CHAPTER 6 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS 6. Changes in concentrations of reactants (or products) as functions of tie are easured to deterine the reaction rate. 6. Rate is proportional to concentration. An increase in pressure will increase the nuber of gas olecules per unit volue. In other words, the gas concentration increases due to increased pressure, so the reaction rate increases. Increased pressure also causes ore collisions between gas olecules. 6.3 The addition of ore water will dilute the concentrations of all solutes dissolved in the reaction vessel. If any of these solutes are reactants, the rate of the reaction will decrease. 6.4 An increase in solid surface area would allow ore gaseous coponents to react per unit tie and thus would increase the reaction rate. 6.5 An increase in teperature affects the rate of a reaction by increasing the nuber of collisions, but ore iportantly the energy of collisions increases. As the energy of collisions increases, ore collisions result in reaction (i.e., reactants products), so the rate of reaction increases. 6.6 The second experient proceeds at the higher rate. I in the gaseous state would experience ore collisions with gaseous H. 6.7 The reaction rate is the change in the concentration of reactants or products per unit tie. Reaction rates change with tie because reactant concentrations decrease, while product concentrations increase with tie. 6.8 a) For ost reactions, the rate of the reaction changes as a reaction progresses. The instantaneous rate is the rate at one point, or instant, during the reaction. The average rate is the average of the instantaneous rates over a period of tie. On a graph of reactant concentration versus tie of reaction, the instantaneous rate is the slope of the tangent to the curve at any one point. The average rate is the slope of the line connecting two points on the curve. The closer together the two points (shorter the tie interval), the ore closely the average rate agrees with the instantaneous rate. b) The initial rate is the instantaneous rate at the point on the graph where tie 0, that is when reactants are ixed. 6.9 The calculation of the overall rate is the difference between the forward and reverse rates. This coplication ay be avoided by easuring the initial rate, where product concentrations are negligible, so the reverse rate is negligible. Additionally, the calculations are siplified as the reactant concentrations can easily be deterined fro the volues and concentrations of the solutions ixed. 6.0 At tie t 0, no product has fored, so the B(g) curve ust start at the origin. Reactant concentration (A(g)) decreases with tie; product concentration (B(g)) increases with tie. Many correct graphs can be drawn. Two exaples are shown below. The graph on the left shows a reaction that proceeds nearly to copletion, i.e., [products] >> [reactants] at the end of the reaction. The graph on the right shows a reaction that does not proceed to copletion, i.e., [reactants] > [products] at reaction end. B(g) Concentration A(g) Concentration A(g) B(g) Tie 6- Tie
6. a) Calculate the slope of the line connecting (0, [C] o ) and (t f, [C] f ) (final tie and concentration). The negative of this slope is the average rate. b) Calculate the negative of the slope of the line tangent to the curve at t x. c) Calculate the negative of the slope of the line tangent to the curve at t 0. d) If you plotted [D] versus tie, you would not need to take the negative of the slopes in (a) (c) since [D] would increase over tie. 6. a) The average rate fro t 0 to t 0.0 s is proportional to the slope of the line connecting these two points: Δ[AX ] ( 0.0088 M 0.0500 M) Rate 0.0003 0.000 M/s Δt ( 0.0 s 0 s) The negative of the slope is used because rate is defined as the change in product concentration with tie. If a reactant is used, the rate is the negative of the change in reactant concentration. The / factor is included to account for the stoichioetric coefficient of for AX in the reaction. b) [AX ] vs tie 0.06 0.05 0.04 [AX ] 0.03 0.0 0.0 0 0 5 0 5 0 5 tie, s The slope of the tangent to the curve (dashed line) at t 0 is approxiately 0.004 M/s. This initial rate is greater than the average rate as calculated in part (a). The initial rate is greater than the average rate because rate decreases as reactant concentration decreases. ( ) Δ[AX ] 0.0088 M 0.049 M 6.3 a) Rate 6.70833 x 0 Δt ( 0.0 s 8.0 s) 4 6.7 x 0 4 ol/l s b) The rate at exactly 5.0 s will be higher than the rate in part (a). The slope of the tangent to the curve at t 5.0 s (the rate at 5.0 s) is approxiately.8 x 0 3 M/s. 6-
