Answers to EXAM III Fall 2014 CCBC-Catonsville (Mon 11/3/14) 1. H 2 O 2 + I H 2 O + IO (slow) H 2 O 2 + IO H 2 O + O 2 + I (fast)

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1 Chem 13 Answers to EXAM III all 014 CCC-Catonsville (Mon 11/3/14) 1. H O + I H O + IO (slow) H O + IO H O + O + I (fast) a) H O H O + O ) Rate law expected would e that of the rate determining step (slowest step) Rate = k [H O ][I ] c) It is first order with respect to H O. d) I is a catalyst. It is added and then consumed efore the end. IO is an intermediate. It is produced in the first step and consumed efore the end.. a) Thermodynamically favored means the product is more stale than the reactant (an exothermic reaction). Diagram A ) The faster reaction would have the smaller activation energy. Diagram A c) The faster reaction would have a larger rate, and therefore, the larger rate constant k. 3. E PE A C D Progress of Reaction 4. Consider the reaction NO (g) + O (g) NO (g). The gas, NO, is colorless and the gas, NO, is reddish rown. The system is allowed to equilirate at a certain temperature to give a mixture with a rown color. If the volume of the container is decreased, what is expected to happen to the color? The pressure would e increased and the equilirium will shift towards the side that would allow the system to e less crowded (to the right). Since NO is reddish rown, the mixture would ecome darker in color.. [] 4 = = = 8 [A] 6. Rememer that sustances existing as a solid do not appear in the equilirium expression. Thus you should consider only the species that are gaseous in this reaction. The right side of the reaction is less crowded, therefore a decrease in volume would cause the equilirium to shift to the right, producing more of the products to restore the equilirium.

2 7. The conjugate ase is the structure after the species acting as an acid, loses one H +. Thus it ecomes H 3 AsO In a rønsted acid/ase reaction, the position of the equilirium goes from a stronger acid to the weaker acid. Thus if the equilirium lies to the left, the stronger acid is on the right side. The two acids eing considered are HS on the left and H SO 3. The stronger acid is H SO H Se is stronger. As a inary acid, the more electronegative central atom pulls electrons closer to itself, allowing the H + to leave more readily. 10. As oxohalo acids, the more oxygen s there are, the strong the acid ecause the anion produced is stailized y having the negative charge spread over more atoms (in ro as compared to ro ) HrO + H O H 3 O + + ro 11. Note that these are not inary acids (as in Question 9) ut oxohalo acids with the same numer of oxygen s. The trend is according to the electronegativity of the halogen (Cl vs. r) rather than considering the anion size. Cl is more electronegative and therefore HOCl is a stronger acid. 1. At first glance none of them seem to e acidic as they do not contain H +. However, the cations are acting as Lewis acids. Therefore we should e comparing the cations. irst figure out the charges of the cations: +3, +3, +1, +, + (ased on nitrate eing 1). The higher the charge of the cation, the stronger is the acid. Thus we eliminate the last three. In comparing the first two (having +3) we now consider the size of the cation. Al 3+ is smaller and therefore the positive charge is confined to a smaller space (more densely positive) and therefore more likely for the hydrated cation to act as an acid. 3+ Al(H O) 6 + OH Al(H O) (OH) + + H O 13. _ The fluoride ion is donating one of its lone pairs to oron. Thus it s acting as a Lewis ase. 14. Since they are all the same concentration, we only need to look at which is the stronger acid (releasing more protons and producing a lower ph). The strongest acid has the lowest pa. 1. A low ph corresponds to an acidic solution. We look at only the cation to decide whether a salt solution is acidic (and look at the anion to decide whether it is asic). NH + 4 is from a weak ase (NH 3 ) and therefore produces a strong acid. NH 4 NO 3 will formaa solution with low ph. () and (E) each has an anion from a weak acid and therefore produces a asic solution. (C) and (D) will form neutral solutions. 16. irst note that HC 8 H 7 O is an acid (with H in front) so the ionization constant is a. The question asks for p of its conjugate ase. irst we calculate the of the conjugate ase: = w/a Then we calculate p from x10 10 = = 8.13x10 1.3x10 10 p = log 8.13x10 = 9.09 _ +

