chemical equilibrium

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chemical equilibrium

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  • 1. 9 Chemical Equilibrium

2. ( Chemical Equilibrium ) ProductsReactants 3. ( PhysicalEquilibrium ) H 2 O(l) H 2 O(g) (ChemicalEquilibrium) N 2 O 4 (g) 2 NO 2 (g) 4. N 2 O 4 (g) 2 NO 2 (g) 4.6310 -3 0.2270 0.0898 0.0204 0.000 0.200 4.6010 -3 0.0880 0.594 0.0523 0.600 0.040 4.6010 -3 0.0967 0.491 0.0475 0.500 0.030 4.6610 -3 0.1020 0.448 0.0457 0.446 0.050 4.6510 -3 0.0851 0.643 0.0547 0.670 0.000 [ N 2 O 4 ] [ NO 2 ] [ N 2 O 4 ] [ NO 2 ] (M) (M) 5. [ NO 2] / [ N 2 O 4] [ NO 2] 2/ [ N 2 O 4] ==4.6310 - 3 2(coefficient)NO 2 (equilibrium constant) =K= =4.6310 - 3 aA+bB cC+dD a , b , cdA , B, CD 6. ( K) K= : K - -( yield ) 7. 1:K >> 12300o C2O 3 (g) 3O 2 (g)K= =2.5410 12 O 2 O 3 O 3 Cl[Cl 2]=0.76 M[Cl] 2 =(0.76)(1.410 38 )=1.110 38 M[Cl]=1.010 19 M ClCl 2 9. 3 :K1 380o C CO(g)+H 2 O(g) H 2 (g)+CO 2 (g) K==5.10 [CO]=0.200 M,[H 2 O]=0.400 M[H 2 ]=0.300 M [CO 2 ]==1.36 M 10. (homogeneousequilibrium) : N 2 O 4 (g) 2 NO 2 (g) -K C (M, mol/L)Kc =-Kp(atm) Kp= 11. K C K p K C K p aA(g) bB(g) K C =K p= P A P B AB P A V=n A RTP B V=n B RT P A =P B = 12. P AP B KpK p = = n A/ Vn B/ Vmol/L[A][B] 13. n=b a = R=0.082L.atm/K.mol aA+bB cC+dD Kc= Kp=Kp==Kc 14. K P=K C n=(c + d) (a + b) 15. 1 PCl 5 (g) PCl 3 (g)+Cl 2 (g)K P=1.05250o CPCl 5 =0.8atm ,PCl 3 =0.4atmCl 2 (250o C) K P =1.05 = =2.10atm 16. 2 N 2 (g)+3H 2 (g) 2NH 3 (g)K P=4.310 4 200o CK C K P =K C T=273 + 200=473 K n=2 (3 + 1)=- 2mol 4.310 4 =K C(0.082473) 2 K C=0.65 17. (HeterogeneousEquilibrium) :(reactants, products)phasegassolidliquid CaCO 3 (s) CaO(s)+CO 2 (g) =K C =[ CO 2]K P= 18. 3 NH 4 HS(s)NH 3 (g)+H 2 S(g)K C, K P partialpressuregas=0.3 K P ==(0.3)(0.3)=0.09K P=K C n=2 0=2mol T=300K 0.09=K C(0.082300) 2 K C =1.49 x 10 4 19. (MultipleEquilibria) products 1)A+B C+D=2)C+D E+F=A+B E+F K C =X = X = 20. 2 Equilibrium constant;K C=K C 4 K C N 2 (g)+3 H 2 (g) 2 NH 3 (g) K 1 =10--------- (1) (1)2 NH 3(g) N 2 (g)+3H 2 (g)K 2 ===0.13(1)3 N 2 (g) + 9 H 2 (g) 6 NH 3 (g)K 3=(K 1 ) 3=(10) 3 21. K -mol/ L -(atm) -(s)(l)K C -K C K P