0.06 0.05 0.04 [AX ] 0.03 0.0 0.0 0 0 5 0 5 0 5 tie, s 6.4 Use Equation 6. to describe the rate of this reaction in ters of reactant disappearance and product appearance. [A] [B] [C] Rate Δ Δ Δ Δt Δt Δt A negative sign is used for the rate in ters of reactant A since A is reacting and [A] is decreasing over tie. Positive signs are used for the rate in ters of products B and C since B and C are being fored and [B] and [C] increase over tie. Reactant A decreases twice as fast as product C increases because olecules of A disappear for every olecule of C that appears. Δ[A] Δ[C] Δt Δt ol A/L i s ( ol C/L i s) ol C/L i s 4 ol/l s The negative value indicates that [A] is decreasing as the reaction progresses. The rate of reaction is always expressed as a positive nuber, so [A] is decreasing at a rate of 4 ol/l s. [D] [E] [F] 6.5 Rate Δ Δ Δ Δ t 3 Δ t 5 Δ t 5/ ol F/L i s ( 0.5 ol E/L i s) 3/ ol E/L i s 0.46667 0.4 ol/l s Δ[A] Δ[B] Δ[C] 6.6 Rate Δ t Δt Δt Rate is defined as the change in product concentration with tie. If a reactant is used, the rate is the negative of the change in reactant concentration. The / factor is included for reactant B to account for the stoichioetric coefficient of for B in the reaction. Δ[A] Δ[B] (0.50 ol/l) 0.5 ol/l s (unrounded) Δ t Δt s [B] decreases twice as fast as [A], so [A] is decreasing at a rate of / (0.5 ol/l s) or 0. ol/l s. 6-3
Δ[D] Δ[E] Δ[F] 6.7 Rate Δ [G] Δ[H] Δt 3 Δt Δ t Δt Δt [H] ΔΔ t Δ[D] ( 0.0 ol/l) 0.05 ol/l s Δt s 6.8 A ter with a negative sign is a reactant; a ter with a positive sign is a product. The inverse of the fraction becoes the coefficient of the olecule: N O 5 (g) 4 NO (g) + O (g) 6.9 CH 4 + O H O + CO 6.0 a) Rate b) Rate c) 0.0033 0.000 ol/l 0.00 0.00 s 0.0055 0.007 ol/l 4.00.00 s 6.7 x 0 4 ol/l s 8.0 x 0 4 ol/l s Initial Rate Δ y / Δ x [(0.0040 0.000) ol/l] / [4.00 0.00) s].5 x 0 3 ol/l s d) Rate at 7.00 s [(0.0030 0.0050) ol/l] / [.00 4.00) s].857 x 0 4.9 x 0 4 ol/l s e) Average between t 3 s and t 5 s is: Rate [(0.0050 0.0063) ol/l] / [5.00 3.00) s] 6.5 x 0 4 ol/l s Rate at 4 s 6.7 x 0 4 ol/l s thus the rates are equal at about 4 seconds. 6. Rate Δ[N ] t Δ Δ[H ] 3 Δt Δ[NH 3] Δt Δ[O ] Δ[O 3] 6. a) Rate 3 Δt Δt b) Use the ole ratio in the balanced equation: 5.7 x 0 ol O /L i s ol O 3 /L i s.45 x 0 5 ol/l s 3 ol O /L i s 6-4
6.3 a) k is the rate constant, the proportionality constant in the rate law. k represents the fraction of successful collisions which includes the fraction of collisions with sufficient energy and the fraction of collisions with correct orientation. k is a constant that varies with teperature. b) represents the order of the reaction with respect to [A] and n represents the order of the reaction with respect to [B]. The order is the exponent in the relationship between rate and reactant concentration and defines how reactant concentration influences rate. The order of a reactant does not necessarily equal its stoichioetric coefficient in the balanced equation. If a reaction is an eleentary reaction, eaning the reaction occurs in only one step, then the orders and stoichioetric coefficients are equal. However, if a reaction occurs in a series of eleentary reactions, called a echanis, then the rate law is based on the slowest eleentary reaction in the echanis. The orders of the reactants will equal the stoichioetric coefficients of the reactants in the slowest eleentary reaction but ay not equal the stoichioetric coefficients in the overall reaction. c) For the rate law rate k[a] [B] substitute in the units: Rate (ol/l in) k[a] [B] Rate ol/l i in ol/l i in k 3 [A] [B] ol ol ol 3 L L L 3 ol L k 3 Lin i ol k L /ol in 6.4 a) Plot either [A ] or [B ] versus tie and deterine the negative of the slope of the line tangent to the curve at t 0. b) A series of experients at constant teperature but with different initial concentrations are run to deterine different initial rates. By coparing results in which only the initial concentration of a single reactant is changed, the order of the reaction with respect to that reactant can be deterined. c) When the order of each reactant is known, any one experiental set of data (reactant concentration and reaction rate) can be used to deterine the reaction rate constant at that teperature. 6.5 a) The rate doubles. If rate k[a] and [A] is doubled, then the rate law becoes rate k[ x A]. The rate increases by or. b) The rate decreases by a factor of four. If rate k[/ x B], then rate decreases to (/) or /4 of its original value. c) The rate increases by a factor of nine. If rate k[3 x C], then rate increases to 3 or 9 ties its original value. 