K-==00 x=0.403-0.403x 1.403x=0.403 x=0.287 Thus,pco=1.00-x=0.713 atm Pco2=X=0.287 (b)Calculate the value of G(kJmol)for the reaction at 1000C.Is this reaction spontaneous at 1000℃？ △G°=-RTIn(Kp) =-8.314 JK mol(1273K)ln(0.403) =+9619Jmo =+9.62kmol Since AG>0,the reaction is not spontaneous at 1000C. 4 (a)Calculate the pH of a 2.00 M solution of trimethylamine,N(CH3)3.Kp for this base is 7.4 x 105. N(CH3)ag)+H2ON(CH3)3H'+OHQ K.-[N(CH,),74x10 [N(CH,)3] N(CH3)3(g)N(CH3)3H'()OH) Initial,M 2.00 0 0 Change,M -X +x +x Equilibrium,M 2.00-x X Here,Kb <<0.001,thus x <<2.00. 55 CO2 p CO p x K 0.403 p 1x x 0.403 0.403x 1.403x 0.403 x 0.287 === − = − = = Thus, pCO = 1.00 – x = 0.713 atm pCO2 = x = 0.287 (b) Calculate the value of ΔGo (kJ mol-1) for the reaction at 1000o C. Is this reaction spontaneous at 1000o C? ΔG0 = -RTln(Kp) = -8.314 J K-1 mol-1 (1273 K)ln(0.403) = +9619 J mol-1 = +9.62 kJ mol-1 Since ΔG0 > 0, the reaction is not spontaneous at 1000o C. 4. (a) Calculate the pH of a 2.00 M solution of trimethylamine, N(CH3)3. Kb for this base is 7.4 x 10-5. N(CH3)3(aq) + H2O(l) ∩ N(CH3)3H+ (aq) + OH- (aq) 3 3 (aq) (aq) 5 b 3 3 [N(CH ) H ][OH ] K 7 [N(CH ) ] + − .4 10− = = × N(CH3)3(aq) N(CH3)3H+ (aq) OH- (aq) Initial, M 2.00 0 0 Change, M -x +x +x Equilibrium, M 2.00 - x x x Here, Kb <<0.001, thus x <<2.00