Energy transformations Chemical Equilibrium Thermodynamics Energy transformations

N2O4 2 NO2 Initially forward reaction rapid As some reacts [N2O4] so rate forward  Initially Reverse reaction slow No products As NO2 forms  Reverse rate Eventually rateforward = ratereverse Equilibrium

Reaction Reversibility Closed system Equilibrium can be reached from either direction Independent of whether it starts with “reactants” or “products” Always have the same composition at equilibrium under same conditions N2O4 2 NO2 

Products Reactants Equilibrium N2O4 2 NO2 For given overall system composition Always reach same equilibrium concentrations Whether equilibrium is approached from forward or reverse direction

If equilibrium is established, concentrations don’t change any more.

Each set of equilibrium concentrations Equilibrium position

Check it yourself!

Relation between Kp and Kc

Learning Check Kc = 11.7 Consider the reaction: 2NO2(g) N2O4(g) If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature? n = nproducts – nreactants = 1 – 2 = –1 Kc = 11.7

Interpreting Kp or Kc Means product rich mixture 2SO2(g) + O2(g)  2SO3(g) Kc = 7.0  1025 at 25 ° C Reaction goes only ~ halfway Means reactant rich mixture H2(g) + Br2(g)  2HBr(g) Kc = 1.4  10–21 at 25 °C

more than phase involved Concentrations of pure liquids and solids do not change with changing the amounts of these substances.

Pure solids and liquids do not appear in the expression for equilibrium constant.

Reaction Quotient Q If Q=K → equilibrium Q ≠ K → not at equilibrium tries to reach equilibrium

For the synthesis of ammonia at 500oC, the equilibrium constant is 6 For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0×10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: Q >> K Q will decrease with time More reactants, less products System shifts to the right Reversed reaction dominates The amount of N2 and H2 will increase The amount of NH3 will decrease

Your turn! N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2(g). N2O4 2 NO2

At a certain temperature a 1. 00-L flask initially contained 0 At a certain temperature a 1.00-L flask initially contained 0.298 mol PCl3(g) and 8.70×10-3 mol PCl5(g). After the system had reached equilibrium, 2.00×10-3 mol Cl2(g) was found in the flask. no change neq ceq 6.70×10-3 0.300 2.00×10-3

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-L flask. co change ceq Shift to the right Q < K x=0.387 mol/L

Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15×10-2 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1.500-L flask. Calculate the equilibrium concentrations of all species. co change ceq Shift to the right Q < K x=1.528 mol/L

Quadratic equation Q > K 3.000 mol H2 and 6.000 mol F2 are mixed in a 3.000-L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15×10-2. Calculate the equilibrium concentration of each component. Quadratic equation Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00×10-2. Suppose HI at 5.000×10-1 atm, H2 at 1.000×10-2 atm, and I2 at 5.000×10-3 atm are mixed in a 5.000-L flask. Calculate the equilibrium pressures of all species. Q > K

1.0 mol NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations? change ceq x=0.010 mol/L

How valid is this approximation? Approximation justified!!

H2 HI I2 I. Concentration Changes If the conc. of a substance is increased, the equilibrium will shift in a way that will decrease the conc. of the substance that was added. Where will the reaction shift? Decrease HI - forward Decrease I2 - backward H2 HI forward Increase H2 - I2 Increase HI - backward

Note: adding/removing a solid/liquid to an equilibrium system will not cause any shift in the position of equilibrium because this doesn’t change their concentration. addition of an inert gas such as He, Ar, Kr, etc. at constant volume, pressure and temperature does not affect the equilibrium because it doesn’t change the concentrations or partial pressures. Increase CO2 : Increase CaO Decrease CO2 Adding Kr - backward - No effect - forward - No effect

The effect of pressure (volume) on an equilibrium system. An increase in pressure (decrease in volume) shifts the position of the equilibrium in such a way as to decrease the pressure by decreasing the number of moles of gaseous component. When the volume is increased (pressure decreased), a net reaction occurs in the direction that produces more moles of gaseous component (increase in system pressure). Decrease in pressure : Decrease in volume: backward forward

Sample Problem 17.12 Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position PROBLEM: For the following reactions, predict the direction of the reaction if the pressure is increased: (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) (a) CO2 is the only gas present. The equilibrium will shift to the direction with less moles of gas. Answer: backward SOLUTION: There are more moles of gaseous reactants than products. Answer: forward (c) There are an equal number of moles of gases on both sides of the reaction. Answer: no effect

Effect of Temperature Changes  T shifts reaction in direction that absorbs heat (enhances the endothermic change)  T shifts reaction in direction that produces heat (enhances the exothermic change) Changes in T change the value K K depends on T T of exothermic reaction makes K smaller More heat (product) forces equilibrium to reactants T of endothermic reaction makes K larger More heat (reactant) forces equilibrium to products

Effect of Change in Temperature Ice water Boiling water Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Reaction endothermic Adding heat shifts equilibrium toward products Cooling shifts equilibrium toward reactants Center – system at equilibrium Fig. 14.7

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O(ℓ) Your Turn! The equilibrium between aqueous cobalt ion and the chlorine ion is shown: [Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O(ℓ) pink blue It is noted that heating a pink sample causes it to turn violet. The reaction is: endothermic exothermic cannot tell from the given information

LE CHATELIER’S PRINCIPLE STRESS SHIFT WHY? increase concentration of a substance away from substance extra concentration needs to be used up decrease concentration of a substance towards substance need to produce more of substance to make up for what was removed increase pressure of system towards fewer moles of gas for gas: pressure increase = volume decrease decrease pressure of system towards more moles of gas for gas: pressure decrease = volume increase increase temperature of system away from heat/ energy exothermic reaction is favored extra heat/ energy must be used up decrease temperature of system towards heat/ energy exothermic reaction is favored more heat/ energy needs to be produced to make up for the loss add a catalyst NO SHIFT The rates of both the forward and reverse reactions are increased by the same amount.

N2(g) + 3H2(g)  2NH3(g) + 22.0 kcal  10. Decrease press. 9. Increase press. 8. Decrease temp 7. Increase temp 6. Remove NH3 5. Remove H2 4. Remove N2 3. Add NH3 2. Add H2 1. Add N2 [NH3] [H2] [N2] Shift Stress

2CO2(s) + heat 2CO(s) + O2(g)   CO2 is added CO is added O2 is added CO is removed pressure is increased pressure is decreased temp. is increased temp. is decreased catalyst is added stress 2CO2(s) + heat 2CO(s) + O2(g)   Do by hand.

Given the reaction: N2(g) + O2(g) + 182.6 kJ  2NO(g)  DO NOW  Given the reaction: N2(g) + O2(g) + 182.6 kJ  2NO(g) Which change would cause an immediate increase in the rate of the forward reaction? increasing the concentration of NO(g) increasing the concentration of N2(g) decreasing the reaction temperature decreasing the reaction pressure 

Which changes occur when O2(g) is added to the system? DO NOW  2POCl3(g) + energy  2PCl3(g) + O2(g)  Which changes occur when O2(g) is added to the system? 1) The equilibrium shifts right and [PCl3] increases. 2) The equilibrium shifts right and [PCl3] decreases. 3) The equilibrium shifts left and [PCl3] increases. 4) The equilibrium shifts left and [PCl3] decreases

2SO2(g) + O2(g)  2SO3(g) + heat QUIZ  2SO2(g) + O2(g)  2SO3(g) + heat  For each of the following stresses, state in which direction the equilibrium will shift and what will happen to the concentration of SO2. Explain. concentration of O2 is decreased pressure is decreased temperature is decreased a catalyst is added