Dr. Zellmer Chemistry 1220 Wednesday Time: 18 mins Spring Semester 2019 February 13, 2019 Quiz IV. Name KEY Rec. TA/time. 2 mol gas 0 mol gas

Transcription

1 Dr. Zellmer Chemistry 1220 Wednesday Time: 18 mins Spring Semester 2019 February 13, 2019 Quiz IV Name KEY Rec. TA/time 1. (9 pts) For the following reaction answer the questions below. PH 3 (g) + BCl 3 (g) W PH 3 BCl 3 (s) a) (2 pts) For the system at equilibrium, what happens to the reaction when the pressure is decreased (by increasing the volume of the container) at constant temperature? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) EXPLAIN! 2 mol gas 0 mol gas PH 3 (g) + BCl 3 (g) W PH 3 BCl 3 (s) Decrease pressure (by increasing volume): Shifts to left (2 mol gas <--- 0 mol gas) Decrease pressure (by increasing volume) : shift to the side with more moles of gas. For Pressure changes (due to volume changes): P inc ==> fewer moles gas (this is equivalent to the Molarity inc., shift to side with fewer moles in soln) P dec ==> more moles gas (this is equivalent to the Molarity dec., shift to side with more moles in soln) Changes in Pressure do not affect liquids, solids or aqueous (their conc. do not change w. P like conc. of gases do) If no change in moles gas changes in P have no effect NOTE: This is really a conc. effect. Pressure is inversely proportional to the volume. It is directly proportional to the molarity (which is inversely prop. to the volume), P = M*RT. When the volume changes the molarity changes as does the pressure. Whether it s a gaseous solution or a liquid solution changing volume can shift the reaction. Volume dec. ==> shifts toward fewer moles of gas or solutes in soln Volume inc. ==> shifts toward more moles of gas or solutes in soln ***** continued on next page *****

2 1. (Cont.) b) (2 pts) For the system at equilibrium, what happens to the reaction when PH 3 is added? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) EXPLAIN! When a reactant is added the rxn shifts to the right (forward direction) to use some of the added reactant. When you add something (other than a pure solid or liquid) the reaction shifts away from added substance to use some of it. Reactant Add Product Remove Shift Right Remove Add Left Move AWAY from what is ADDED Move TOWARD what is REMOVED NOTE: K remains constant for changes in conc. & pressure (Numerical value of K changes only for temp. changes NOTE: Adding or Removing Pure SOLIDS of LIQUIDS (as long as some is present) does NOT affect the equil. However, if the reaction takes place in solution increasing the volume of the solution by adding a liquid (usually solvent) can shift the rxn toward more moles in solution. This is because the conc. of pure solids and liquids are essentially constant and do NOT appear in K. NOTE: As a rxn shifts the concentrations and amounts change. In this example, PH 3 is added and the reaction shifts to the right. As it shifts to the right the conc. of PH 3 goes back down (but the conc. winds up greater than what was present in the original equilibrium). The conc. of BCl 3 dec. so it s new conc. is less than the original. The conc. of the PH 3 BCl 3 solid does NOT change. However, it s amount (mass) does change. When the rxn shifts to the right some more solid forms so it s amount does change. c) (2 pts) For the system at equilibrium, what happens to the reaction when a little PH 3 BCl 3 is removed? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) EXPLAIN! No shift. Normally if you remove something the equil. would shift toward what s removed to replace it. However, this is a solid. The conc. of a solid is constant so adding some or removing part of it does not change it s conc. so it has no affect on the equil. The solid does not appear in Q or K so it has no affect on the equilibrium if some more is added or some is removed (as long as some is present). The amount (mass) of the solid changes but it s conc. doesn t so the reaction doesn t shift. ***** continued on next page *****

3 1. (Cont.) d) (3 pts) Assuming the above reaction is exothermic, what happens when the temperature increases? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) Also, what happens to the value of K? EXPLAIN! PH 3 (g) + BCl 3 (g) W PH 3 BCl 3 (s) + Heat For temperature changes you treat heat as a reactant or product. Also, changes in temperature are the only changes that cause a numerical change in the equilibrium constant, K. Since the rxn. is exothermic heat is a product. You inc. temp by adding heat. Move AWAY from ADDED. That means the reaction shifts to the left (reverse direction). As it does, the conc. of products decreases and conc. of reactants increases. These changes actually take place because the numerical value of the equilibrium constant, K, decreases, due to the temp. inc. You can think of it as the rxn shifts left and results in less product and more reactant and since temp. or heat are not explicitly in the expression for K the numerical values of K had to dec. in order to get less product and more reactant (a shift to the left). Remember: ΔH < 0 (!); exothermic; heat is a product - inc T (add heat, a product), shift left, K dec - dec T (remove heat, a product), shift right, K inc ΔH > 0 (+); endothermic; heat is a reactant - inc T (add heat, a reactant), shift right, K inc - dec T (remove heat, a reactant), shift left, K dec ΔH < 0 ΔH > 0 T 8, 7, K 9 T 8, 6, K 8 T 9, 6, K 8 T 9, 7, K 9 T & K move in opposite direction Product favored by LOW T T & K move in same direction Product favored by HIGH T

