Chapter 15 Chemical Equilibrium PDF

Chapter 15 Chemical Equilibrium Copyright © 2023 Pearson Education, Inc. All Rights Reserved 15.1 The Concept of Chemical Equilibrium Equilibrium is a state of balance. Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. In the figure...

Chapter 15 Chemical Equilibrium Copyright © 2023 Pearson Education, Inc. All Rights Reserved 15.1 The Concept of Chemical Equilibrium Equilibrium is a state of balance. Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. In the figure above, Copyright © 2023 equilibrium is Inc. Pearson Education, finally reached All Rights in Reserved the third picture. 15.1 The Concept of Chemical Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate (vertical dashed line). At equilibrium, the amount of each reactant and product remains constant. Copyright © 2023 Pearson Education, Inc. All Rights Reserved 15.2 The Equilibrium Constant In a system at equilibrium, both the forward and reverse reactions are being carried out. The equation is written with a double arrow: N2O4 (g) 2NO2 (g) For the forward reaction: N2O 4 ( g ) ® 2 NO 2 ( g ) Rate = k f éëN2O 4 ùû For the reverse reaction: 2 NO 2 ( g ) ® N2O 4 ( g ) 2 Rate = k r éëNO 2 ùû Copyright © 2023 Pearson Education, Inc. All Rights Reserved Equilibrium Constant Expression At equilibrium Rate f = Rate r 2 k f éëN2O 4 ùû = k r éëNO 2 ùû Rewriting, it becomes the expression for the equilibrium constant (K): 2 éëNO 2 ùû kf = = a constant éëN2O 4 ùû k r Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Equilibrium Constant Consider the generalized reaction aA + bB dD +eE The equilibrium expression for this reaction would be [D] [E] d e ¬ products Kc = [ A ] [B] ¬ reactants a b Also, since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written ( ) ( ) P d P e Kp = D E ( PA ) ( PB ) a b How does the rate law figure into all of this? Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Equilibrium Constant Consider the generalized reaction aA + bB dD +eE The equilibrium expression for this reaction would be [D] [E] d e ¬ products Kc = [ A ] [B] ¬ reactants a b Also, since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written ( PD ) ( PE ) d e Kp = ( A) ( B) a b P P How does the rate law figure into all of this? Copyright © 2023 Pearson Education, Inc. All Rights Reserved It doesn’t!!!!!!!!!!!!  The Haber Process Consider the Haber process, which is the industrial preparation of ammonia: N2(g) + 3H2(g) 2NH3(g) The equilibrium constant depends on stoichiometry: 2 éëNH3 ùû Kc = 3 éëN2 ùû éëH2 ùû Copyright © 2023 Pearson Education, Inc. All Rights Reserved Achieving Equilibrium it doesn’t matter where you start; The same equilibrium is reached. éëNH3 ùû 2 N2(g) + 3H2(g) 2NH3(g) K c = éN ù éH ù 3 ë 2û ë 2û Relative amount of change is based on stoichiometry. Copyright © 2023 Pearson Education, Inc. All Rights Reserved  Evaluating K c K sub c (1 of 2) Table 15.1 Initial and Equilibrium Concentrations of N2O 4 ( g ) and NO 2 ( g ) at 100 °C Initial Initial Equilibrium Equilibrium éëN2O 4 ùû ( M ) éëNO 2 ùû ( M ) éëN2O 4 ùû ( M ) éëNO 2 ùû ( M ) Kc Experiment left bracket N sub 2 O sub 4, right bracket, in moles left bracket N O sub 2, right bracket, in moles left bracket N sub 2 O sub 4, right bracket, in moles left bracket N O sub 2, right bracket, in moles K sub c 1 0.0 0.0200 0.00140 0.0172 0.211 2 0.0 0.0300 0.00280 0.0243 0.211 3 0.0 0.0400 0.00452 0.0310 0.213 4 0.0200 0.0 0.00452 0.0310 0.213 Each experiment can begin and end with different concentrations. But all end with the same equilibrium. