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Unformatted text preview: 3 = 5.9 x10-13 (0.89) 2 x = 4.9 x10-5 Verify x/0.89 = 4.9E-5/0.89 is less than 1x10-4 so it is valid to ignore 2x. At equilibrium: p_NO 2 = 0.89 atm p_NO = 9.8 x10-5 atm p_O 2 = 4.9 x10-5 atm 14.40 Both gases come from ammonium carbamate. So at equilibrium, p_NH 3 = 2p_CO 2 . Also, p_NH 3 + p_CO 2 = 0.115 atm. So 2p_CO 2 + p_CO 2 = 0.115 atm At equilibrium: p_CO 2 = 0.0383 atm and p_NH 3 = 2p_CO 2 = 0.0767 atm b) Use equilibrium partial pressures: K = (p_NH 3 ) 2 p_CO 2 = (0.0767) 2 (0.0383) = 2.25x10-4 14.54 a) Left. b) Left c) Left. d) Pumping an inert gas at constant T and P increases V. Increasing V shifts equilibrium to right. e) Partial pressures of reactive gases stay constant. They are still at equilibrium. So equilibrium does not shift. 14.56 a) K increases with T means delta H is positive => endothermic reaction b) This means reactants give more gas molecules than products. This is a net decrease in the number of gas molecules when we convert reactants to products....
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- Spring '08