Equilibrium Lecture Examples Problem 1: Consider the reaction: J R 8.3145 T 98.15K K PCl 5 PCl 3 + Cl a.) Compute the standard Gibbs Free energy for this system at 98.15K. In which direction will the reaction proceed if the pressure are all at 1 bar, or standard conditions? b.) Produce an expression for their reaction quotient.if the partial pressures of the components are as follows, compute Q and determinein which direction will the reaction proceed. c.) Compute a value for the equilibrium constant, K p. d.) At equilibrium, the pressures of PCl 3 and Cl were measured to be 0.5 bar. What is the equilibrium pressure of PCl 5.? e.) Compute K c for this reaction system. P PCl5 0.105bar P PCl3 0.0175bar P Cl 0.040bar 1bar ΔG PCl5 78.36 ΔG PCl3 67.78 ΔG Cl 0 ΔG rxn_std ΔG PCl3 ΔG Cl ΔG PCl5 10.456 Reaction will proceed to the left because ΔG > 0. Q P PCl3 P Cl P o P PCl5 6.667 10 3 Note: For this document to insure integrity of the units, I included the standard pressure in the expression. You will not be required to do this although you are required to know that it is there.
0.0175bar 0.040bar 1bar 1bar ΔG rxn ΔG rxn_std RT ln 1.965 0.105bar 1bar Reaction will proceed to the right because ΔG < 0. ΔG rxn_std K p e RT 0.0147 P PCl3 0.5bar P Cl 0.5bar K p 0.0147 P PCl3 P Cl P o P PCl5 P PCl3 P Cl P PCl5 Solving... P PCl5 P PCl3 P Cl K p 4.43bar RT K p K c M o 1 L Δn In this reaction system, Δn -1. Solving... K c K p RT M o 1 0.365 Note!! This is true for gases only!!
Problem : For the following system, the equilibrium constant is K p 0.074 at 18 o C. Compute the partial pressures of each component and the total pressure at equilibrium. NH 4 SH(s) NH 3 (g) + H S(g) NH 4 SH(s) NH 3 (g) + H S(g) ICE Diagram Initial 0 0 Change x x Equilibrium 0 x x 0 x x K p P NH3 P HS P NH3 P HS x 0.074 Solving.. K p 0.074 x Kp 0.7bar P NH3 0.7bar P HS P Total x P Total 0.544bar
Problem 3 The equilibrium reaction that produces noxious NO gas is given as NO <> N O 4 with a K c 16 at 5 o C a.) Compute K p for the reaction of N O 4 <> NO b.) Compute the partial pressures of NO and N O 4 at equilibrium of the initial pressure of N O 4 1. bar R u 0.0805 T u 98.15 Note: (The subscripts denote that units have been removed.) K p K c ( RT ) Δn Δn 1 1 1 K c K 16 p K c R u T u K p 0.113 P init 1. N O 4 <> NO Initial 0 Change -x +x Equilibrium - x x P NO K p P NO4 ( x ) P init x 4x rearranging K P init x p P init x K p P init K p x 4x 4x K p x K p P init 0 which has the solution.. or x K p K p 4( 4) K p P init x 0.171 ( 4) P NO4 1. 0.171 1.09 P NO x 0.341
Problem 4 - For the PCl 5 equilibrium system presented above, compute the equilibrium pressures of all species if an initial amount of 0.500 bar of PCl 5 is placed in a reaction vessel From above K p 0.0147 1.0bar PCl 5 PCl 3 + Cl ICE Diagram Initial Change 0.50 bar 0 0 x x x Equilibrium 0.50 x 0 x 0 x K p 0.0147 P PCl3 P Cl P o P PCl5 P PCl3 P Cl P PCl5 x ( 0.50 x) Putting in standard quadratic form.. K p ( 0.50bar x) x or x K p x K p 0.50bar a 1 b K p 0.015bar c K p 0.50bar 7.35 10 3 bar x K p K p 4K p 0.50bar 0.079bar P PCl5 0.50bar x 0.41bar P PCl3 x 0.079bar P Cl x 0.079bar Checking.. P PCl3 P Cl P PCl5 0.0147
Problem 5 - In a previous lecture, we computed the Gibbs Free Energy for the Haber process and found that when the temperature is raised, the reaction becomes less spontaneous. The Free energies computed are given below. From these, compute the % conversion of hydrogen and nitrogen into ammonia at temperatures of 98.15 K and 398.15 K if 1 bar of N and 3 bar of H are introduced into a reaction vessel and allowed to come to equilibrium. ΔG rxn_98 34.155 ΔG rxn_398 14.68 From these, let's first compute the equilibrium constants. ΔG rxn_98 ΔG rxn_398 R( 98.15K) K 98 e 9.631 10 5 R( 398.15K) K 398 e 84.309 Note that we can do this only because the ΔG values are given at the respective temperatures Let us now solve this equilibrium system for a given K. 3 H (g) + N (g) NH 3 (g) Initial Change 3 bar 1 bar 0-3 x -x + x ICE Diagram Equilibrium 3-3 x 1 - x x 3 (1 - x) K p P H P NH3 3 P N P NH3 Note: for this problem, I removed 3 PH P N division by for simplification.
