AP Chapter 13

# AP Chapter 13 - Reactions are reversible Equilibrium A B C...

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1 Equilibrium Reactions are reversible c A + B C + D ( forward) c C + D A + B (reverse) c Initially there is only A and B so only the forward reaction is possible c As C and D build up, the reverse reaction speeds up while the forward reaction slows down. c Eventually the rates are equal Reaction Rate Time Forward Reaction Reverse reaction Equilibrium What is equal at Equilibrium? c Rates are equal c Concentrations are not. c Rates are determined by concentrations and activation energy. c The concentrations do not change at equilibrium. c or if the reaction is verrrry slooooow. Law of Mass Action c For any reaction c j A + k B l C + m D c K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power c K is called the equilibrium constant. c is how we indicate a reversible reaction Playing with K c If we write the reaction in reverse. c l C + m D j A + k B c Then the new equilibrium constant is c K ’ = [A] j [B] k = 1/K [C] l [D] m

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2 Playing with K c If we multiply the equation by a constant c nj A + nk B nl C + nm D c Then the equilibrium constant is c K’ = [C] nl [D] nm ([C] l [D] m ) n = K n [A] nj [B] nk = ([A] j [B] k ) n The units for K c Are determined by the various powers and units of concentrations. c They depend on the reaction. K is CONSTANT c At any temperature. c Temperature affects rate. c The equilibrium concentrations don’t have to be the same, only K. c Equilibrium position is a set of concentrations at equilibrium. c There are an unlimited number. Equilibrium Constant One for each Temperature Calculate K c N 2 + 3H 2 2NH 3 c Initial At Equilibrium c [N 2 ] 0 =1.000 M [N 2 ] = 0.921M c [H 2 ] 0 =1.000 M [H 2 ] = 0.763M c [NH 3 ] 0 =0 M [NH 3 ] = 0.157M Calculate K c N 2 + 3H 2 2NH 3 c Initial At Equilibrium c [N 2 ] 0 = 0 M [N 2 ] = 0.399 M c [H 2 ] 0 = 0 M [H 2 ] = 1.197 M c [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M c K is the same no matter what the amount of starting materials
3 Equilibrium and Pressure c Some reactions are gaseous c PV = nRT c P = (n/V)RT c P = CRT c C is a concentration in moles/Liter c C = P/RT Equilibrium and Pressure c 2SO 2 (g) + O 2 (g) 2SO 3 (g) c Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) c K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] Equilibrium and Pressure c K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) c K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 c K = Kp (1/RT) 2 = Kp RT (1/RT) 3 General Equation c j A + k B l C + m D c K p = (P C ) l (P D ) m = (C C xRT) l (C D xRT) m (P A ) j (P B ) k (C A xRT) j (C B xRT) k c K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k c K p = K (RT) ( l+m)-(j+k) = K (RT) n c n=(l+m)-(j+k)=Change in moles of gas Homogeneous Equilibria c So far every example dealt with reactants and products where all were in the same phase. c

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AP Chapter 13 - Reactions are reversible Equilibrium A B C...

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