6.6 The order for each reactant is the exponent on the reactant concentration in the rate law. The orders with respect to [BrO 3 ] and to [Br ] are both. The order with respect to [H + ] is. The overall reaction order is the su of each reactant order: + + 4. + first order with respect to BrO 3, first order with respect to Br, second order with respect to H, fourth order overall 6.7 second order with respect to O 3 ( ) order with respect to O first order overall 6.8 a) The rate is first order with respect to [BrO 3 ]. If [BrO 3 ] is doubled, rate k[ x BrO 3 ], then rate increases to or ties its original value. The rate doubles. b) The rate is first order with respect to [Br ]. If [Br ] is halved, rate k[/ x Br ], then rate decreases by a factor of / or / ties its original value. The rate is halved. c) The rate is second order with respect to [H + ]. If [H + + ] is quadrupled, rate k[4 x H ], then rate increases to 4 or 6 ties its original value. 6-5
6.9 a) The rate increases by a factor of 4. b) The rate decreases by a factor of. c) The rate increases by a factor of. 6.30 The order for each reactant is the exponent on the reactant concentration in the rate law. The order with respect to [NO ] is, and the order with respect to [Cl ] is. The overall order is the su of the orders of individual reactants: + 3 for the overall order. 6.3 The order with respect to [HNO ] is 4, and the order with respect to [NO] is. The overall order is the su of the orders of the individual reactants 4 for the overall order. 6.3 a) The rate is second order with respect to [NO ]. If [NO ] is tripled, rate k[3 x NO ], then rate increases to 3 or 9 ties its original value. The rate increases by a factor of 9. b) The rate is second order with respect to [NO ] and first order with respect to [Cl ]. If [NO ] and [Cl ] are doubled, rate k[ x NO ] [ x Cl ], then the rate increases by a factor of x 8. c) The rate is first order with respect to [Cl ]. If Cl is halved, rate k[/ x Cl ], then rate decreases to ½ ties its original value. The rate is halved. 6.33 a) Doubling the HNO concentration changes the rate by a factor of [] 4 6. b) Doubling the NO concentration changes the rate by a factor of / [] / 4, thus, the rate decreases by a factor of 4. c) One half the HNO concentration changes the rate by a factor of [/] 4 / 6, thus the rate decreases by a factor of 6. 6.34 a) To find the order for reactant A, first identify the reaction experients in which [A] changes but [B] rate exp [A] exp is constant. Set up a proportionality:. Fill in the values given for rates and rate exp [A] exp concentrations and solve for, the order with respect to [A]. Repeat the process to find the order for reactant B. Use experients and (or 3 and 4 would work) to find the order with respect to [A]. rate exp [A] exp rate exp [A] exp 45.0 ol/l i in 0.300 ol/l 5.00 ol/l i in 0.00 ol/l 9.00 (3.00) log (9.00) log (3.00) Using experients 3 and 4 also gives nd order with respect to [A]. Use experients and 3 with [A] 0.00 M or and 4 with [A] 0.300 M to find order with respect to [B]. n rate exp [B] exp rate exp 3 [B] exp 3 0.0 ol/l i in 0.00 ol/l 5.00 ol/l i in 0.00 ol/l.00 (.00) n log (.00) n log (.00) n The reaction is first order with respect to [B]. b) The rate law, without a value for k, is rate k[a] [B]. c) Using experient to calculate k (the date fro any of the experients can be used): rate k [A] [B] 5.00 ol/liin 5.00 x 0 3 L /ol in [0.00 ol/l] [0.00 ol/l] n 6-6
6.35 a) The rate law is rate k [A] [B] n [C] p. Use experients and to find the order with respect to [A]. rate exp [A] exp rate exp [A] exp -.5 x 0 ol/l i in 0.000 ol/l -3 6.5 x 0 ol/l in 0.0500 ol/l i.00 (.00) log (.00) log (.00) The order is first order with respect to A. Use experients and 3 to find the order with respect to [B]. rate rate [B] [B] exp 3 exp 3 exp exp n n 5.00 x 0 ol/l i in 0.000 ol/l.5 x 0 ol/l in 0.0500 ol/l i 4.00 (.00) n log (4.00) n log (.00) n The reaction is second order with respect to B. Use experients and 4 to find the order with respect to [C]. p rate exp 4 [C] exp 4 rate exp [C] exp 3 p 6.5 x 0 ol/l i in 0.000 ol/l 3 6.5 x 0 ol/l in 0.000 ol/l i.00 (.00) p log (.00) p log (.00) p 0 The reaction is zero order with respect to C. b) Rate k [A] [B] [C] Rate k [A][B] 3 rate 6.5 x 0 ol/l i in c) k 50.0 L /ol s [A][B] [0.0500 ol/l][0.0500 ol/l] 6.36 a) A first order rate law follows the general expression, rate k[a]. The reaction rate is expressed as a change in concentration per unit tie with units of M/tie or ol/l tie. Since [A] has units of M (the brackets stand for concentration), then k has units of tie : Rate k[a] ol ol L i tie tie L b) Second order: Rate k[a] ol ol k L i tie L ol k L i tie L/ol tie or M tie ol L or ol tie L i tie ol L 6-7
