The Equilibrium Constant (K eq)

Size: px

Start display at page:

Download "The Equilibrium Constant (K eq)"

Brett Chapman
6 years ago
Views:

1 The Constant (K eq) 1. The Law: K eq is called the. Keq = [PRODUCTS] [REACTANTS] eg. aa + bb + xx + yy + where a, b, x, y are of the balanced equation. A,B,X & Y are the and in the equation Keq = [X] x [Y] y [A] a [B] b The value of K eq does not change for a given rxn at a given. The factor representing any substance whose conc cannot change is always 1 and there fore not included (i.e. a solid or a pure liquid). o i.e. if H 2 O is produced then [X ] x = 1 and therefore not written 2. The Meaning of K eq : A small K eq value means that the rxn is mostly in the form of A large. Keq = [PRODUCTS] Comparison [REACTANTS] Small K eq [1]/[10] = 0.1 reactants > products Large K eq [10]/[1] =10 reactants < products K eq = 1 [10]/[10] =1 reactants = products The value of K eq changes when we a) rxns b) change the of a given rxn c) change the of the equation. 3. Building K eq equations 1

2 Reaction 1 N 2 (g) + 3H 2 (g) 2NH 3 (g) Keq = [NH 3 ] 2 [N 2 ] [H 2 ] 3 Reaction 2 (reaction 1 reversed) 2NH 3 (g) N 2 (g) + 3H 2 (g) Keq = [N 2 ] [H 2 ] 3 [NH 3 ] 2 Reaction 3 (reaction 1 doubled) 2N 2 (g) + 6H 2 (g) 4NH 3 (g), Keq = [NH 3 ] 4 [N 2 ] 2 [H 2 ] 6 Therefore Whenever a rxn is reversed, its K eq value becomes the of the original value doubled, 4. K eq with Solids and Liquids When we write the K eq expression for a rxn with solids or pure liquids, we simply leave them E.g. CaCO 3(s) CaO (s) + CO 2(g) K eq = [CaO (s) ] [CO 2(g) ] therefore = [CO 2(g) ] [CaCO 3(s) ] Definition p. 59 e.g. CaCO 3(s) + 2HF (g) CaF 2(s) + H 2 O (l) + CO 2(g) K eq = 2

3 5. Effect of Temperature on Keq For a rxn at 0 C, K eq = 5. Same rxn at 25 C, K eq = 200. We see two changes due to the temperature increase 1. Rxn has shifted right because K eq. 2. Fwd rxn is thermic, because products increase When the temperature changes, the value of K eq also changes When the temperature increases in an endothermic rxn, the equilib will shift to the right and the value of K eq will. When the temperature decreases in an endothermic rxn, the equilib will shift to the left and the value of K eq will. When the temperature increases in an exothermic rxn, the equilib will shift to the left and the value of K eq will When the temperature decreases in an exothermic rxn, the equilib will shift to the right and the value K eq will 6. Effect of Concentration on K eq IMPORTANT: temperature is the only factor that affects K eq If T remains constant, changing the conc of reactants or products does not change the value for K eq Why? Eg. (at eqilib) A + B C + D K eq = [C] [D] = 4.0 [A] [B] Now add C, so [C] is greater and now K eq 4 but equilib will shift left new equilib also has K eq = 4 Recall: Whatever we do, nature tries to undo Therefore As long as T is not changed, the equilib will always shift just enough to keep the ratio equal to the value of the equilib constant (K eq )! 7. Catalyst effect on Keq 3

4 Recall, a catalyst does NOT shift equilib Therefore No shift, Therefore NO change in K eq 8. Effect of Pressure or/ Volume on K eq if Volume or Pressure causes equilib shift, we see the same effect as conc. on K eq if no T, no K eq A change in total volume or total pressure does not change the value of the equilib constant K eq. The equilib will shift to keep the ratio equal to K eq You will have to know how to calculate 5 different types of K eq questions Type 1 - Find K eq, given equilib concs K eq CALCULATIONS Given the equilib system: PCl 5(g) PCl 3(g) + Cl 2(g) At equilib: [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M. K eq = [PCl 3 ][Cl 2 ] = = [PCl 5 ] Note : K eq does not include any Type 2 Find Equilib. Concs, given K eq N 2(g) + 3H 2(g) 2NH 3(g) K eq = 625 Calculate [H 2 ] if at equilib [N 2 ] = M and [NH 3 ] = 0.12 M a) Write K eq expression: b) Plug in known values and solve: K eq = = [H 2 ] = 4

5 Type 3 Find equilib concs or K eq, given initial concs and one equilib conc The ICE method [I] = [C] = [E] = Initial concentrations Change in concentration Equilib concentrations N 2(g) + 3H 2(g) 2NH 3(g) e.g. Put only N 2 and H 2 into container. [I] : [N 2 ] = 0.32 M and [H 2 ] = 0.66 M. wait for a while. [E] : [H 2 ] = 0.30 M. [C] : [H 2 ] = M Question: a) Find the equilib [N 2 ] and [NH 3 ]. b) Calculate K eq a) use ICE table and write what you know Initial conc. [I] (change in conc.) [C] conc [E] N 2 + 3H 2 2NH 3 b) calculate change in conc using mole ratios [N 2 ] = ( ) x = since [H 2 ] and [N 2 ], [NH 3 ] must [NH 3 ] = ( ) x = c) Do the math: d) Find value of K eq K eq = = = Type 4 Find Equilib shift, given initial concs and K eq 5