  • K
  • Kc Kc
  • - KcKc
  • KcKc

22. 1. 1elementary step ( ) A+2B AB 2 Forwardrate;rate f =k f[A][B] 2 Reverserate;rate r =kr [AB 2 ] rate f =rate r k f[A][B]2 =k r [AB 2 ] = =K C K f k r 23. 5 Cu 2+ Fe 2+ Cu 2+ +2 Fe 2+ Cu+2 Fe 3+ Kc2.2elementarysteps (2 ) :2BB 2 :A+B 2 AB 2 :A+2B AB 2 24. == == K/K// = =.= singlemultiple step K C = =K C 25. K C ? 1.( )Q C K C Q C =(reaction quotient)=** K C = 26. Q C =K C Q C >K C product Q CK C ()(reverse)( ) 29. 7 350o CN 2 (g)+3 H 2 (g) 2NH 3 (g)Kc =2.3710- 3 (M)0.6838.801.05[ NH 3]3.65 M(Q C, K C)[NH 3 ][ NH 3] Q C ===2.8610 -2 Q C >K C Q C=K C 30. 8 430o C2NO(g)+O 2 (g) 2NO 2 (g)K P= 1.510 5 ( atm )0.0010.020.50 Q P Q P ===1.810 5 Q P >K P 31. 9 A B [A] 0 =0.5mol/LK C= 1.5[A] , [B]A B (M)0.50 (M)-x +x (M)0.5 x x K C =1.5=x =1.5(0.5 - x)x =0.75 - 1.5x 32. 2.5x =0.75 x =0.3 [A]=0.5 0.3=0.2M [B]=0.3M *check ; K C ==1.5 10 A 2 (g)+B 2 (g) 2 AB(g) (mol)0.50.51430o CK C=49[A 2 ], [B 2 ][AB] 33. A 2 (g)+B 2 (g) 2AB(g) (M)0.50.50(M)-x -x 2 x (M)(0.5 x )(0.5 x )2 x K C=49=49= 34. 7 =2 x =7(0.5 x) 2 x =3.5 7 x 9 x =3.5x =0.39M [A 2 ]=0.5 0.39=0.11M [B 2 ]=0.5 0.39=0.11M [AB]=2(0.39)=0.78M 35. 11 A 2 +B 2 2 ABK C =49 (M)1.01.01.0 (M)-x -x + 2x (M)(1.0 x )(1.0 x )(1.0 + 2 x ) K C=49=49= 36. 7 = 7(1.0 x ) =1.0 + 2x 9x =6 x =0.67 [A 2 ]=1.0 0.67=0.33 M [B 2 ]=1.0 0.67=0.33 M [AB]=1.0+2(0.67)=2.34M 37. (Factors that affect chemical equilibrium) 1. 3. 2. 4. ( Le Chateliers Principle) 38. 1. Fe(CNS) 3 FeCNS 2+ (aq) Fe 3+ (aq)+CNS- (aq) -Fe(NO 3 ) 3 Fe 3+ :[ Fe 3+ ] -NaCNS CNS- :[ CNS - ] -H 2 C 2 O 4(C 2 O 4 2- Fe 3+) Fe 3+ :[ Fe 3+ ] 39. 2. solid , liquidP , V gasP , V PV=nRT P=RT P 1 / VP=V(n/ V) 40. 12 N 2 O 4 (g) 2NO 2 (g)Q C = P=V= [NO 2], [N 2 O 4] [NO 2] 2 [N 2 O 4] Q C >K C P=V=[NO 2] 2[N 2 O 4] Q C PCl 5 (g) PCl 3 (g)+Cl 2 (g)

  • Q C =
  • [PCl 3][Cl 2][PCl 5]Q C >K C

42.

  • H 2 (g)+CO 2 (g) H 2 O(g)+CO(g)
  • Q C =
  • gasQ C =K C

14 2 NOCl(g) 2 NO(g)+Cl 2 (g) Q C =product>reactant (3)(2)P= Q C