122ch15b - 29 15.51) 30 15.51) (cont.) 31 15.51) (cont.) 32...

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Unformatted text preview: 29 15.51) 30 15.51) (cont.) 31 15.51) (cont.) 32 15.51) (cont.) 33 15.52) 9 mol gas 4 NH3 (g) + 5 O2 (g) º 10 mol gas 4 NO (g) + 6 H2O (g) + heat )H = ! 904.4 kJ (exothermic, heat is a product) Change Shift Yield of NO effect on value of K a) [NH3] inc -------> inc NO change b) [H2O] inc <------- dec NO change c) [O2] dec <------- dec NO change AWAY from ADDED, d) Ptot inc <------- TOWARD REMOVED dec NO change Pressure changes due to volume changes: P inc. ===> fewer moles of gas P dec. ===> more moles of gas Inc. total pressure by adding an inert gas (a gas that’s not part of the reaction or doesn’t react with anything) has no effect since the partial pressure of a gas in the equil. doesn’t change. e) Add catalyst No chg No chg No chg A catalyst does not change the position of equilibrium (shift reaction). It simply speeds up reaching equilibrium. f) T inc <------- dec Decreases Changing temperature is the only thing that effects the numerical value of K. Temp. is not explicitly in the expression for K. Treat heat as a reactant (endothermic) or product (exothermic). 34 15.53) How do the following effect the value of K for an exothermic rxn. ()H < 0)? a) remove reactant or product: NO effect Changing conc. of reactants or products has NO effect on the numerical value of K. b) dec. vol. : NO effect Changing pressure or concentration by changing volume has NO effect on the numerical value of K. c) dec. T : K inc Treat heat as reactant (endothermic) or product (exothermic). Remember, AWAY from ADDED, Reactants TOWARD REMOVED º Products + heat Dec. T (remove heat), (Exothermic, )H < 0) shifts right, K inc As rxn. shifts to right, [Products] inc, [Reactants] dec., K inc. [Products] 8 K = -----------------[Reactants] 9 8 K inc as numerator inc & denominator dec This happens because T is not an explicit part of K. d) Add catalyst: NO effect A catalyst does not change the position of equilibrium (shift reaction). It simply speeds up reaching equilibrium. 35 15.56) 36 15.57) 37 15.61) 15.64) 38 15.64) (cont.) 39 15.64) (cont.) b) Initial PNH = y 3 but y ! 2x = 2.899 atm y = (2.899 + 2x) atm = 2.899 + 2 (2.910) = 8.719 atm = 8.72 atm PNH , initial = 8.72 atm 3 Moles and grams of NH3 initially present PV (8.719 atm) ( 1.00L) ? mol NH3 = --------- = ----------------------------------------- = 1.854 x 10!1 nRT (0.08206 LCatm/molCK) (573 K) mol 17.03 g NH3 ? g NH3 = 0.1854 mol NH3 x ----------------- = 3.1578 g = 3.16 g NH3 1 mol NH3 c) PT = PNH + PH + PNH = 2.910 + 8.730 + 2.899 3 2 3 = 14.539 = 14.54 atm 15.66) PH3BCl3 (s) a) º PH3 (g) + BCl3 (g) Kp = Kc (RT))n Kp = 0.052 at 60 °C )n is the change in moles of gases Kc = Kp (RT)!)n )n = (1 + 1) ! 0 = 2 Kc = (0.052) {(0.08206 LCatm/molCK)(333K)}!2 Kc = 6.9638 x 10!5 Kc = 7.0 x 10!5 40 15.66) (cont.) ...
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