6 Trial K eq = K eq using concentrations If Trial K eq < K eq rxn shifts If Trial K eq > K eq rxn shifts. If Trial K eq = K eq, the reaction will. Trial K eq = Q ( the reaction quotient) When system may not be at EQB eg: CO (g) + H 2 O (g) CO 2(g) + H 2(g) Keq = 10.0 How will equilib shift if the initial concentrations = 0.80M CO, 0.050M H 2 O, 0.50M CO 2, and 0.40M H 2.? K eq = = = but actual K eq = 10.0 therefore, Trial K eq < K eq and equilib shifts Type 5 Find [E] of all species, given [I] of all species and K eq H 2 + I 2 2HI The Keq = 55.6 [I]: [H 2 ] = M and [I 2 ] = M Step 1. Which way will rxn shift? Start with only H 2 and I 2 so Trial K eq = = = Actual K eq = Therefore K eq, so rxn must shift! let [H 2 ] = x and ([C] for H 2 = -x so -x -x +2x H 2 + I 2 2HI Fill in first 3 rows on table on next page [I] H 2 + I 2 2HI 6

7 [C] [E] formula [E] Plug into K eq equation: K eq = = [H 2 ][I 2 ] = = = x = Therefore at eqb [HI] = 2x = 2 ( ) = M [H 2 ] = [I 2 ] = = = M Answer check : K eq = = = 55.6 (close enough) The Constant (K eq) 7

8 1. The Law: K eq is called the equilib constant. Keq = [PRODUCTS] [REACTANTS] eg. aa + bb + xx + yy + where a, b, x, y are coefficients of the balanced equation. A,B,X & Y are the products and reactants in the equation Keq = [X] x [Y] y [A] a [B] b The value of K eq does not change for a given rxn at a given temperature. The factor representing any substance whose conc cannot change is always 1 and there fore not included (i.e. a solid or a pure liquid). o i.e. if H 2 O is produced then [X ] x = 1 and therefore not written 2. The Meaning of K eq : A small K eq value means that the rxn is mostly in the form of reactants A large products. Keq = [PRODUCTS] Comparison [REACTANTS] Small K eq [1]/[10] = 0.1 reactants > products Large K eq [10]/[1] =10 reactants < products K eq = 1 [10]/[10] =1 reactants = products The value of K eq changes when we d) switch rxns e) change the temperature of a given rxn f) change the form of the equation. 3. Building K eq equations 8

9 Reaction 1 N 2 (g) + 3H 2 (g) 2NH 3 (g) Keq = [NH 3 ] 2 [N 2 ] [H 2 ] 3 Reaction 2 (reaction 1 reversed) 2NH 3 (g) N 2 (g) + 3H 2 (g) Keq = [N 2 ] [H 2 ] 3 [NH 3 ] 2 Reaction 3 (reaction 1 doubled) 2N 2 (g) + 6H 2 (g) 4NH 3 (g), Keq = [NH 3 ] 4 [N 2 ] 2 [H 2 ] 6 Therefore Whenever a rxn is reversed, its K eq value becomes the reciprocal of the original value doubled, square 5. K eq with Solids and Liquids When we write the K eq expression for a rxn with solids or pure liquids, we simply leave them out E.g. CaCO 3(s) CaO (s) + CO 2(g) K eq = [CaO (s) ] [CO 2(g) ] therefore = [CO 2(g) ] [CaCO 3(s) ] Definition p.59 e.g. CaCO 3(s) + 2HF (g) CaF 2(s) + H 2 O (l) + CO 2(g) K eq = [CO2] [ HF2] 2 5. Effect of Temperature on Keq For a rxn at 0 C, K eq = 5. Same rxn at 25 C, K eq = 200. We see two changes due to the temperature increase 9

10 3. Rxn has shifted right because K eq increases. 4. Fwd rxn is endothermic, because it decreases When the temperature changes, the value of K eq also changes When the temperature increases in an endothermic rxn, the equilib will shift to the right and the value of K eq will increases. When the temperature decreases in an endothermic rxn, the equilib will shift to the left and the value of K eq will decreases. When the temperature increases in an exothermic rxn, the equilib will shift to the left and the value of K eq will decrease When the temperature decreases in an exothermic rxn, the equilib will shift to the right and the value K eq will increases 6. Effect of Concentration on K eq IMPORTANT: temperature is the only factor that affects K eq If T remains constant, changing the conc of reactants or products does not change the value for K eq Why? Eg. (at eqilib) A + B C + D K eq = [C] [D] = 4.0 [A] [B] Now add C, so [C] is greater and now K eq 4 but equilib will shift left new equilib also has K eq = 4 Recall: Whatever we do, nature tries to undo Therefore As long as T is not changed, the equilib will always shift just enough to keep the ratio equal to the value of the equilib constant (K eq )! 7. Catalyst effect on Keq Recall, a catalyst does NOT shift equilib Therefore No shift, Therefore NO change in K eq 8. Effect of Pressure or/ Volume on K eq 10