Applying the ICE diagram results... K p ( x ) [ 3( 1 x) ] 3 ( 1 x) 4x 7( 1 x) 4 rearranging x ( 1 x) 4 K p 7 C where for convenience C K p 7 4 4 C K 98 7 4 549.688 Note that in this instance, the left hand side is a perfect square. Taking the square root will allow us to solve the system.. x x x C rearranging ( 1 x) 4 ( 1 x) 1 x x C 1 x x C Cx Cx x finally (whew), putting into standard quadratic form... Cx ( C 1) x C 0 Applying the quadratic formula.. a C 549.7 b ( C 1) 5100.4 c C 549.7 x ( C 1) (( C 1) ) 4C C Now let us produce some numbers!!
K 98 7 At 98.15 K C 98 4.55 10 3 C 98 1 4C 98 C 98 1 x 98 C 98 98.039 % C 98 1 4C 98 C 98 1 x 98 C 98 10 % which makes no sense of course. Lets define the % conversion as: x is the amount of N reacted, so is a reasonable value for the % conversion. %_conversion x 98 98.039 % K 398 7 At 398.15 K C 398 4 3.855 C 398 1 4C 398 C 398 1 x 398 C 398 0.815 %_conversion x 398 81.515 %
Problem 6 In the hydrocarbon industry, an important reaction is used to produce hydrogen gas from methane called the methane-steam reaction. Given as CH 4 (g) + H O(g) CO(g) + 3 H (g) At 900 K, the equilibrium constant is about K p 0.183. For this reaction, compute the % conversion of methane if equal pressures of methane and steam are reacted. Do this for initial pressures of 1 atm and 10 atm and determine whether the reaction should be carried out at high or low pressures. Also, is the reaction favored at lower or higher temperatures? ΔH f_ch4 74.6 ΔH f_ho 41.8 ΔH f_co 110.5 ΔH rxn ΔH f_co ΔH f_ch4 ΔH f_ho ΔH rxn 05.9 This is endothermic so the reaction favors high heat. Note also that ΔG f_ch4 50.5 ΔG f_ho 8.6 ΔG f_co 137. ΔG rxn ΔG f_co ΔG f_ch4 ΔG f_ho ΔG rxn 141.9 ΔG rxn RT So K p e K p 1.381 10 5 at 98.15 K
Now, let's solve the system CH 4 (g) + H O(g) CO(g) + 3 H (g) - x - x x 3x x( 3x) 3 K p 0.183 K p z or 7 K p x 4 x this is a perfect square, so 7 x P K o x or p 7 x K p x 0 7 a 1.147 b 1 P K o 1bar or 10bar p x 1 1 4a a x1 ( ) 0.49 x10 ( ) 0.867 x1 ( ) 1 4.87 % x10 ( ) 10 8.671 %
Aqueous Solutions: Acids and Bases Problem 1: Compute the ph of a 0.005 M solution of a.) the strong acid HBr b.) the weak acid HF, K a 3.53 x 10-4 a.) HBr dissociates completely in water, this at the completion of the reaction, H + is equal to the original concentration of HBr, or 0.005 M. Thus ph log( 0.005).301 b.) HF is weak and so we need to perform an equilibrium calculation. I'll do this in general for a monoprotic weak acid. Let the initial concentration of HF be C o K a 3.5310 4 C o 0.005 HF(aq) H + (aq) + Cl - (aq) ICE Diagram Initial Change C o 0 0 x x x Equilibrium C o x 0 x 0 x K a ( H) ( F) ( HF) x C o x Putting in standard quadratic form.. K a C o K a x x or x K a x C o K a 0 by the quadratic equation. x K a K a 4K a C o 1.164 10 3 Thus,, ph log( x).934
Problem - Sodium fluoride, NaF is the conjugate base of HF. Compute the ph of a 0.005 M solution of NaF For this system, we must realize that the sodium ion is a spectator and we can thus ignore it. Secondly, the F - will combine with water to form the conjugate base, HF. Writing.. F - (aq) + H O (l) HF (aq) + OH - (aq) The K b for this base is obtained from the K a of its' conjugate using K w K a 3.53 10 4 K w 110 14 K w K b.833 10 11 K a Executing an ICE diagram on this system reveals an interesting result. F - (aq) + H O (l) HF (aq) + OH - (aq) C o Initial 0 0 Change x x x Equilibrium C o x 0 x 0 x Notice that water is ignored as it is a liquid K b ( HF) ( OH) ( F) x C o x!!!!!! It is identical to the general result we solved for previously! Thus we know the result. x K b K b 4K b C o 3.763 10 7 poh log( x) 6.44 ph 14.00 poh 7.576
