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Chapter-7

# Chapter-7 - Chapter 7 Chemical equilibrium Phase...

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Chapter 7: Chemical equilibrium H 2 O( l ) H 2 O(g) H 2 O( l ) H 2 O(g) Phase equilibrium, Stoichiometric coefficients tell us the proportions in which products form from reactants. That is, they tell us that to form 1 mol of H 2 O(g), we have to use up 1 mol H 2 O( l ). In the phase equilibrium reaction given above, it does not tell us how much of H 2 O( l ) actually left the liquid state and went over to gaseous state. For such information, one needs to know the equilibrium constant . Double headed arrow indicates equilibrium For H 2 O( l ) H 2 O(g) Equilibrium constant, K = Pressure of product gas ___________________ Pressure of reactant gas = P H2O(g) But K does not have units, so we define the pressures relative to a reference pressure of 1 atm. So, K = P H2O(g) ______ ref H2O activity

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Chemical equilibrium …contd For the reaction equilibrium, 2 NO 2 (g) N 2 O 4 (g), P N2O4(g) /1 atm K= ___________ [P NO2(g) /1 atm] 2 = P N2O4(g) ______ [P NO2(g) ] 2 For a general reaction, a A(g) + b B(g) c C(g) + d D(g), Law of mass action How is equilibrium constant used ? Suppose we have a vessel of 1 L volume and 1 mL of H 2 O( l ) is placed in it at 50 o C. Then equilibrium, H 2 O( l ) H 2 O(g) is established. Equilibrium constant, K at 50 o C = 0.1217 atm. K = P H2O /1 atm = 0.1217 (a). (b). V occupied by gas = 1000 mL – 1 mL = 999 mL (c). n H2O(g) = 0.1217 atm x 0.999 L _________________ (0.08206 L atm) x 323.15 K ____ mol K = 4.585 x 10 -3 mol (d). mass of H 2 O(g) = 4.585 x10 -3 mol x 18.0148 g/mol = 0.0826 g (e). Volume of liquid that is moved to gas phase ~ 0.0826 mL K = (P A /1 atm) a (P B /1 atm) b __________________ (P C /1 atm) c (P D /1 atm) d
2 NO 2 (g) N 2 O 4 (g) (a). This reaction indicates that NO 2 and N 2 O 4 are in equilibrium and that two molecules of NO 2 are consumed in forming one N 2 O 4 molecule. (b). But how many molecules each of NO 2 and N 2 O 4 are present at equilibrium? N N O O O O N N O O O O N N O O O O N O O N O O N O O N O O N O O N O O ? ? This question is answered by equilibrium constant, K = P N2O4 (P NO2 ) 2 _____ , which determined by measuring P N2O4 and P NO2 at equilibrium For the above reaction, at 298 K, K=8.8. Hence P N2O4 = 8.8 x (P NO2 ) 2 (c). Chemical equilibrium is a dynamic process. Forward and backward reaction rates are equal. (d). At a given T, equilibrium reaches the same point irrespective of 2 2 4

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A(g) B(g) Initial 30 torr 0 Change -X +X ------------------------------------------- At equilib 30-X X ------------------------------------------ K= 2.0 = P B ------ P A = x 30-x ------ 2.0(30-X) = X or 60-2X = X 60 = 3X X = 20 Suppose that 30 torr of B(g) was introduced initially. What are the pressures at equilibrium ? A(g) B(g) Initial 0 torr 30 Change X -X ------------------------------------------- At equilib X 30-X ------------------------------------------ K= 2.0 = P B --- P A = 30-X X ------ 2X = 30- X 3X = 30 X=10 Example : The equilibrium constant for the reaction A(g) B(g) is 2.0 at 25 0 C. If 30 torr of A(g) is introduced into a vessel at 25 0 C, what are the pressures of A(g) and B(g) at equilibrium.
Equilibrium constant Examples: PCl 5 (g)PCl 3 (g) + Cl 2 (g) K = P PCl3 P Cl2 ________ P PCl5 2NOCl(g) 2NO(g) + Cl 2 (g) K= (P NO ) 2 P Cl2 ________ (P NOCl ) 2 N 2 (g) + 3 H 2 (g) 2NH 3 (g) Calculation of Equilibrium Constants : A mixture of CO(g) and Cl 2 (g), with initial pressure of 0.60 atm for CO(g) and 1.10 atm for Cl 2 (g), is prepared at 600 o C and constant volume. At equilibrium, the partial pressure of COCl 2 (g) is found to be 0.10 atm. What is the equilibrium constant for the reaction, CO(g) + Cl 2 (g) COCl 2 (g)?

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