Introduction to Equilibrium

A Little Review What would be the product of the following reaction? H2 (g) + I2(g) 

A Little More Review If 1.000x10-3 mol of H2 reacts with 2.000x10-3 mol of I2, how much HI should be produced? In reality, this reaction produces less than 1.000x10-3 mol of product? What happened?

A closer look at the reaction Davidson College Applet What happened to the rate of formation of the reactants as time progresses? What happened to the rate of formation of the product as time progresses?

The Reaction Is there a point at which no more molecules are reacting? What how does the rate of the forward reaction compare to the rate of the reverse reaction at the right side of the graph

Equilibrium Equilibrium occurs when the opposing reactions occur at equal rates. To show a reaction at equilibrium: N2O4 (g) 2 NO2 (g)

From a kinetics perspective N2O4 (g) 2 NO2 (g) Forward N2O4  2 NO2 Ratef =kf[N2O4] Reverse 2 NO2  N2O4 Rater = kr[NO2]2 What is true of the rate of forward and reverse reactions at equilibrium? Ratef = Rater

Equilibrium So if Ratef = Rater, then kf[N2O4] = kr[NO2]2 Which means that The ratio of reactants to products at equilibrium is constant no matter what the initial concentrations are at a given temperature and pressure.

Equilibrium Expressions For the reaction a A + b B d D + e E The equilibrium expression is Equilibrium constant Expressed in terms of concentration (mol dm-3)

Write the equilibrium expressions for the following reactions N2 (g) + 3 H2 (g) 2 NH3 (g) Fe3+ (aq) + SCN- (aq) FeSCN2+(aq) Cd2+ (aq) + 4 Br- (aq) CdBr42- (aq)

Sample Problem Given the equilibrium system: PCl5(g)  PCl3(g) + Cl2(g) The system is analyzed at a certain temperature and the equilibrium concentrations areas follows: [PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M. Calculate the Kc for this reaction at the temperature this was carried out.

Equilibrium Expression Using Partial Pressures For the reaction a A + b B d D + e E

Converting Between Kp and Kc Δn is the change in the number of moles of gas in the equation.

What does Kc mean? If Kc >>> 1, then the reaction favors ________________ at equilibrium. If Kc <<< 1, then the reaction favors ________________ at equilibrium.

What about pure substances? Pure substances such as solids and liquids are not shown in Kc equations. CaCO3  Al2(SO4)3 (s) + H2O(l)  2 Al3+(aq) + 3 SO42-(aq)

Calculating K from initial and equilibrium concentrations A closed system initially containing 1.000x10-3 M H2 (g) and 2.000x10-3 M I2 (g) at 448 oC is allowed to reach equilibrium. At equilibrium, the concentration of HI (g) is 1.87x10-3 M. Calculate Kc for the reaction.

The steps Write out the chemical equation if not given. Write out an equilibrium expression Create an ICE chart Fill in information available. Arithmetically or algebraically solve for unknown values.

H2 (g) + I2 (g)  2 HI (g) . H2 (g) + I2 (g)  2 HI (g) I 1.000x10-3 M C E 1.87x10-3 M

H2 (g) + I2 (g)  2 HI (g) . I 1.000x10-3 M 2.000x10-3 M 0 M C E 0.065x10-3 M 1.065x10-3 M

Predicting Product Amounts In the Haber process, N2 + 3 H2  2 NH3 at 500 oC with a Kp = 1.45x10-5. In an equilibrium mixture of the gases at 500 oC, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this system? X^2/(.432)(.928)^3 = 1.45x10-5 X^2 = (1.45x10-5)(.432)(.928)^3 X = sqrt ((1.45x10-5)(.432)(.928)^3) X= sqrt(5.01x10-6) X = 2.24x10-3 atm

Calculating Equilibrium Concentrations Given Initial Concentrations A 1.000 L flask is filled with 1.000 mol of H2 (g) and 2.000 mol of I2 (g) at 448 oC. The Kc for the reaction is 50.5. What are the equilibrium concentrations of each of the reactants and products.

Write out the chemical equation if not given. H2 (g) + I2 (g)  2 HI (g) Write out an equilibrium expression Create an ICE chart I 1.000 M 2.000 M 0 M C E

Fill in information available. Arithmetically or algebraically solve for unknown values. Note: Problems of this type require the quadratic equation. You will not have to do this in IB. Yipee!!! I 1.000 M 2.000 M 0 M C X + 2X E (1.000-X) M (2.000-X) M 2X M Kc = [HI]^2/[H2][I2] = 2x^2/(1.000-x)(2.000-x) = 50.5 4x^2 = 50.5 (x^2-3.000x+2.000) 4x^2 = 50.5x^2- 151.5 x + 101.0 46.5x^2 - 151.5x + 101.0 = 0 Quadratic equation ax^2 + bx + c = 0 X = -b + sqrt(b2-4ac)/2a X = -(-151.5) + sqrt[(-151.5)^2 – 4(46.5)(101.0)]/2(46.5) X = 2.323 or 0.935 X cannot = 2.323 b/c it would result in a negative concentration X = 0.935 [H2] = 1.000- x = 0.065 [I2] = 2.000 – x = 1.065 [HI] = 2x = 1.87

Q vs. K K describes only systems that are at equilibrium Q (the reaction quotient) describes all chemical systems whether they are at equilibrium or not. When the system is at equilibrium, Q = K

Question If Q > K, will the system increase the concentration of reactants or products to reach equilibrium? If Q < K, will the system increase the concentration of reactants or products to reach equilibrium? Q > K reactants Q < K products