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Evaluating K c K sub c (2 of 2) 2 The ratio of éëNO 2 ùû to éëN2O 4 ùû remains constant at 100°C no matter the initial concentrations of NO 2 and N2O 4 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Equilibrium Constants in Terms of Pressure, K p K sub p For gases, PV = nRT (the ideal-gas law). Rearranging, P = ( n / V ) RT ; ( n / V ) is [ ]. P = MRT ( PD ) ( PE ) d e [D] [E] d e ¬ products Kc = Kp = [ A ] [B] ( PA ) ( PB ) a b a b ¬ reactants Substituting into K p : K p = K c ( RT ) Dn where Dn = (moles of gaseous product) - (moles of gaseous reactant) Copyright © 2023 Pearson Education, Inc. All Rights Reserved 15.2 Using Equilibrium Constants Magnitude of K If K >> 1, the reaction favors products; products predominate at equilibrium. If K K, nature will make the reaction proceed to reactants. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Calculating Equilibrium Concentrations If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry). You will set up a table similar to the ones used to find the equilibrium concentration, but the “change in concentration” row will be a factor of “x” based on the stoichiometry. Copyright © 2023 Pearson Education, Inc. All Rights Reserved  An Equilibrium Concentration Calculation Example A 1.000 L flask is filled with 1.000 mol of H2 (g) 2.000 mol of I2 (g) at 448 °C. Given Kc= 50.5 at 448 °C, what are the equilibrium concentrations of H2 (g), I2 (g) and HI (g)? H2 (g) + I2 (g) 2HI (g) Initial concentration (M) 1.000 2.000 0 Change in concentration (M) -x -x +2x negative x negative x Equilibrium concentration (M) 1.000 - x 2.000 - x 2x 1.000 minus x 2.000 minus x Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Example (2 of 3) Set up the equilibrium constant expression, filling in equilibrium concentration expressions from the table. [HI] ( 2x ) 2 2 Kc = = = 50.5 éëH2 ùû éëI2 ùû (1.000 - x )( 2.000 - x ) 50.5(2 – 3x + x2) = 4x2 −𝑏 ± 𝑏! − 4𝑎𝑐 (2 – 3x + x2) = 4/50.5x2 𝑥= 2𝑎 0.9207x2 – 3x + 2 = 0 Solving for x is done using the quadratic formula, resulting in x = 2.323 or 0.935. Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Example (3 of 3) Since x must be subtracted from 1.000 M, 2.323 makes no physical sense. (It results in a negative concentration!) The value must be 0.935. So éëH2 ùû eq = 1.000 - 0.935 = 0.065 M éël 2 ùû eq = 2.000 - 0.935 = 1.065 M [Hl]eq = 2(0.935) = 1.87 M Copyright © 2023 Pearson Education, Inc. All Rights Reserved 15.7 Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” After the shift, the system will return to equilibrium. Concentration and pressure will change but the value of the equilibrium constant remains the same. Copyright © 2023 Pearson Education, Inc. All Rights Reserved How Conditions Change Equilibrium We use Le Châtelier’s Principle qualitatively to predict shifts in equilibrium based on changes in conditions. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Change in Reactant or Product Concentration If the system is in equilibrium – adding a reaction component will result in some of it being used up. – removing a reaction component will result in some of it being produced. – magnitude of equilibrium constant remains the same. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Change in Volume or Pressure When gases are involved in an equilibrium, a change in pressure or volume will affect equilibrium: Higher volume or lower pressure favors the side of the equation with more moles (and vice versa). K remains the same. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Change in Temperature Is the reaction endothermic or exothermic as written? Decide, then assess as you would for a concentration change. – Unlike concentration or pressure changes, K changes in value. Endothermic: Heat acts like a reactant; adding heat drives a reaction toward products. K increases. – Reactants plus heat forms products Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants. K decreases. – Reactants form products plus heat Copyright © 2023 Pearson Education, Inc. All Rights Reserved Equilibrium Affected by Temperature Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Effect of Catalysts Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower temperatures. Copyright © 2023 Pearson Education, Inc. All Rights Reserved