9.4 The response of equilibrium to temperature (continued)

Transcription

1 9.4 The esponse of equilibium to tempetue (continued) In the lst lectue, we studied how the chemicl equilibium esponds to the vition of pessue nd tempetue. At the end, we deived the vn t off eqution: d ln dt (9.4.1) T Fo the exothemic ection, ө is negtive, theefoe d(ln)/dt is negtive, which mens tht the incese of tempetue will hve negtive impct on the chemicl equilibium. This is in ccod with the nlysis bsed on Le Chtelie s pinciple. Thee is nothe expession of the vn t off eqution. We cn ty to find wy to eliminte the T fom the eqution. Such tsk cn be ccomplished by the following lgeb: We clculte the deivtive of 1/T with espect to T d(1/t)/dt -1/T d ln 1 T d( ) T T enge the bove eqution, we get d ln 1 d( ) T (9.4.) With this type of diffeentil eqution, we cn esily get n nlyticl solution by integting it ove cetin tempetue intevl: Plese keep it in mind tht the stndd enthlpy of fomtion is lso function of tempetue!!! Theefoe, when we integte the bove eqution fom T 1 to T, we should ssume tht ө is constnt within tht intevl. 1 d ln( ) T d 1 ( ) T T1 1 1 so ln() ln(1) ( ) T T1 (9.4.);

2 Fom eqution 9.4., it is obvious tht if we know the stndd ection enthlpy nd the chemicl equilibium constnt t one tempetue, we would be ble to clculte the chemicl equilibium unde nothe tempetue. Exmple 9.4, The be ection N (g) (g) N (g) At 98 degee, equilibium constnt 6.1x10 5. The stndd enthlpy of fomtion fo N equls kj mol -1. Wht is the equilibium constnt t 500? Answe: Fist, clculte the stndd ection enthlpy ө ө * f ө (N ) - * f ө ( ) - f ө (N ) *(-46.1) *0-1*0-9. kj mol -1 then ln( ) ln(6.1*10 5 ) ln( ) *(-9.*1000 J mol -1 ) (1/500 1/98) 0.18 Exmple 9.4b, The equilibium constnt of the ection SO (g) O (g) SO (g) (sulfu dioxide to sulfu tioxide) is 4.0* 10 4 t 00.5*10 10 t 500.0*10 4 t 700 Estimte the stndd ection enthlpy t 500 Answe: Fist, we ssume tht the stndd ection enthlpy does not chnge within the bove tempetue intevl. Then, solvingthe eqution 9.4., ln( ) (ln( 1) 1 1 T T1 Plug in the bove numbes, we get ө -00kJ mol -1 ee, we ctully only need two dt in ode to clculte the stndd ection enthlpy. Anothe ppoch to esolve the bove question is vi gphic method (hint: plot ln() s function of (1/T), the slope is -.

3 9.5 The esponse of equilibium to p Definition of p: An indiction of hydonium ion concenttion p - logα( O ) whee, O is hydonium, epesenttion of the stte of poton in queous solution 9.5 Acid-bse equilibium in wte Thee equilibium constnts need to be known in ode to chcteize the influence of p on equilibi: 1) cidity constnt, which is defined s A(q) O (l) O (q) A - (q) O A A (9.5.1) Quite often, we epot with p, which is defined s log( ). ) Bsicity constnt b B(q) O (l) B (q) O - (q) B O b (9.5.) B we cn lso use p b to epesent b, which is p b -log b ) Autopotolysis constnt w O(l) O (q) O - (q) w α( O )*α(o - ) ( 9.5.) pw -log(w) -log(α( O )*α(o - )) -logα( O ) - log α(o - ) becuse p -logα( O ), nd now intoduce po - log α(o - ) we get pw p po (9.5.4) 9.5b The p of cids nd bses Assuming the solutions e idel solution, the ctivities in the eqution cn be eplced by the concenttion of [A], [ O ] nd [A - ], espectively. Becuse of the 1:1 stoichiometic eltionship, we hve [ O ] [A - ]. Anothe ssumption is tht the concenttion of A o B e unchnged fom its nominl vlue due to the smll poton tnsfe. Afte ll, the eqution cn be simplified to

4 [ O ] [ A] o [ O ] (* [A]) 1/ p -log[ O ] - (1/)log - (1/)log[A] (1/)p (1/)log[A] (9.5.5) The p vlue fo bses cn be deived fom the simil wy. Bsed on 1:1 stoichiometic eltionship, we hve [B ] [O - ], Thus b [ O ] [ B] [O] ( b *[B]) 1/ po -log [O] -(1/)log b - (1/)log[B] (p b -log b!!!) (1/)p b (1/)log[B] Becuse p pw po So, p p w (1/)p b (1/)log[B] (9.5.6) 9.5c Acid-bse tittion Ou im hee is to develop quntittive desciption of the p vlue t ny stge of the tittion. Cse 1. Using stong bse to titte wek cid. While the exct solution of the p vlue cn be obtined, fo simplicity we hee e only inteested in simplified sitution, whee numbe of ppoximtion e invoked. Fist step: exmine the souce of O : Souce 1: A(q) O (l) O (q) A - (q) Souce : O(l) O (q) O - (q) Appoximtion 1: Becuse it is wek cid, the mount of A- fom souce is vey smll comping with the mount of A. Appoximtion : The mount of hydonium ion coming fom souce is negligible,

5 Though A is wek cid. Assuming the initil concenttion of [A] A 0, nd volume V A Fom eqution 9.5.4, we know the initil p vlue of the wek cid cn be clculted s p (1/)p (1/)log[A 0 ] Afte cetin time intevl, volume V B of titnt is dded, the totl volume become V V A V B ; Afte the ddition of stong bse, the following ection occu A (q) O - (q) A - (q) O(l) Note tht the concenttion of [A - ] comes fom the slt which is fomed due to the cid nd bse ection. The contibution fom souce 1 is negligible. We epesent it with [A - ] S. ecll eqution O A A [ O ][ A [ A] ] Afte dding V B mount of bse, the concenttion of A becomes A We get p p log(a /S) (this expession is clled endeson-sselblch eqution ) In genel fom, the bove eqution cn be expessed s p p log ([cid]/[bse])