c) Third order: rate k[a] 3 3 ol ol k L i tie L ol k L i tie L /ol tie or M tie 3 ol L 3 d) 5/ order: rate k[a] 5/ 5/ ol ol k L i tie L ol k L i tie L 3/ /ol 3/ tie or M 3/ tie 5/ ol L 5/ 6.37 a) zero order b) first order c) / order d) 7/ order 6.38 a) Rate k [CO] [Cl ] n Use experients and to find the order with respect to [CO]. rate exp [CO] exp rate exp [CO] exp 9.9 x 0 ol/liin.00 ol/l 30.33 x 0 ol/l in 0.00 ol/l i 9.699 (0.0) log (9.699) log (0.0) 0.9867 The reaction is first order with respect to [CO]. Use experients and 3 to find order with respect to [Cl ]. n rate exp 3 [Cl ] exp 3 rate exp [Cl ] exp 9 n.30 x 0 ol/l i in.00 ol/l 30.33 x 0 ol/l in 0.00 ol/l i 9.774 (0.0) n log (9.774) n log (0.0) n 0.990 The reaction is first order with respect to [Cl ]. Rate k [CO][Cl ] b) k Rate / [CO][Cl ] Exp : k (.9 x 0 9 ol/l s) / [.00 ol/l][0.00 ol/l].9 x 0 8 L/ol s Exp : k (.33 x 0 30 ol/l s) / [0.00 ol/l][0.00 ol/l].33 x 0 8 L/ol s Exp 3: k 3 (.30 x 0 9 ol/l s) / [0.00 ol/l][.00 ol/l].30 x 0 8 L/ol s Exp 4: k 4 (.3 x 0 3 ol/l s) / [0.00 ol/l][0.000 ol/l].3 x 0 8 L/ol s k avg (.9 x 0 8 +.33 x 0 8 +.30 x 0 8 +.3 x 0 8 ) L/ol s / 4.3 x 0 8 L/ol s 6-8
6.39 The integrated rate law can be used to plot a graph. If the plot of [reactant] versus tie is linear, the order is zero. If the plot of ln [reactant] versus tie is linear, the order is first. If the plot of inverse concentration (/[reactant]) versus tie is linear, the order is second. a) The reaction is first order since ln[reactant] versus tie is linear. b) The reaction is second order since / [reactant] versus tie is linear. c) The reaction is zero order since [reactant] versus tie is linear. 6.40 The half-life (t / ) of a reaction is the tie required to reach half the initial reactant concentration. For a first-order process, no olecular collisions are necessary, and the rate depends only on the fraction of the olecules having sufficient energy to initiate the reaction. 6.4 The rate expression indicates that this reaction is second order overall (the order of [AB] is ), so use the second order integrated rate law to find tie. ([AB] /3 [AB] 0 /3 (.50 M) 0.500 M) kt AB [ AB ] t [ ] 0 [ AB] [ AB] t t t k 0.500 M.50 M t 0. L/ol i s t 6.6667 7 s 6.4 [ AB ] kt t [ AB] 0 [ AB ] kt + t [ AB] 0 [ AB ] (0. L/ol s) (0.0 s) + t.50 M [ AB ].66667 t M [AB] t 0.375 0.4 M 6.43 a) The given inforation is the aount that has reacted in a specified aount of tie. With this inforation, the integrated rate law ust be used to find a value for the rate constant. For a first order reaction, the rate law is ln [A] t ln [A] 0 kt. Using the fact that 50% has decoposed, let [A] 0 M and then [A] t 50% of M 0.5 M: ln [A] t ln [A] 0 kt ln [0.5] ln [] k(0.5 in) 0.69347 0 k(0.5 in) 0.69347 k(0.5 in) k 0.0660 in Alternatively, 50.0% decoposition eans that one half-life has passed. Thus, the first order half-life equation ay be used: ln ln ln t / k 0.06604 0.0660 in t 0.5 in k / 6-9
b) Use the value for k calculated in part a. Let [A] 0 M; since 75% of A has decoposed, 5% of A reains and [A] t 5% of [A] 0 or 0.5{A] 0 0.5[] 0.5 M ln [A] t ln [A] 0 kt ln[a] t ln[a] 0 t k ln[0.5] ln[] t -0.0660 in t.0045.0 in If you recognize that 75.0% decoposition eans that two half-lives have passed, then t (0.5 in).0 in. 6.44 a) t / ln / k ln / (0.00 yr ) 577.6 5.8 x 0 yr b) ln ([A] 0 / [A] t ) kt [A] o 00% [A] t.5% k 0.00 yr t ln 00%.5% ) t t 73.86795.7 x 0 3 yr If the student recognizes that.5% reaining corresponds to three half-lives; then siply ultiply the answer in part (a) by three. 6.45 In a first order reaction, ln[nh 3 ] versus tie is a straight line with slope equal to k. A new data table is constructed below. (Note that additional significant figures are retained in the calculations.) x-axis (tie, s) [NH 3 ] y-axis (ln[nh 3 ]) 0 4.000 M.3869.000 3.986 M.3879.000 3.974 M.37977.388.386 ln[nh 3 ].384.38.38.378 0 0.5.5 Tie, s k slope rise/run (y y ) / (x x ) k (.37977.3869) / (.000 0) (0.0065) / (.) 3.60 x 0 3 s 3 x 0 3 s (Note that the starting tie is not exact, and hence, liits the significant figures.) b) t / ln / k ln / (3.60 x 0 3 s ).6 x 0 s E a /RT 6.46 k Ae The Arrhenius equation indicates a negative exponential relationship between teperatures and the rate constant, k. In other words, the rate constant increases exponentially with teperature. 6-0