11 if Volume or Pressure causes equilib shift, we see the same effect as conc. on K eq if no T, no K eq A change in total volume or total pressure does not change the value of the equilib constant K eq. The equilib will shift to keep the ratio equal to K eq. You will have to know how to calculate 5 different types of K eq questions Type 1 - Find K eq, given equilib concs K eq CALCULATIONS Given the equilib system: PCl 5(g) PCl 3(g) + Cl 2(g) At equilib: [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M. K eq = [PCl 3 ][Cl 2 ] = [0.40][0.40] = 0.50 [PCl 5 ] 0.32 Note : K eq does not include any units Type 2 Find Equilib. Concs, given K eq N 2(g) + 3H 2(g) 2NH 3(g) K eq = 625 Calculate [H 2 ] if at equilib [N 2 ] = M and [NH 3 ] = 0.12 M a) Write K eq expression: b) Plug in known values and solve: K eq = [NH 3 ] = [0.12] 2 [H 2 ] 3 [N 2 ] [H 2 ] 3 [0.030] [H 2 ] = M Type 3 Find equilib concs or K eq, given initial concs and one equilib conc The ICE method [I] = Initial concentrations 11

12 [C] = Change in concentration [E] = Equilib concentrations N 2(g) + 3H 2(g) 2NH 3(g) e.g. Put only N 2 and H 2 into container. [I] : [N 2 ] = 0.32 M and [H 2 ] = 0.66 M. wait for a while. [E] : [H 2 ] = 0.30 M. [C] : [H 2 ] = M Question: a) Find the equilib [N 2 ] and [NH 3 ]. b) Calculate K eq a) use ICE table and write what you know N 2 + 3H 2 2NH 3 Initial conc. [I] (change in conc.) [C] M M M conc [E] = = 0.24 b) calculate change in conc using mole ratios [N 2 ] = ( M H 2 ) x 1 mol N 2 = M H 2 3 mol H 2 since [H 2 ] and [N 2 ], [NH 3 ] must [NH 3 ] = (+0.36 M H 2 ) x 1 mol NH 3 = M NH 3 3 mol H 2 c) Do the math: d) Find value of K eq K eq = [NH 3 ] 2 = [0.24] 2 = 10.7 [H 2 ] 3 [N 2 ] [0.20] 3 [0.30] Type 4 Find Equilib shift, given initial concs and K eq Trial K eq = K eq using initial concentrations If Trial K eq < K eq rxn shifts right If Trial K eq > K eq rxn shifts left. If Trial K eq = K eq, the reaction will not shift at all. Trial K eq = Q ( the reaction quotient) When system may not be at EQB 12

13 eg: CO (g) + H 2 O (g) CO 2(g) + H 2(g) Keq = 10.0 How will equilib shift if the initial concentrations = 0.80M CO, 0.050M H 2 O, 0.50M CO 2, and 0.40M H 2.? Trial K eq = [CO 2 ] [H2] = [0.50][0.40] = 5.0 [CO] [H 2 O] [0.80][0.050] but actual K eq = 10.0 therefore, Trial K eq < K eq and equilib shifts right Type 5 Find [E] of all species, given [I] of all species and K eq [I]: [H 2 ] = M and [I 2 ] = M H 2 + I 2 2HI The Keq = 55.6 Step 1. Which way will rxn shift? Start with only H 2 and I 2 so Trial K eq = HI] 2 = [0] 2 = 0 [H 2 ][I 2 ] [0.200][0.200] Actual K eq = Therefore K eq increases, so rxn must shift right! let [H 2 ] = x and ([C] for H 2 = -x so -x -x +2x H 2 + I 2 2HI Fill in first 3 rows on table on next page H 2 + I 2 2HI [I] [C] -x -x +2x [E] formula x x 0 + 2x = 2x [E] Plug into K eq equation: 13

14 K eq = HI] 2 = 55.6 [H 2 ][I 2 ] = [2x] 2 = 55.6 [0.200-x][0.200-x] = [2x] 2 = 55.6 ][0.200-x] 2 = 2x = x x = Therefore at eqb [HI] = 2x = 2 (0.1577) = M [H 2 ] = [I 2 ] = x = = M Answer check : K eq = HI] 2 = [0.315] 2 = [H 2 ][I 2 ] [0.0423] 2 (close enough) 14

Chapter 2 - Practice Test #1

Chapter 2 - Practice Test #1 6. Chemical systems tend to move toward positions of A. minimum enthalpy and maximum entropy. B. maximum enthalpy and minimum entropy. C. minimum enthalpy and minimum entropy.