Chemical Equilibrium Reactions Don’t Just Stop, They find Balance

Chemical Equilibrium Reactions Don’t Just Stop, They find Balance

Equilibrium is Attained When the Rates of the Forward and Reverse Reactions are the Same. Forward Rate = kf [A] Reverse Rate = kr [B] kf [A] = kr [B]

Equilibrium : The Haber Process and Nitrogen Fixation N2(g) + 3H2(g)  2NH3(g) Chemistry at work, p. 521

Equilibrium : The Haber Process and Nitrogen Fixation Note that equilibrium can be reached from either the forward or reverse direction

jA + kB  pR + q S The LAW OF MASS ACTION allows us to express the relative concentrations of reactants and products at equilibrium in terms of a quantity called the equilibrium constant, ‘K’ such that: [R]p [S]q K = [A]j [B]k

The Haber Process and the Law of Mass Action N2(g) + 3H2(g)  2NH3(g) [NH3]2 K = [N2][H2]3 The equilibrium constant expression depends only on the stoichiometry of the reaction, not on its mechanism.

Suppose we discover that the equilibrium concen- trations of NO2and N2O4are 0.0172 M and 0.00140 M, respectively N2O4  2 NO2 [0.0172]2 [NO2]2 Kc = = 0.221 = [N2O4] [0.00140]

Heterogeneous Equilibria a reaction which may possess reactants or products which are in different phases. CaCO3(s)  CaO(s) + CO2(g)

The density of a pure liquid or solid is a constant at any given temperature and changes very little with temperature. Thus the effective concentration of a pure liquid or solid is constant regardless of how much pure liquid or solid is present. Given: CaCO3(s)  CaO(s) + CO2(g) [CaO(s)][CO2] K = [CO2] so K = [CaCO3(s)] Even though they do not appear in the equilibrium constant expression,pure solids and liquids must be present for equilibrium to be established

The Magnitude of Equilibrium Constants What does it mean when the constant equals K = 1 x 1040 or K = 1 x 10-40

In one of their experiments, Haber and co-workers introduced a mixture of hydrogen and nitrogen into a reaction vessel and allowed the system to attain chemical equilibrium at 472 °C. The equilibrium mixture of gases was analyzed and found to contain 0.01207 M H2, 0.0402 M N2, and 0.00272 M NH3. From these data, calculate the equilibrium constant for: N2(g) + 3H2(g)  2NH3(g) (.00272)2 (.0402)(.01207)3 = 104.7

Converting from Kc to Kp Solutions are Understood in terms of Molarities Gas Pressure is Understood in Terms of Atmospheres Kp = Kc(RT)n, n = (mol. of prod – mol. of react) Using the value of Kc = 0.105 for the reaction: N2(g) + 3H2(g)  2NH3at 472 °C Convert to Kp Kp = 0.105 (0.0821 L-atm/mol-K)(745 K)-2 0.105 Kp= (0.0821 L-atm/mol-K)(745 K)2 Kp= 2.81 x 10-5 atm

Predicting the Direction of Equilibrium Le Châteliers principle : If a system at Equilibrium is disturbed by a change in temperature, pressure or the concentration of a component, the system will shift its equilibrium position so as to counteract the effect of the disturbance.

Effects of Pressure and Volume Changes If a system is at equilibrium and the total pressure is increased by the application of an external pressure by a change in volume, the system will respond by a shift in equilibrium in the direction that reduces the pressure by shifting to the side with less moles. N2O2(g) 2NO2(g)

What would happen if 1 atm of argon gas were added to the following reaction already at equilibrium. N2 + 3H2 2 NH3

Effects of Temperature Changes When heat is added to a system, the equilibrium shifts in the direction that absorbs heat. In an endothermic reaction reactants are converted to products, and K increases. In an exothermic reaction, the opposite occurs. heating cooling Room temperature Co(H2O)62+(aq)+ 4Cl-(aq)CoCl42-(aq)+6H2O(l) H > 0

Predicting the Direction of Equilibrium Given, N2(g) + 3H2(g)  2NH3(g) [NH3]2 K = and Kc = 0.105 [N2][H2]3 If the equilibrium concentrations were to start at: [2.00]2 Q = = 0.500 [1.00][2,00]3 What must happen in order for the 0.500 value, lets now call it the reaction quotient ‘Q’, to return to Kc= 0.105?

Predicting the Direction of Equilibrium [NH3]2 Q = = 0.500 and Kc = 0.105 [N2][H2]3 Equilibrium will re-emerge if the concentration of NH3 decreases and the concentrations of N2 and H2 increase. In a closed system, this would require that the reaction favor the formation of reactants N2(g) + 3H2(g) 2NH3(g) If Q > K, the reaction will shift to the reactants If Q = K, the reaction is at equilibrium If Q < K, the reaction will shift to the products

ICEBOXA mixture of 5.00 x 10-3 mol of H2 and 1.00 x 10-2 mol of I2 is placed in a 5.00 L container at 448°C and allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 °C for the reaction: H2(g) + I2(g)  2HI H2(g) + I2(g)  2HI Initial 2.00 x 10-3 M O M 1.00 x 10-3 M Change + .935 x 10-3 1.87 x 10-3 M - .935 x 10-3 - Equilib. .065 x10-3 1.065 x 10-3 1.87 x 10-3 M [HI]2 (1.87 x 10-3 )2 (1.065 x 10-3)(.065 x10-3) Kc = = [H2] [ I2]

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol I2 at 448°C. The value of the equilibrium constant for the following reaction is 50.5. What are the equilibrium concentrations of H2, I2, and HI? H2(g) + I2(g)  2HI H2(g) + I2(g)  2HI Initial 1.000 M 2.000 M 0 M Change Equilibrium

H2(g) + I2(g)  2HI Initial 0 M 1.000 M 2.000 M Change - x M - x M + 2x M (2.000 - x) (1.000 - x) Equilibrium 2x [HI]2 (2x)2 50.5 = = [H2][I] (1.000 - x) (2.000 - x) 4x2 = 50.5(x2 - 3.00 x + 2.00) 46.5x2 - 151.5x +101.0 = 0 x =-(-151.1) + (-151.4)2 - 4(46.5)(101.0) [H2] = 0.065 M [I2 ] = 1.065 M [HI] = 1.870 M 2(46.5) x= 2.323 or 0.935

Haber Process

Chemical Equilibrium Reactions Don’t Just Stop, They find Balance