Lecture 14 Chater 16, Sectins 3-4 Equilibrium Nifty K math Q and K Cnnectin with G Le Chatelier
Remember In general fr a reactin like aa + bb dd + ee K [ ] d D [ E] e [ ] a A [ ] b B
K s can be cmbined Suse we have a -ste reactin ste 1: 3 + ste : 3 O +. Overall: + O Add reactins t get verall reactin multily K s K K 1 ( ) O 3 3 K, tt 3 ( ) O 3 O ( ) ( )
Cmare Q and K Q and K are cnstructed the same way Q is true at all cncentratins/ressures At uilibrium, Q is ual t the value f K Cmaring Q and K tells us which directin a will g If Q < K (numeratr t small) make rducts If Q K (just right) uilibrium If Q > K (numeratr t big) make reactants
S if Q > K, can we say anything abut G at these cnditins? 5% 1. Yes, G must be < 0 5%. Yes, G must be 0 5% 3. Yes, G must be > 0 5% 4. N, dn t talk t me abut G 1 3 4 5
K and G are cnnected Recall frm Chater 14 G G + RT ln Q If we are uilibrium, then G Q K 0 0 G G + RT ln RT ln K Since yu knw hw t calculate G, yu can calculate K! K
What is K fr making ATP? ADP + H PO 4 ATP + H O G +30.6 kj/ml 5% 1. 0 5%. 1.41 10 6 5% 3. 4.36 10 6 5% 4. 0.988 1 3 4 5
What abut H and S? Of curse, if K is related t G, it is als related t H and S. G H ln K r RT ln T S H RT K RT ln + S R K K e H RT e S R S, if yu had H and S, yu culd calc K at many tems
Disturbances in Equilibrium Let s say we are making rcket fuel CH 4 (g) + H O(g) Q CH CO(g) + 3H (g) ( ) What haens if we are at uilibrium, then intrduce mre steam? Q instantly becmes smaller Nw Q < K, s we frm mre rducts CO 4 H H O 3 Intrducing reactant, ushed reactin tward rducts We shifted the uilibrium
Equilibrium is Dynamic CH 4 (g) + H O(g) CO(g) + 3H (g)
Le Chatelier s Princile Resnse f a system at uilibrium t utside influence is described by Le Chatelier s rincile System at uilibrium resnds t reduce the change We intrduced extra H O, system tried t reduce H O by using sme u, which make mre rduct. True fr all rducts and reactants Chemicals Heat Pressure/Vlume (fr gas-hase reactants/rducts)
Le Chatelier Examle CaCl (s) Ca + (aq) + Cl (aq) If this reactin is endthermic ( H +585 kj) will heat ush left r right? Endthermic means that heat is like a reactant S, adding heat ushes tward rducts mre salt disslves
Le Chatelier examle What haens t the fllwing uilibrium if we reduce the vlume f the cntainer? CaCO 3 (s) CaO (s) + CO (g) 33% 1. It shifts left tward reactants 33%. If shifts right tward rducts 33% 3. N change 1 3 4 5
Effect f Catalyst Catalyst changes E A Catalyst des nt effect rducts r reactants, s it des nt effect G Therefre catalyst has n effect n K Thugh it des hel system reach uilibrium mre quickly
Change What Occurs Effect n Equilibrium Effect n K Additin f Reactant Added reactant cnsumed Shift t right N Change Additin f Prduct Added rduct cnsumed Shift t left N Change Decrease V, Increase P Pressure Decreases Shift t fewer gas mlecules N Change Increase V, Decrease P Pressure Decreases Shift t mre gas mlecules N Change Increase T Heat is cnsumed Shift in end directin Change Decrease T Heat is generated Shift in ex directin Change
Equil. frm Initial Cnditins Often given initial cncentratins/ressures and asked t find cncentratins at uilibrium ADP + H PO 4 ATP + H O Initially: [ADP] 0.10 M [ATP] 0 M [H PO 4 ] 0.4 M K 4.36 10 6 What are cncentratins at uilibrium?
Equil. frm Initial Cnditins ADP + H PO 4 ATP + H O [ADP] [H PO 4 ] [ATP] Initial 0.10 0.4 0 Change x x + x Equilibrium 0.10 x 0.4 x + x 4.36 10 4.36 10 4.36 10 6 6 6 [ ATP] x [ ADP][ H PO ] ( 0.10 x)( 0.4 x) (.04 0.34x + x ) x K 1.00x + 1.046 10 x 4 7 0
Tricks 4.36 10 6 x 1.00x + 1.046 10 YUCK, hw d we slve that? 7 0 Quadratic uatin Plynmial slver in calculatr (r excel) In this case we knw x is small Because K EQ is much less than initial cncentratins S, x is way, way smaller than x negligible 1.00x + 1.046 10 7 0 x 1.046408 10 7 Slving exactly gives x 1.046398 10 7
Tday G t Neckers lecture 4:00 VDW10 Mnday Wrk n K wrksheet (excet art C) Finish CAPA #8