Chemical Equilibrium

Transcription

1 0.110/5.60 Fall 005 Lecture #10 age 1 Chemical Equilibrium Ideal Gases Questin: What is the cmsitin f a reacting miture f ideal gases? e.g. ½ N (g, T, ) + 3/ H (g, T, ) = NH 3 (g, T, ) What are N,, and H NH 3 at equilibrium? Let s lk at a mre general case ν A A(g, T, ) + ν B B(g, T, ) = ν C C(g, T, ) + ν D D(g, T, ) The ν i s are the stichimetric cefficients. Let s take a miture f A, B, C, and D with artial ressures = X, = X, = X, and = X A A B A C C D D Is this miture in equilibrium? We can answer by finding G if we allw the reactin t rceed further. We knw µ i ( T, ) fr an ideal gas in a miture and we knw that G = n iµ i i ε = ε ν µ ( ) + ν µ ( ) ν µ ( ) + ν µ ( G( ) C C g, T, D D g, T, A A g, T, B B g, T, )

2 0.110/5.60 Fall 005 Lecture #10 age where ε is an arbitrary small number that allws t let the reactin rceed just a bit. We knw that µ ( ) µ ( ) where µ i ( T ) i g, T, = T + RT ln imlied 1 bar i i i is the standard chemical tential f secies i at 1 bar and in a ure (nt mied) state. C D G( ) = + + C C T D D T A A T B B T + RT ln ν ν A B A B ε ε ν µ ( ) ν µ ( ) ν µ ( ) ν µ ( ) G = G + RT lnq where G = ν µ ( T ) + ν µ ( T ) ν µ ( T ) + ν µ ( ) C C D D A A B B T and = is the reactin qutient C C Q ν A A ν B B G is the standard change in free energy fr taking ure reactants int ure rducts. G = G = H T S rn rn rn r G = G ( rducts) G ( reactants) frm frm If G ( ε) < 0 then the reactin will rceed sntaneusly t frm mre rducts G ( ε) > 0 then the backward reactin is sntaneus G ( ε) = 0 N sntaneus changes Equilibrium

3 0.110/5.60 Fall 005 Lecture #10 age 3 At Equilibrium G ( ε) = 0and this imlies G = RT lnq rn eq Define Qeq = K the equilibrium cnstant K = = = C C ν XC XC νa νb νa νb X X A B eq A B eq ν K X and thus G = RT lnk, rn K = G RT e Nte frm this that K ( T ) is nt a functin f ttal ressure. It is K ν K which is K T,. X = ( ) X Recall that all i values are divided by 1 bar, s K and K X are bth unitless. Eamle: H (g) + CO (g) = H O(g) + CO(g) T = 98 K =1 bar H (g) CO (g) H O(g) CO(g) Initial # a b 0 0 f mles # mles a- b- at Eq. Ttal # mles at Eq. = (a ) + (b ) + = Mle fractin at Eq. a b

4 0.110/5.60 Fall 005 Lecture #10 age 4 ( G kj/ml frm ) ,600 kj/ml ( J/K-ml)( 98 K) G = K = e = e = rn 8.6 kj/ml and HO CO XHOX CO K = = = X X a b H CO H CO Let s take a = 1 ml and b = ml We need t slve ( 1)( ) ( )( ) = A) Using arimatin methd: K << 1, s we eect << 1 als. Assume 1 1, = ml (indeed << 1) ( )( ) 6 B) Eactly: = K = ( ) ( ) ( ) = 0 6 ( ) ( ) ( 6 ) ( ) ( ) ( ) = ± The - sign gives a nnhysical result (negative value) Take the + sign nly = ml (same)

5 0.110/5.60 Fall 005 Lecture #10 age 5 Effect f ttal ressure: eamle N O 4 (g) = NO (g) Initial ml # n 0 # at Eq. n- Ttal # mles at Eq. = n + = n + n X i s at Eq. n + n + X n NO X 4 NO n 4 n + NO NO 4 n + K = = = = 4α K = where α = n is the fractin reacted 1 α K K K K ( 1 α ) = α α 1 + = α = = α = K K K 1 If increases, α decreases Le Chatelier s Princile, fr ressure: An increase in ressure shifts the equilibrium s as t decrease the ttal # f mles, reducing the vlume. In the eamle abve, increasing shifts the equilibrium tward the reactants

6 0.110/5.60 Fall 005 Lecture #10 age 6 Anther eamle: NO(g) + O (g) = NO (g) Initial ml # 1 0 K = at 98 K # at Eq Ttal # mles at Eq. = X i s at Eq. 1 ( ) 1 3 = 3-3 NO NO 1 NO 1 NO O NO O NO O ( 3 ) 3 ( 1 ) X X K = = = = X X X X 3 K >> 1 s we eect K r ( 1 ) = 1 3 ( 1 ) K K 13 In this case, if then as eected frm Le Chatelier s rincile.