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Chapter 13 CHEMICAL EQUILIBRIUM - 13 CHEMICAL EQUILIBRIUM...

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13 CHEMICAL EQUILIBRIUM 13.1 The concept of equlibrium 13.1.1 Reversible reaction : Reversible reaction can go in both direction. As soon as some product molecules are formed, the reverse process begins to take place and reactant are formed from product. The reversible sign : Sulfur dioxide and sulfur trioxide exist in equilibrium as indicated in the following equation: 2SO 2 (g) + O 2 (g) 2SO 3 (g) The net concentration of SO 2 ,O 2 and SO 3 do not change. However, SO 2 and O 2 combining to form SO 3 at the same time SO 3 decomposed to form SO 2 and O 2 Non-reversible reactions proceed in only one direction .The reaction proceeds toward the formation of product. In the following reaction,the formation of KCIO 3 from the mixture of KCI and O 2 does not take place. 2KClO 3 (g) 2KCl (s) + 3O 2 (g) 13.1.2 State of equilibrium A system reached an equilibrium when a forward and a reverse reaction proceed at equal rates, and the concentrations of reactants and products remain constant over time.The system is said to be in dynamic equilibrium . C and D (product) A and B (reactants) t a Time A + B C + D Figure 13.1.1 Concentration versus time graph for the reversible reaction. The concentration of C and D increase with increasing of time(Figure 13.1.1). In contrast, the concentration of A and B decrease with increasing of time. Up to a point, (t a ) both reaction has no further change in the concentration of products or reactant. Concentration
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Time Time Figure 13.1.2 Rate versus time graph for the reversible reaction. In reversible reaction, the rate of forward reaction equals to rate of reverse reaction (Figure 13.1.2) 13.1.2 Law of chemical equilibrium The equilibrium concentration of reactants and products are related by law of mass action/ law of chemical equilibrium. Consider general reaction aA + bB cC + dD At equilibrium, [ C ] c . [ D ] d = K [ A ] a . [ B ] b where: K = equilibrium constant [ ] = concentration Example 13.1.1 : The following equilibrium concentrations are measured at 483K; [CO]=1.03M, [H 2 ]=0.322M and [CH 3 OH]=1.56M . Determine the equilibrium constant. CO(g) + 2H 2 (g) CH 3 OH(g) K = [CH 3 OH] = 1.56 [CO][H 2 ] 2 1.03(0.322) 2 = 14.5 Rate A + B C + D Time Time
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13.1.4 The equilibrium constant expression The concentrations in the equilibrium constant expression are usually given in moles per liter (M), and so the symbol K is often written with a subscript c for “concentration” as in K c . aA(g) + bB(g) cC(g) + dD(g) K c = [ C ] c [ D ] d [ A ] a [ B ] b When the reactants and products in a chemical equation are gaseous, we can formulate the equilibrium expression in terms of partial pressures instead of molar concentrations. aA(g) + bB(g) cC(g) + dD(g) K p = (P C ) c (P D ) d (P A ) a (P B ) b where : P A = partial pressure of gas A, etc. K p = equilibrium constant for the reaction in term of pressure of reactants. The values and units for K p and K c are depend on the way in which the chemical equation is written.
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