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) ______ P ref (=1 atm) = a 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 N O O N O O N O O ? ? This question is answered by equilibrium constant, K = P N2O4 (P NO2 ) 2 _____ , which is 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 whether you started the reaction with NO or N O .

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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)
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## This note was uploaded on 04/13/2008 for the course CHEM 102a taught by Professor Hanusa during the Fall '06 term at Vanderbilt.

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Chapter 7 - Chapter 7 Chemical equilibrium Phase...

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