6.47 The Arrhenius equation, k Ae E /RT a, can be used directly to solve for activation energy at a specified teperature if the rate constant, k, and the frequency factor, A, are known. However, the frequency factor is usually not known. To find E a without knowing A, rearrange the Arrhenius equation to put it in the for of a linear plot: ln k ln A E a / RT where the y value is ln k and the x value is / T. Measure the rate constant at a series of teperatures and plot ln k versus / T. The slope equals E a / R. 6.48 a) The value of k increases exponentially with teperature. b) A plot of ln k versus / T is a straight line whose slope is E a / R. (a) (b) 6.49 Substitute the given values into the Arrhenius equation and solve for k. k 4.7 x 0 3 s T 5 C (73 + 5) 98 K k? T 75 C (73 + 75) 348 K E a 33.6 kj/ol 33600 J/ol ln k Ea k R T T k ln 4.7 x 0 s 3 k ln 4.7 x 0 s 3 33600 J/ ol 8.34 J/ol i K 348 K 98 K.94855 (unrounded) Raise each side to e x 6-
k 7.08577 3 4.7 x 0 s k (4.7 x 0 3 s ) (7.08577) 0.039858 0.033 s 6.50 Substitute the given values into the Arrhenius equation and solve for E a. k 4.50 x 0 5 L/ol s T 95 C (73 + 95) 468 K k 3.0 x 0 3 L/ol s T 58 C (73 + 58) 53 K E a? ln k Ea k R T T 3 k J 3.0 x 0 L/ol i s Rln 8.34 ln 5 E a k ol K 4.50 x 0 L/ol s i T T 53 K 468 K E a.3984658 x 0 5 J/ol.40 x 0 5 J/ol 6.5 Substitute the given values into the Arrhenius equation and solve for E a. k 0.76/s T 77 C (73 + 77) 000. K k 0.87/s T 757 C (73 + 757) 030. K E a? ln k Ea k R T T k J 0.87/ s Rln E a k 8.34 ln ol K i 0.76 / s T T 030. K 000. K E a 3.8585 x 0 4 J/ol 3.9 x 0 4 J/ol 6.5 The central idea of collision theory is that reactants ust collide with each other in order to react. If reactants ust collide to react, the rate depends on the product of the reactant concentrations. 6.53 No, collision frequency is not the only factor affecting reaction rate. The collision frequency is a count of the total nuber of collisions between reactant olecules. Only a sall nuber of these collisions lead to a reaction. Other factors that influence the fraction of collisions that lead to reaction are the energy and orientation of the collision. A collision ust occur with a iniu energy (activation energy) to be successful. In a collision, the orientation, that is, which ends of the reactant olecules collide, ust bring the reacting atos in the olecules together in order for the collision to lead to a reaction. 6.54 At any particular teperature, olecules have a distribution of kinetic energies, as will their collisions have a range of energies. As teperature increases, the fraction of these collisions which exceed the threshold energy, increases; thus, the reaction rate increases. 6.55 a) rate increases b) rate increases 6.56 No, for 4 x 0 5 oles of EF to for, every collision ust result in a reaction and no EF olecule can decopose back to AB and CD. Neither condition is likely. In principle, all reactions are reversible, so soe EF olecules decopose. Even if all AB and CD olecules did cobine, the reverse decoposition rate would result in an aount of EF that is less than 4 x 0 5 oles. 6-
6.57 Collision frequency is proportional to the velocity of the reactant olecules. At the sae teperature, both reaction ixtures have the sae average kinetic energy, but not the sae velocity. Kinetic energy equals / v, where is ass and v velocity. The ethylaine (N(CH 3 ) 3 ) olecule has a greater ass than the aonia olecule, so ethylaine olecules will collide less often than aonia olecules, because of their slower velocities. Collision energy thus is less for the N(CH 3 ) 3 (g) + HCl(g) reaction than for the NH 3 (g) + HCl(g) reaction. Therefore, the rate of the reaction between aonia and hydrogen chloride is greater than the rate of the reaction between ethylaine and hydrogen chloride. The fraction of successful collisions also differs between the two reactions. In both reactions the hydrogen fro HCl is bonding to the nitrogen in NH 3 or N(CH 3 ) 3. The difference between the reactions is in how easily the H can collide with the N, the correct orientation for a successful reaction. The groups (H) bonded to nitrogen in aonia are less bulky than the groups bonded to nitrogen in triethylaine (CH 3 ). So, collisions with correct orientation between HCl and NH 3 occur ore frequently than between HCl and N(CH 3 ) 3 and NH 3 (g) + HCl(g) NH 4 Cl(s) occurs at a higher rate than N(CH 3 ) 3 (g) + HCl(g) (CH 3 ) 3 NHCl(s). Therefore, the rate of the reaction between aonia and hydrogen chloride is greater than the rate of the reaction between ethylaine and hydrogen chloride. 6.58 Each A particle can collide with three B particles, so (4 x 3) unique collisions are possible. 6.59 [(.0 ol A) (6.0 x 0 3 A/ol A)] x [(. ol B) (6.0 x 0 3 B/ol B)] 7.76495 x 0 47 7.76 x 0 47 unique collisions 6.60 The fraction of collisions with a specified energy is equal to the e Ea/RT ter in the Arrhenius equation. f e Ea/RT (5 C 73 + 5 98 K) 3 00 kj/ ol 0 J E a / RT ( 8.34 J/ ol i K)( 98 K) kj E a / RT 40.36096 Fraction e Ea/RT e 40.36096.9577689 x 0 8.96 x 0 8 6.6 The fraction of collisions with a specified energy is equal to the e Ea/RT ter in the Arrhenius equation. 3 f e Ea/RT 00 kj/ ol 0 J (50. C 73 + 50. 33 K) E a / RT ( 8.34 J/ol K)( 33 K) kj E a / RT 37.38095 Fraction e Ea/RT e 37.38095 6.753 x 0 7 The fraction increased by (6.753 x 0 7 ) / (.9577689 x 0 8 ).73775.7 6.6 a) The reaction is exotheric, so the energy of the reactants is higher than the energy of the products. E a (fwd) Energy E a (rev) ABC + D ΔH rxn Reaction coordinate AB + CD b) E a(rev) E a(fwd) ΔH rxn 5 kj/ol ( 55 kj/ol).70 x 0 kj/ol 6-3
