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Chemical Equilibrium Reactions Dont Just Stop, They find Balance Reactions Dont Just Stop, They find Balance.

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Presentation on theme: "Chemical Equilibrium Reactions Dont Just Stop, They find Balance Reactions Dont Just Stop, They find Balance."— Presentation transcript:

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2 Chemical Equilibrium Reactions Dont Just Stop, They find Balance Reactions Dont Just Stop, They find Balance

3 Equilibrium is Attained When the Rates of the Forward and Reverse Reactions are the Same. Forward Rate = k f [A] Reverse Rate = k r [B] k f [A] = k r [B]

4 Equilibrium : The Haber Process and Nitrogen Fixation N 2 (g) + 3H 2 (g) 2NH 3 (g) Chemistry at work, p. 521

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

6 jA + kB pR + q S LAW OF MASS ACTION 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 [A] j [B] k K =

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

8 Suppose we discover that the equilibrium concen- trations of NO 2 and N 2 O 4 are M and M, respectively [NO 2 ] 2 [N 2 O 4 ] K c = [0.0172] 2 [ ] = = N 2 O 4 2 NO 2

9 Heterogeneous Equilibria a reaction which may possess reactants or products which are in different phases. CaCO 3 (s) CaO(s) + CO 2 (g)

10 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. K = [CaO(s)][CO 2 ] [CaCO 3 (s)] Given: CaCO 3 (s) CaO(s) + CO 2 (g) so K = [CO 2 ] Even though they do not appear in the equilibrium constant expression,pure solids and liquids must be present for equilibrium to be established

11 The Magnitude of Equilibrium Constants What does it mean when the constant equals K = 1 x or K = 1 x

12 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 M H 2, M N 2, and M NH 3. From these data, calculate the equilibrium constant for: N 2 (g) + 3H 2 (g) 2NH 3 (g) (.00272) 2 (.0402)(.01207) 3 = 104.7

13 Converting from K c to K p Solutions are Understood in terms of Molarities Gas Pressure is Understood in Terms of Atmospheres K p = K c (RT) n, n = (mol. of prod – mol. of react) Using the value of K c = for the reaction: N 2 (g) + 3H 2 (g) 2NH 3 at 472 °C Convert to K p K p = ( L-atm/mol-K)(745 K) -2 Kp=Kp= ( L-atm/mol-K)(745 K) K p = 2.81 x atm

14 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. Predicting the Direction of Equilibrium

15 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. 2NO 2 (g) N 2 O 2 (g)

16 What would happen if 1 atm of argon gas were added to the following reaction already at equilibrium. N 2 + 3H 2 2 NH 3

17 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. Co(H 2 O) 6 2+ CoCl 4 2- Co(H 2 O) 6 2+ (aq) + 4Cl - (aq) CoCl 4 2- (aq) +6H 2 O (l) H > 0 heating cooling Room temperature

18 Predicting the Direction of Equilibrium Given, N 2 (g) + 3H 2 (g) 2NH 3 (g) [NH 3 ] 2 [N 2 ][H 2 ] 3 K = and K c = If the equilibrium concentrations were to start at: [2.00] 2 [1.00][2,00] 3 Q = = What must happen in order for the value, lets now call it the reaction quotient Q, to return to K c = 0.105?

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

20 K c = [HI] 2 [H 2 ] [ I 2 ] ICEBOX A 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 H 2 (g) + I 2 (g) 2HI Initial Change Equilib x M 2.00 x M O M 1.87 x M x x x10 -3 = ( 1.87 x ) 2 (1.065 x )(.065 x10 -3 )

21 Initial Change Equilibrium A L flask is filled with mol of H 2 and mol I 2 at 448°C. The value of the equilibrium constant for the following reaction is What are the equilibrium concentrations of H 2, I 2, and HI? H 2 (g) + I 2 (g) 2HI H 2 (g) + I 2 (g) 2HI M M 0 M

22 Initial Change Equilibrium H 2 (g) + I 2 (g) 2HI M M 0 M - x M + 2x M ( x) ( x) 2x 50.5 = [HI] 2 [H 2 ][I] = (2x) 2 ( x) ( x) 4x 2 = 50.5(x x ) 46.5x x = 0 x =-(-151.1) + (-151.4) 2 - 4(46.5)(101.0) 2(46.5) x= or [H 2 ] = M [I 2 ] = M [HI] = M

23 Haber Process

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