4 2. (3 pts) What is(are) the difference(s) between the Arrhenius and Bronsted-Lowry definitions of a base? Arrhenius base: A substance that has OH! (hydroxide ion), in the formula and when dissolved in water, increases the concentration of OH! ions. Restricted to aqueous solutions. Although the book states NH 3 is an Arrhenius base this is technically NOT truly an Arrhenius base in the fundamental sense of the definition. It has no OH in the formula. When put NH 3 in water it does increase the conc. of OH!. The book uses this looser definition to state NH 3 is an Arrhenius base (it is not). Part of the reason is Arrhenius himself considered it a base. However, in order to do so he had to come up with the substance ammonium hydroxide, NH 4 OH, which does NOT actually exist. If you put NH 3 in H 2 O you do get NH 4 + and OH! ions. However, if you evaporate the water you do not get solid NH 4 OH. Instead as the water evaporates the NH 3 comes out as a gas. Still to this day you may see the name ammonium hydroxide on a bottle of aqueous ammonia (NH 3 dissolved in H 2 O). Bronsted-Lowry acid: A substance (molecule or ion) that can accept a proton, H +, from another substance. This theory is about proton transfer reactions. It is NOT restricted to aqueous solutions, whereas Arrhenius theory is. This theory also introduced the idea about conjugate acid-base pairs. Proton Acceptor This theory covers all Arrhenius acids and bases. Lewis base: A substance (molecule or ion) that can donate a pair of electrons to another substance (although the electrons themselves are not transferred). It is NOT restricted to aqueous solutions, whereas Arrhenius theory is. Electron-pair Donor This is the most comprehensive theory of the three. It covers all BL acids and bases (and Arrhenius acids and bases) and reactions which can explain why things w/o hydrogens in the formula can cause a solution to be acidic. The LB is not actually losing free electrons, it combines with something that needs a pair of electrons because it s electron deficient. NOTE: The e - pair is not actually transferred from the base to the acid. They combine and share the pair of e -.

5 3. (8 pts) D 2 O(R) is deuterium oxide, or better known as heavy water. D 2 O(R) has an equilibrium dissociation constant, K D (along the lines of K w for water) of 1.35 x 10!15 at 25.0EC. Answer the following using this information. a) (2 pts) Calculate [D + ] and [OD & ] for D 2 O at this temperature. Analogous to H 2 O we would have, K D = [D + ] [OD & ] = 1.35 x 10 &15 For the autoionization of pure D 2 O, [D + ] = [OD & ] = x x 2 = 1.35 x 10 &16 x = [D + ] = [OD & ] = x 10 &8 = 3.67 x 10 &8 b) (1 pt) What is the pd of this liquid? (Hint: pd is a px function, like ph)? pd =!log [D + ]; [D + ] = 10 & pd ; pod =!log [OD & ]; [OD & ] = 10 & pod pd =!log [3.67 x 10 &8 ] = pod =!log [3.67 x 10 &8 ] = (3 decimal paces, 3 s.f.) c) (1 pt) What is the pk D of D 2 O? pk D =!log K D =!log (1.35 x 10 &15 ) =!log (1.35 x 10 &15 ) = = pd + pod = pk D = (3 s.f., the 3 decimal places) d) (4 pts) Calculate the pd of a solution of M Sr(OD) 2 (a strong base in D 2 O)? H 2 O Sr(OD) > Sr 2+ (aq) + 2 OD & (aq) (strong base - get 2 OD & for every Sr(OD) 2 ) pd + pod = pk D = [OD & ] = 2 [Sr(OD) 2 ] = 2 (0.200 M) = M pod =!log [OD & ] =!log [0.400] = (3 sig. fig., to the right of the decimal point) pd = ! pod = = = or K D = [D + ] [OD & ] = 1.35 x 10 &15 K D 1.35 x 10 &15 [D + ] = = = x 10 &15 M [OD & ] pd =!log [D + ]; pd =!log [D + ] =!log [3.375 x 10 &15 ] = NOTE: The significant figures in ph and poh are to the right of the decimal point (as explained in class and in Appendix A of the textbook.

6 4. (1 pts) What is the conjugate acid of H 2 P 2 O 7 2!? H 2 P 2 O 7 2! is amphoteric - can act as an acid or a base. You are told to treat it as an acid in this case. To get the conjugate base of an acid you simply REMOVE an H + from the acid: 2! H 2 P 2 O 7 v H + + 3! HP 2 O 7 acid conj. base To get the conjugate acid of a base you simply ADD an H + to the base: 2! H 2 P 2 O 7 + H +! v H 3 P 2 O 7 base conj. acid 5. (4 pts) A solution of HBr has a ph = What are the [H 3 O + ], [OH! ], and poh and what was the initial [HBr] in the solution at 25EC? HBr is a strong acid meaning it completely ionizes: HBr (aq) ----> H + (aq) + Br! (aq) ph =!log[h + ] [H + ] = 10!pH = 10!5.50 = x 10!6 M = 3.2 x 10!6 M (2 s.f. - really only 2 s.f., in the ph - the ones to the right of the decimal pt.) For any of the six monoprotic strong acids [H + ] = [acid] 0 ([acid] 0 is the original conc. of acid in the problem) [H + ] = [H 3 O + ] = [HBr] = 3.2 x 10!6 M Find [OH! ] and poh - two different ways: K w = [H + ] [OH! ] = 1.0 x 10!14 or K w 1.0 x 10!14 [OH & ] = = = 3.16 x 10!9 M poh =!log [OH! ] [H + ] 3.16 x 10!6 = 3.2 x 10!9 M =!log(3.16 x 10!9 ) = 8.50 (2 s.f.) ph + poh = pk w = poh = & 5.50 = 8.50 [OH! ] = 10!pOH = 10!8.50 = 3.2 x 10!9 M NOTE: The significant figures in the ph, poh and pk are to the right of the decimal point.