c) bond foring A B C D bond weakening 6.63 a) b) ΔH rxn E a(fwd) E a(rev) 5 kj/ol 85 kj/ol 40 kj/ol c) A..... B A..... B 6.64 a) The reaction is endotheric since the enthalpy change is positive. E a (rev) NOCl + Cl Energy NO + Cl ΔΗ +83 kj E a (fwd) +86 kj Reaction coordinate b) Activation energy for the reverse reaction: E a (rev) E a (fwd) ΔH 86 kj 83 kj 3 kj. 6-4
c) To draw the transition state, look at structures of reactants and products: Cl Cl + N O Cl + Cl N O The collision ust occur between one of the chlorines and the nitrogen. The transition state would have weak bonds between the nitrogen and chlorine and between the two chlorines. Cl Cl N O 6.65 The rate of an overall reaction depends on the slowest step. Each individual step s reaction rate can vary widely, so the rate of the slowest step, and hence the overall reaction, will be slower than the average of the individual rates because the average contains faster rates as well as the rate-deterining step. 6.66 An eleentary step is a single olecular event, such as the collision of two olecules. Since an eleentary step occurs in one-step, its rate ust be proportional to the product of the reactant concentrations. Thus, the exponents in the rate of an eleentary step are identical to the coefficients in the equation for the step. Since an overall reaction is generally a series of eleentary steps, it is not necessarily proportional to the product of the overall reactant concentrations. 6.67 Yes, it is often possible to devise ore than one echanis since the rate law for the slowest step deterines the rate law for the overall reaction. The preferred echanis will be the one that sees ost probable, where olecules behave in their expected fashion. 6.68 Reaction interediates have soe stability, however liited, but transition states are inherently unstable. Additionally, unlike transition states, interediates are olecules with noral bonds. 6.69 A biolecular step (a collision between two particles) is ore reasonable physically than a terolecular step (a collision involving three particles) because the likelihood that two reactant olecules will collide with the proper energy and orientation is uch greater than the likelihood that three reactant olecules will collide siultaneously with the proper energy and orientation. 6.70 No, the overall rate law ust contain reactants only (no interediates) and is deterined by the slow step. If the first step in a reaction echanis is slow, the rate law for that step is the overall rate law. 6.7 If the slow step is not the first one, the faster preceding step produces interediates that accuulate before being consued in the slow step. Substitution of the interediates into the rate law for the slow step will produce the overall rate law. 6.7 a) The overall reaction can be obtained by adding the three steps together: () A(g) + B(g) X(g) fast () X(g) + C(g) Y(g) slow (3) Y(g) D(g) fast Total: A(g) + B(g) + X(g) + C(g) + Y(g) X(g) + Y(g) + D(g) A(g) + B(g) + C(g) D(g) b) Interediates appear in the echanis first as products, then as reactants. Both X and Y are interediates in the given echanis. Interediate X is produced in the first step and consued in the second step; interediate Y is produced in the second step and consued in the third step. Notice that neither X nor Y were included in the overall reaction. c) Step: Molecularity Rate law A(g) + B(g) X(g) biolecular rate k [A][B] X(g) + C(g) Y(g) biolecular rate k [X][C] Y(g) D(g) uniolecular rate 3 k 3 [Y] 6-5