3 17. irst calculate Q and compare to. [SO 3] (0.04) Q = = = 0.30 [SO ] [O ] (0.00) (0.008) Since Q smaller than, the equilirium will shift to the product side (to the right). 18. Let = (CH 3 ) 3 N Since is given, we write the ionization of as a ase. + H O H + + OH I C x +x +x E x x x x (0.040 x) = = 7.4x10 Assume x to e negligile. x = 7.4x x = 7.4x x = 7.4x = 1.7x10 ( sig. fig.) rom the ICE tale, we see x = [OH ]. 3 poh = log1.7x10 =.764 ( decimal places) ph = = 11.3 ( decimal places) 19. ( pts) A uffer contains equimolar concentration of NaNO and HNO. Write the net ionic equation to show how the uffer acts when HCl is added to it. irst of all, the active ingredients of this uffer are NO and HNO. Na + plays not part in the uffer action. HCl is a strong acid and therefore will dissociate 100% into H + and Cl. It is the H + that is essential to this reaction. H + would react with the conjugate ase NO. H + + NO HNO or H 3 O + + NO HNO + H O Either equation is acceptale. They mean exactly the same thing. Cl is also a spectator ion and should not e included. 0. Hydration of the Mn + refers to the cation ecoming surrounded y 6 H O: Mn H O Mn(H O) 6 1. Mn(H O) OH Mn(H O) (OH) + + H O. Question did not ask for C ut p. The expression must e written in terms of the partial pressures, not the molar concentrations: TiCl4 CO P = p p p Cl Proof validity of assumption: x = = 0.038

4 3. p = c (RT) n where n = 1 3 = p 3.8x10 3.8x10 c = = = = 7.7 = 8 n (RT) [(0.0806)(47)] This is the equation we are aiming for: NO (g) N (g) + O (g) c =? (1) ½ N (g) + ½ O (g) NO (g) c = 4.8x10 10 (This eqn needs to e reversed) () NO (g) ½ N (g) + ½ O (g) c = 1/(4.8x10 10 ) This eqn needs to e multiplied y w and c is squared. NO (g) N (g) + O (g) c = {1/(4.8x10 10 )} Now we add this equation to equation (). NO (g) N (g) + O (g) c = {1/(4.8x10 10 )} NO (g) NO (g) + O (g) c = 1.1x10 to give... NO (g) N (g) + O (g) c = 1/(4.8x10 10 )} x (1.1x10 ) = 4.8x (8 pts) In a 0.73 M solution, a weak monoprotic acid is 1.% ionized. Calculate a of the acid. Show your ICE tale and calculations. (Hint: Use HA as a generic formula for the acid.) HA + H O H 3 O + + A I C x + x +x E 0.73 x x x Since the acid is 1.% ionized, that means 1.% of 0.73 exists in the dissociated form at equilirium x = 1.% x 0.73 = 9.187x10 x (9.18 x10 ) 8.44 x10 ( x10 ) a = = = = 1.31x10 (3 sig.fig.) (0.73-x)

5 6. I (g) + r (g) Ir (g) c =.0x10 I C x x +x E x x x (x) c = =.0x10 (0.000 x) x is not negligile with that large, ut this equation can e simplified y finding the square root of oth sides. (x) (0.000 x) =.0x10 (x) = 1.8 (0.000 x) x = 1.8 (0.000 x) x = x x + 1.8x = x = x = = = Answer: [Ir] = x = (0.0444) = M Ir HNO 3 H SO 4 HClO 4 HCl Hr HI nitric acid sulfuric acid perchloric acid hydrochloric acid hydroromic acid hydroiodic acid