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Chemical Equilibrium.pdf

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University of Alberta
Sai Yiu

UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102) Chemical Equilibrium Introduction • Many physical /chemical processes are reversible. • If the reversible process is in a closed vessel (neither the reactant nor the product can escape), a state of equilibrium will develop. For example: I in water 2 I in di chl oro m eth ane Water Saturated salt vsolution 2 NaCl H 2 (l) H 2 (g) NaCl (s) Na (aq) + Cl (aq) I2(aq) I2(CH C2 ) 2 Chemical equilibrium In the following reversible reaction: N 2 (4) 2NO (g2 Colorless gas Brown color gas Reaction starts from NO 2 Reaction starts from N O 2 4 Equilibrium is reached 1 Characteristics ofequilibrium state: 1. The equilibrium is dynamic (i.e. both the forward and backward reactions are still reacting but there is no net change) 2. Rate of forward reaction = rate of reversedreaction 3. The concentrations of the reactants and products are constant The Equilibrium Constant Expression For a chemical reaction: a A + b B c C + d D Both the forward and reversed reactions are elementary steps and so, Rate = k [A] x [B] b (forward) (forward) k (forward)rate constant of the forward reaction. Rate = k [C] x [D] d (reversed) (reversed) k (reversed)rate constant of the reversedreaction. Since at equilibrium, rate = rate (forward) (reversed) k [A] x [B] = k [C] x [D] d (forward) (reversed) c d [C] x [D] k (forward)k (reversed) = K c a b [A] x [B] • K is called the equilibrium constant and it is dependent only on the temperature. c • The ratio is called the equilibrium constant expression. • In most cases, the values of K are shown as dimensionless numbers. This is because in thermodynamics, the concentration is expressed as the ‘activity’ which is without unit. For the reaction: N 2O4(g) 2NO (2) 2 [NO 2] Kc= [N O ] 2 4 2 Significance of K K is an indication of how far a reaction proceeds toward product at a given temperature. -30 • Small K: N 2g) + O (2) 2NO (g) K = 1 x 10 Very little product is formed. There is actually no reaction. • Large K: 2CO (g) + O (2) 2CO (2) K = 2.2 x 10 22 High concentration of product is formed. The reaction goes to completion • Intermediate K: 2BrCl (g) Br (g) + Cl (g) K = 5 2 2 The reaction mixture contains a considerate amount of both the reactant and product. Equilibrium position Equilibrium position is the concentrations of the reactants and the products at equilibrium. For the chemical reaction: N 2g) + 3H (2) 2NH (3) The equilibrium constant expression is: [NH 3 2 K c= 3 [N2][ H2] o The following are two different sets of equilibrium positions at 127 C: [0.203] 2 K = = 0.0602 c 3 [0.399][1.197] 2 [1.82] K c= = 0.0602 [2.59][2.77]3 Thus, t here is only one equilibrium constant for a particular system at a particular temperature but you can have an infinite number of equilibrium positions. 3 Equilibria Involving Gases The ideal gas law: PV = nRT n P = RT V Concentration Thus partial pressure of a gas can be used to express concentration if T is kept constant. 2SO 2g) + O 2g)  2SO (g3 2 pSO3 Kp= p 2x p SO2 O2 pSO3, pSO2 pO2are the partial pressures of3SO ,2SO and2O respectively. Reaction Quotient (Q) – Direction of a chemical reaction How does a reaction behave when it is not at equilibrium? Determine the reaction quotient, Q of the reaction. a A + b B  c C + d D C D[ ]d [C] eq]D deq Q c K c A B[ ]b [A] eq]B beq If Q < K System is not at equilibrium, it needs to form more products. Reaction system is shifting to the right (forward reaction favored) If Q > K System is not at equilibrium, it needs to form more reactant s. Reaction system is shifting to the left (reverse reaction favored) If Q = K at a given T, the system is at equilibrium. 4 For reaction: N 2 4g) 2NO (2) Q > K Q = K Q < K Q = K Reaction Quotient changes with time until equilibrium is reached Example At 100 oC, K = 60.6 for the reaction p 2NOBr (g)  2NO (g) + Br (g)2 In a given experiment, 0.10 atm of each component is placed in a container. Is the system at equilibrium? If not, in which direction will it proceed? Relationships Involving Equilibrium Constants or Reaction Quotients 1. The concentrations of pure solids, pure liquids and solvents of dilute solutions never appear in the equilibrium constant expressions. H2(g) + S(s) H 2(g) [H 2S] K = c [H2][S] 5 As long as there is some solid present, its ‘concentration’ is a constant and so, we can take it out from the equilibrium expression [H 2] K c = [H ] 2 Another example: 2H 2(l) 2H 2g) + O2(g) 2 K = [H 2 [O 2 The above reactions are called heterogeneous equilibriabecause two phases were involved. 2. If an overall reaction is the sum of two or more reactions, the overall equilibrium constant is the product of the equilibrium constants for the steps: K overallK1x K 2 K 3... The following sequence of individual steps has been proposed for the overall reaction between H2
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