d) Yes, the echanis is consistent with the actual rate law. The slow step in the echanis is the second step with rate law: rate k [X][C]. Since X is an interediate, it ust be replaced by using the first step. For an equilibriu, rate forward rxn rate reverse rxn. For step then, k [A][B] k [X]. Rearranging to solve for [X] gives [X] (k / k )[A][B]. Substituting this value for [X] into the rate law for the second step gives the overall rate law as rate (k k / k )[A][B][C] which is identical to the actual rate law with k k k / k. e) Yes, The one step echanis A(g) + B(g) + C(g) D(g) would have a rate law of rate k[a][b][c], which is the actual rate law. 6.73 a) () ClO (aq) + H O(l) HClO(aq) + OH (aq) fast () I (aq) + HClO(aq) HIO(aq) + Cl (aq) slow (3) OH (aq) + HIO(aq) H O(l) + IO (aq) fast (overall) ClO (aq) + I (aq) Cl (aq) + IO (aq) b) HClO(aq), OH (aq), and HIO(aq) c) () Biolecular; Rate k [ClO ] [H O] () Biolecular; Rate k [I ][HClO] (3) Biolecular; Rate 3 k 3 [OH ][HIO] d) For the slow step: Rate k[i ][HClO] However, HClO is an interediate, and should be replaced. Fro step (), leaving out the water, [HClO] [ClO ] / [OH ]. Replacing [HClO] in the slow step rate law gives: Rate k[i ][ClO ] / [OH ]. This is not the observed rate law. The echanis is not consistent with the actual rate law. 6.74 Nitrosyl broide is NOBr(g). The reactions su to the equation NO(g) + Br (g) NOBr(g), so criterion (eleentary steps ust add to overall equation) is satisfied. Both eleentary steps are biolecular and cheically reasonable, so criterion (steps are physically reasonable) is et. The reaction rate is deterined by the slow step; however, rate expressions do not include reaction interediates (NOBr ). Derive the rate law. The slow step in the echanis is the second step with rate law: rate k [NOBr ][NO]. Since NOBr is an interediate, it ust be replaced by using the first step. For an equilibriu like Step, rate forward rxn rate reverse rxn. Solve for [NOBr ] in Step : Rate (forward) Rate (reverse) k [NO][Br ] k [NOBr ] [NOBr ] (k / k )[NO][Br ] Rate of the slow step: Rate k [NOBr ][NO] Substitute the expression for [NOBr ] into this equation, the slow step: Rate k (k / k )[NO][Br ][NO] Cobine the separate constants into one constant: k k (k / k ) Rate k[no] [Br ] The derived rate law equals the known rate law, so criterion 3 is satisfied. The proposed echanis is valid. 6.75 I. NO(g) + O (g) NO (g) Rate k[no] [O ] II. a) NO(g) N O (g) b) N O (g) + O (g) NO (g) slow Rate k[n O ][O ] k b (k a / k a ) [NO] [O ] k[no] [O ] 6-6
III. a) NO(g) N (g) + O (g) b) N (g) + O (g) NO (g) slow Rate k[n ][O ] k b (k a / k a ) [NO] [O ] [O ] k[no] [O ] a) All the echaniss are consistent with the rate law. b) The ost reasonable echanis is II, since none of its eleentary steps are ore coplicated than being biolecular. 6.76 a) Gold is a heterogeneous catalyst. b) 6.77 No, a catalyst changes the echanis of a reaction to one with lower activation energy. Lower activation energy eans a faster reaction. An increase in teperature does not influence the activation energy, but instead increases the fraction of collisions with sufficient energy to react. 6.78 a) No, by definition, a catalyst is a substance that increases reaction rate without being consued. The spark provides energy that is absorbed by the H and O olecules to achieve the threshold energy needed for reaction. b) Yes, the powdered etal acts like a heterogeneous catalyst, providing a surface upon which the reaction between O and H becoes ore favorable because the activation energy is lowered. 6.79 Catalysts decrease the aount of energy required for a reaction. To carry out the reaction less energy ust be generated. The generation of less energy eans that fewer by-products of energy production will be released into the environent. 6.80 a) Rate k [NO ] b) Rate k [NO] [O 3 ] c) The concentration of NO is one-half of the original when the reaction is 50% coplete. Rate k i [ NO ] Reaction : 4 Rate50% k 0.5( NO ) Rate k i [ NO] Reaction : O3 4 Rate50% k 0.5( NO) 0.5( O3 ) 6-7
d) [NO ] i [CO] i, when the reaction is 50% coplete, [CO] / [CO] i and [NO ] 0.75[NO ] i. [ NO ] ( ) Rate k i.7778.8 Rate50% k 0.75 NO e) [NO] i [O 3 ] i, when the reaction is 50% coplete, [O 3 ] / [O 3 ] i and [NO] 0.75[NO] i. Rate k i [ NO] O3.6667.7 Rate50% k 0.75( NO) 0.5( O3 ) 6.8 a) (two peaks) b) the second step (higher peak) c) exotheric (end below beginning) 6.8 a) Water does not appear as a reactant in the rate-deterining step. Note that as a solvent in the reaction, the concentration of the water is assued not to change even though soe water is used up as a reactant. This assuption is valid as long as the solute concentrations are low ( M or less). So, even if water did appear as a reactant in the rate-deterining step, it would not appear in the rate law. See rate law for step below. b) Rate law for step (): rate k [(CH 3 ) 3 CBr] Rate law for step (): rate k [(CH 3 ) 3 C + ] Rate law for step (3): rate 3 k 3 [(CH 3 ) 3 COH + ] c) The interediates are (CH 3 ) 3 C + and (CH 3 ) 3 COH +. d) The rate-deterining step is the slow step, (). The rate law for this step agrees with the actual rate law with k k. 6.83 a) Add the two