Controlling ph Problem 3 - One can produce an acid/base system of a given ph. However, often the ph is required to be relatively stable as reactions are going on or to maintain a level of toxicity. This is done through "buffers". Calculate the ph of a solution prepared with 300 ml of 0. M HF and 50 ml of 0.3 M NaF. (K a 3.53 x 10-4 ) M HF 0.0M V HF 300mL M F 0.30M V F 50mL Note: Na + is a spectator ion and has no effect on the acidity. Thus it is left out. m HF M HF V HF 60.000m m F M F V F 75.000m K a 3.5310 4 pk a log K a 3.45 m F ph pk a log m 3.549 HF
Problem 4 - You wish to prepare a buffer for a skin medicinal product that will keep a of ph 7.0 a.) A list of acids an their pk a 's are provided below. Suggest the best acid to use and a conjugate base to add to it to produce the desired ph. b.) Compute the amount of the base needed to add to 500 ml of 0.45 M of the selected acid to produce the desired ph. Acid pka HOCl 7.53 HSeO -1 4 1.9 HOCN 3.45 H PO -1 4 7.0 a.) We want an acid that has a pk a as close to the desired ph as possible. For the desired ph 7.00, H PO -1 4 is the best choice. The conjugate base of this acid is HPO - 4, thus dissolving Na HPO 4 salt would work. b.) In order to obtain the desired ph, we need to know the correct base/acid ratio. pk a 7.0 ph 7.00 M Acid 0.45M V Acid 500mL m Acid M Acid V Acid 5m ph pk a log Base rearranging Ratio 10 ph pk a 0.631 Acid Thus we need 0.631 times the acid of base or m Base m Acid ( Ratio) 141.965m mg MW NaHPO4 141.96 m Mass NaHPO4 m Base MW NaHPO4 0.153gm
Problem 5-5.0 ml of a 1.0 M strong acid HCl are added to the buffer prepared. a.) Calculate the ph change for the buffer. b.) Calculate the ph change that would occur if the acid were added to 500 ml of unbuffered water. M HCl 1.0M V HCl 5.0mL m HCl M HCl V HCl 5m The acid attacks the base, HPO - 4 and converts it into the acid, H PO -1 4. Thus we add 0.5 m to the acid and remove 0.5 m from the base. Now 14.0 m ph pk a log 5m 6.97499 5.0m 5m ph_change 7.00 ph 0.05007 In water we only need the concentration of the acid when added and thus diluted. M HCl 500mL m HCl 5.0mL 9.901 10 3 M M HCl ph log.004 ph_change 7.00 3.004 3.9960 M o Quite a difference!!
Solubility Equilibrium Problem 6 - Certain salts have limited solubility and the solubility is governed by the same equilibrium principles. Consider a saturated solution of Mg(OH). K sp 1.5 x 10-11. a.) Compute the solubility of Mg(OH) in pure water at equilibrium. In addition, since this is a base, compute the ph. b.) Compute the solubility and ph of Mg(OH) in a solution of 0.05 M Mg(NO 3 ) which is completely soluble in water. a.) We follow the same rules as before. Write an ICE diagram and solve. K sp 1.510 11 Mg(OH) (s) Mg + (aq) + OH -1 (aq) ICE Diagram Initial 0 0 Change x x Equilibrium 0 x x 0 x x K sp [Mg + ][OH -1 ] (x)(x) 4 x 3 solving x K sp 4 1 3 1.554 10 4 This is the saturated solution arity. Note the magnitude which is small OH x 3.107 10 4 poh log( OH) 3.508 ph 14.00 poh 10.49
b.) In a solution with a common ion, we need to include it into the ICE table..doing so yields.. Mg(OH) (s) Mg + (aq) + OH -1 (aq) ICE Diagram Initial 0.05 0 Change x x Equilibrium 0.05 x x K sp [Mg + ][OH -1 ] (x + 0.05)(x) Here we have a cubic polynomial to solve which is difficult. However, we know two things.. 1.) The Mg + concentration present to begin with is larger than the Mg + equilibrium value in pure water..) The concentration will only get smaller. We will thus make the assumption that x << 0.05 Applying.. K sp [Mg + ][OH -1 ] (x + 0.05)(x) (0.05)(x) 0.10x Now we solve easily as: x K sp 0.10 1.5 10 5 We confirm that x is indeed much smaller than 0.05 so our assumption is justified. Finally.. OH x.449 10 5 poh log( OH) 4.611 ph 14.00 poh 9.389