects of Volume and Pressure Changes

If the equilibrium involves gaseous products or reactants, the concentration of these species will be changed if we change the volume of the container. For example, if we decrease the volume of the container, the partial pressures of each gaseous species will increase. Le Châtelier’s principle predicts that if pressure is increased, the system will shift to counteract the increase. –the system shifts to remove gases and decrease pressure..

An increase in pressure favors the direction that has fewer moles of gas. A decrease in pressure favors the direction that has more moles of gas. In a reaction with the same number of moles of gas in the products and reactants, changing the pressure has no effect on the equilibrium. No change will occur if we increase the total gas pressure by the addition of a gas that is not involved in the reaction because the partial pressures of the gases will stay constant.

Effect of Temperature Changes Temperature affects the equilibrium constant: For an endothermic reaction, heat can be considered a reactant. –Adding heat causes a shift to the right –Removing heat causes a shift to the left For an exothermic reaction, heat can be considered a product. –Adding heat causes a shift to the left –Removing heat causes a shift to the right

At room temperature, an equilibrium mixture (light purple) The mixture is placed in a beaker of warm water. –The mixture turns deep blue - shift toward products - CoCl 4 2– The mixture is placed in ice water. –The mixture turns bright pink - shift toward reactants - Co(H 2 O) 6 2+.

The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst DOES NOT effect the composition of the equilibrium mixture

#6 Consider the following equilibrium, for which ΔH < 0: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g). How will each of the following changes affect an equilibrium mixture of the three gases? (a) O 2 (g) is added to the system (b) the reaction mixture is heated (c) the volume of the reaction vessel is doubled (d) a catalyst is added to the mixture (e) the total pressure of the system is increased by adding a noble gas (f) SO 3 (g) is removed from the system (a)shifts right (b)shifts left (c)shifts left – increasing volume decreases pressure  favors the side with more moles of gas (d)no effect (e)no effect (f) shifts right + heat

#7 How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant or product (b) decrease in the volume (c) decrease in the temperature (d) addition of a catalyst (a)no effect (b)no effect (c)Temperature affects Keq: exothermic: inverse relationship endothermic: direct relationship increase Keq (d)no effect

#8 For the reaction, PCl 5 (g)  PCl 3 (g) + Cl 2 (g) ΔH rxn = +111 kJ. Fill in the following table: Change to reactionReaction shiftsChange in K Add PCl 5 Remove Cl 2 Add Ar Decrease P Increase T Add a catalyst Decrease V of container right none right none no shiftnone right none right increases no shift none left none

Review! Gibbs Free Energy: ΔG –Combines enthalpy and entropy to tell us whether a reaction will be spontaneous or not If ΔG is (+) the reaction is nonspontaneous If ΔG is (–) the reaction is spontaneous If ΔG = 0 the system is at equilibrium

19.7 Free Energy and the Equilibrium Constant ΔG o applies under standard conditions, but most reactions don’t happen under standard conditions! ΔG and Q apply to any conditions. It is useful to determine whether substances will react under specific conditions: ΔG = ΔG o + RT lnQ At equilibrium, Q = K eq and ΔG = 0 0 = ΔG o + RT ln K eq ΔG o = – RT ln K eq On equation sheet!

ΔG o = – RT ln K eq From the above we can conclude that at 298K: If ΔG o is negative, then K eq is greater than 1 (-) ΔG means spontaneous rxn, favors products If ΔG o = 0, then K eq = 1 If ΔG o is positive, then K eq is less than 1 (+) ΔG means nonspontaneous rxn, favors reactants

#9Explain qualitatively how ΔG changes for each of the following reactions as the partial pressure of O 2 is increased (a) 2 CO(g) + O 2 (g)  2 CO 2 (g) (b) 2 H 2 O 2 (l)  2 H 2 O(g) + O 2 (g) ΔG = ΔG o + RT lnQ (a) ΔG becomes smaller (or more negative) (b) ΔG becomes larger (or more positive)

#10Consider the reaction: 2 NO 2 (g)  N 2 O 4 (g). (a) Using data from Appendix C, calculate ΔG o at 298 K. (b) Calculate ΔG at 298 K if the partial pressures of NO 2 and N 2 O 4 are 0.40 atm and 1.60 atm respectively. (a) ΔG o = ΔG o N 2 O 4 (g) - 2 ΔG o NO 2 (g) = 98.28 - 2(51.84) = - 5.40 kJ

#11Use data from Appendix C to calculate K p at 298 K for each of the following reactions: (a) H 2 (g) + I 2 (g)  2 HI(g)

20.6 Effect of Concentration on Cell EMF Review: E is positive – spontaneous E is negative - nonspontaneous E =0 - equilibrium A voltaic cell is functional until E = 0 at which point equilibrium has been reached. (The cell is then “dead.”) The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction.

The Nernst Equation We can calculate the cell potential under nonstandard conditions. Recall that: ΔG = ΔG o + RT lnQ and ΔG = - nFE We can substitute in our expression for the free energy change: - nFE = - nFE o + RT lnQ Rearranging, we get the Nernst equation: The Nernst equation can be simplified by collecting all the constants together and using a temperature of 298 K: (n is the # of e transferred) On AP Equation Sheet! R=8.31

#12 At 298K we have the reaction: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) [Cu 2+ ] = 5.0 M and [Zn 2+ ] = 0.050M: Q = [Zn +2 ] / [Cu +2 ] (a) Cu +2 (aq) + 2e  Cu(s)E o red = +.34 V Zn(s)  Zn +2 (aq) + 2eE o ox = + 0.763V E o = + 1.10 V

Cell EMF and Chemical Equilibrium A system is at equilibrium when Q=K and ΔE = 0. Thus, if we know the E o, we can calculate the equilibrium constant. On AP Equation Sheet!

#13What is the effect on the emf of the cell which has the overall cell reaction: Zn(s) + 2 H +1 (aq)  Zn +2 (aq) + H 2 (g) for each of the following changes? (a) the pressure of the H 2 is increased in the cathode compartment (b) zinc nitrate is added to the anode compartment (c) sodium hydroxide is added to the cathode compartment, decreasing [H +1 ] (d) the surface area of the anode is doubled. (a) Q increases, E decreases (b) [Zn +2 ] increases, Q increases, E decreases (c) [H +1 ] decreases, Q increases, E decreases (d) No effect – does not appear in the Nernst equation

#14A voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn(s) + Ni +2 (aq)  Zn +2 (aq) + Ni(s) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Ni +2 ] = 3.00 M and [Zn +2 ] = 0.100 M? (a) Ni +2 (aq) + 2e  Ni(s)E o red = - 0.28 V Zn(s)  Zn +2 (aq) + 2eE o ox = + 0.763V Ni +2 (aq) + Zn(s)  Ni(s) + Zn +2 (aq)E o = + 0.48 V

#15 Using the standard state reduction potentials listed in Appendix E, calculate the equilibrium constant for the following reaction at 298 K: (a) Zn(s) + Sn +2 (aq)  Zn +2 (aq) + Sn(s)

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