Exam Chemistry 4, Spring 005 March, 005 Part. Answer 7 of the following 8 multiple choice questions. If you answer more than 7 cross out the one you wish not to be graded, otherwise only the first 7 will be graded. 4 points each.. The value of K eq for the following reaction is 0.5: SO (g) + NO (g) SO 3 (g) + NO(g) The value of K eq at the same temperature for the reaction below is. SO (g) + NO (g) SO 3 (g) + NO(g) A) 0.50 B) 0.06 C) 0. D) 0.5 E) 6. Acetic acid is a wea acid that dissociates into the acetate ion and a proton in aqueous solution: HC H 3 O (aq) C H 3 O - (aq) + H + (aq) At equilibrium at 5 C a 0.00 M solution of acetic acid has the following concentrations: [HC H 3 O ] = 0.0990 M, [C H 3 O - ] =.33 0-3 M, and [H + ] =.33 0-3 M. The equilibrium constant, K eq, for the ionization of acetic acid at 5 C is. A) 5.7 0 4 B) 0.00 C).75 0-7 D).79 0-5 E) 5.7 0 6 3. Consider the following reaction at equilibrium. CO (g) CO(g) + O (g) H = -54 J Le Châtelier s Principle predicts that the partial pressure of CO(g) can be maximized by carrying out the reaction. A) at high temperature and high pressure B) at high temperature and low pressure C) at low temperature and low pressure D) at low temperature and high pressure E) in the presence of solid carbon
4. Which energy difference in the energy profile below corresponds to the activation energy for the forward reaction? Energy x y A) x B) y C) x + y D) x y E) y x Reaction pathway 5. Which one of the following will change the value of an equilibrium constant? A) changing temperature B) adding other substances that do not react with any of the species involved in the equilibrium C) varying the initial concentrations of reactants D) varying the initial concentrations of products 6. Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: NO NO + O In a particular experiment at 300 C, [NO ] drops from 0.000 to 0.00650 M in 00s. The rate of disappearance of NO for this period is M/s. A) 0.35 B) 3.5 0-3 C) 3.5 0-5 D) 7.0 0-3 E).8 0-3
7. The reaction CH 3 -N C CH 3 -C N is a first-order reaction. At 30.3 C, = 6.9 0-4 s -. If [CH 3 -N C] is.00 0-3 initially, [CH 3 -N C] is after.000 0 3 s. A) 5.33 0-4 B).34 0-4 C).88 0-3 D) 4.7 0-3 E).00 0-6 8. The effect of a catalyst on a chemical reaction is to. A) lower the activation energy B) accelerate the forward reaction only C) mae reactions more exothermic D) react with product, effectively removing it and shifting the equilibrium to the right Part. Answer 4 of the following 5 short answer questions. If you answer more than 4 cross out the one you wish not to be graded, otherwise only the first 4 will be graded. 6 points each. 9. In the following reaction label the two sets of conjugate acid-base pairs. To do this draw a line between both members of each pair and label each as an acid or a base: NH 3 + H O + NH 4 + OH - Base Acid Acid Base 0. Predict the products for the following reactions. Does the equilibrium position lie to the left or right for each? (a) O - + H O OH - Equilibrium lies to the right (O - is a strong base). Also do the "tug-of-war": O -..H +.. - OH, the oxide ion has the larger charge, so it will be the stronger base. However, OH - is the strongest base that can exist in aqueous solution. (b) HOAc + H O H 3 O + + OAc - Equilibrium lies to the left (HOAc is a wea acid) where HOAc is acetic acid (or CH 3 COOH).
. HF (K a = 6.8 0-4 ) is a stronger acid than HCN (K a = 6. 0-0 ). Which is the stronger conjugate base, F - or CN -? CN - is the conjugate base of the weaer acid HCN (relative to HF), and is therefore a stronger conjugate base (relative to F - ).. For the reaction NO + CO NO + CO the experimental rate law is rate = [NO ] The following mechanism has been proposed: NO NO 3 + NO NO 3 + CO NO + CO a. Which step is the rate determining step? Step (see rate law) b. Is there a reactive intermediate in this reaction? If so what is it? Yes. The reactive intermediate is NO 3 (produced in step and consumed in step ) 3. The rate law for a third order reaction is [A] - = t [A] o To mae a straight line plot to verify third order behavior, a. what would you plot on the vertical axis? /[A] b. what would you plot on the horizontal axis? t c. what would the slope equal?
Part 3. Answer all of the following questions. points each. 4. At 00 C, the equilibrium constant for the reaction below is.40 0 3. NO(g) N (g) + O (g) A closed vessel is charged with 36. atm of NO. At equilibrium, what is the partial pressure of O? NO(g) N (g) + O (g) Initial (atm) 36. 0 0 Changes -x +x +x Equilibrium (atm) 36. x x x K eq P = P N PNO x (36. x) Tae the square root of both sides Thus, the partial pressure of O x = 48.99 36.- x x = 768.5 97.98x 98.98x = 768.5 x = 7.9 O =.40 0 =.40x0 3 3 at equilibrium is7.9 atm.
5. Determine the rate law for the reaction NO (g) + Br (g) NOBr (g) from the following initial rate study. (Write the full rate law in the form: rate =?) [NO] (mol L - ) [Br ] (mol L - ) Rate (mol L - sec - ).00.00.00 0-6.00.00.00 0-6.00.00 4.00 0-6 Rate = [NO] x [Br ] y Order with respect to [NO]: By inspection we see that as we double the concentration of NO (experiments and ) while holding [Br ] constant, the initial rate of the reaction doubles therefore, we can say that the reaction is first-order with respect to [NO]. Order with respect to [Br ]: By inspection we see that as we double the concentration of Br (experiments and 3) while holding [NO] constant, the initial rate of the reaction increases by a factor of 4 therefore we can say that the reaction is second-order with respect to [Br ]. Rate = [NO][Br ] 6. For the first-order reaction N O 5 N O 4 + O, the activation energy is 06 J/mol. How many times faster will the reaction go at 00 C than at 5 C? ln T T T ln T E R = 8.34 0 T = 373. K, T = 98. K 06 J/mol = -3 8.34 0 J/mol K 373. K 98. K a ln E a = R T T = 06 J/mol 373. K 98. K = 5.38 0 J/mol K = 8.59 3-3 373. K times as fast 98. K
7. A reaction that is second order in reactant A has [A] o = 0.00 M. The half-life is 45.6 sec. What is [A] after 3.00 min?. Find the rate constant, : For a second-order reaction the half-life, t /, is related to as follows: / = = 0.00M 45.6s t o - = 0.09 M s. [A] after 3 minutes (80 s) is related to [A] o as follows (for a second-order process): 80s 80s 80s 80s = + t - = + (0.09 M s 0.00M = 4.6 M - = 0.0406 M o )(80s) TOTAL EXAM OUT OF 00 EXAM % 0.95 TOTAL HW OUT OF HW % 0.05 FINAL CORRECTED EXAM
R = 0.0806 L atm mol - K - = 8.34 J mol - K - T(O C) = 73.K K w =.0x0-4 x = -b ± b - 4ac a ln [A] t [A] = - t ln [A] t = - t + ln [A] o [A] t = [A] o o e - t - = t [A] t [A] o t ½ = -ln ½ = 0.693 t ½ = [A] o = A e -E a/rt ln = - E a R T + ln A ln T T = - E a R T - or ln T = E a T T R T - T