equations: () X(g) + O 3 (g) XO(g) + O (g) () XO(g) + O(g) X(g) + O (g) Overall O 3 (g) + O(g) O (g) b) Step () rate k [X][O 3 ] Step () rate k [XO][O] c) X acts as a catalyst, and XO acts as an interediate. d) Step () is the rate-deterining (slow) step, so use its rate law with NO X. rate k [NO][O 3 ] (6 x 0 5 c 3 /olecule s) (.0 x 0 9 olecule/c 3 ) (5 x 0 olecule/c 3 ) 3 x 0 7 olecule/s 6.84 Rearrange t / ln / k to k ln / t / k ln / 5730 yr.0968 x 0 4 yr (unrounded) Use the first-order integrated rate law: 4 C t ln kt 4 C 0 [ 5.5% ] [ ] t ln 00% 0 (.0968 x 0 4 yr ) t t.5476 x 0 4.54 x 0 4 yr 6.85 The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k and k, are not expressly stated, the relative ties give an idea of the rate. The reaction rate is proportional to the rate constant. At T 0 C (73 + 0) 93 K, the rate of reaction is apple/4 days while at T 0 C (73 + 0) 73 K, the rate is apple/6 days. Therefore, rate apple/4 days and rate apple/6 days are substituted for k and k, respectively. k /4 T 93 K k /6 T 73 K E a? 6-8
ln k k Ea R T T ( ) k J /6 Rln 8.34 ln E a k ol K i ( / 4) T T 73 K 93 K E a 4.609666 x 0 4 J/ol 4.6 x 0 4 J/ol The significant figures are based on the Kelvin teperatures. 6.86 Rearrange t / ln / k to k ln / t / k ln / 9.8 x 0 3 d 7.0793 x 0 5 d (unrounded) Use the first-order integrated rate law with BP benzoyl peroxide: [ BP] ln kt BP 6.87 a) [ ] t 0 [ 95% ] [ ] t 0 ln (7.0793 x 0 5 d ) t 00% t 75.057 7.3 x 0 d Energy CO + NO ΔΗ -6kJ CO + NO Reaction coordinate b) Yes, the alternative echanis is consistent with the rate law since the rate of the echanis is based on the slowest step, NO (g) N (g) + O (g). The rate law for this step is rate k [NO ], the sae as the actual rate law. The alternative echanis includes an eleentary reaction (step ) that is a terolecular reaction. Thus, the original echanis given in the text is ore reasonable physically since it involves only biolecular reactions. 6.88 a) The general rate law is: Rate k[a] x [B] y [C] z Rate Rate k k [ A] [ B] [ C] [ A] [ B] [ C] Rate Rate x y z x y z [B] [B] [C] [C] x k [ A] x k A [ ] 5 9.6 x 0 ol/l i s k 6 6.0 x 0 ol/ L i s k 6 4 x x [ 0.096 ol/l] [ 0.04 ol/l] x x 6-9
Rate Rate Rate Rate 4 4 [ A] [ B] [ C] [ A] [ B] [ C] k k [C] [C] 4 k A B k A B y z y z 4 4 4 [ ] [ ] [ ] [ ] y y 4 4 5 9.6 x 0 ol/l i s k 6.5 x 0 ol/l i s k 64 (8) (0.5) y y 0 Rate k A B C Rate3 k A B C [A] [A] 3 0 z Rate k [ B] [ C] 0 z Rate3 k B C [ ] [ ] [ ] [ ] [ ] [ ] 0 z 0 z 3 3 3 [ ] [ ] 3 3 6 6.0 x 0 ol/l s k 5.5 x 0 ol/l s k [ ] [ [ ] [ ] ] y y 4 4 0.096 ol/l 0.085 ol/l 0.0 ol/l 0.70 ol/l [ 0.085 ol/l] [ 0.03 ol/l] [ 0.034 ol/l] [ 0.080 ol/l] 0 z 0 z 3 3 Since y 0, the B ter ay be ignored, it is only shown here for copleteness. 0.4 0.4 z z b) You can use any trial to calculate k, with the rate law: Rate k[a] [B] 0 [C] k[a] [C] Using Experient : k Rate / [A] [C] (6.0 x 0 6 ol/l s) / [(0.04 ol/l) (0.03 ol/l)] 0.355 0.33 L /ol s This value will need to be divided by the coefficient of the substance to which the initial rate refers. If the initial rate refers to the disappearance of A or B then the constant (k') is: k' k / 0.355 / 0.676 0.6 L /ol s If the initial rate refers to the disappearance of C or the foration of D then the constant (k') is: k' k 0.33 L /ol s If the initial rate refers to the foration of E then the constant (k') is: k' k / 3 0.355 / 3 0.08507 0. L /ol s c) Rate k[a] [C] (substitute the appropriate k value fro part (b).) d) Rate [ A] Δ Δt [ B] Δ Δt [ C] Δ Δ t [ D] Δ Δt [ E] Δ 3 Δt 6.89 Use the given rate law, and enter the given values: rate k[h + ] [sucrose] [H + ] i 0.0 M [sucrose] i.0 M The glucose and fructose are not in the rate law, so they ay be ignored. a) The rate is first-order with respect to [sucrose]. The [sucrose] is changed fro.0 M to.5 M, or is increased by a factor of.5/.0 or.5. Then the rate k[h + ][.5 x sucrose]; the rate increases by a factor of.5. b) The [sucrose] is changed fro.0 M to 0.5 M, or is decreased by a factor of 0.5/.0 or 0.5. Then the rate k[h + ][0.5 x sucrose]; the rate decreases by a factor of ½ or half the original rate. c) The rate is first-order with respect to [H + ]. The [H + ] is changed fro 0.0 M to 0.000 M, or is decreased by a factor of 0.000/0.0 or 0.0. Then the rate k[0.0 x H + ][sucrose]; the rate decreases by a factor of 0.0. Thus, the reaction will decrease to /00 the original. 6-0
d) [sucrose] decreases fro.0 M to 0. M, or by a factor of (0. M /.0 M) 0.. [H + ] increases fro 0.0 M to 0. M, or by a factor of (0. M / 0.0 M) 0. Then the rate will increase by k[0 x H + ][0. x sucrose].0 ties as fast. Thus, there will be no change. 6.90 a) Enzyes stabilize the transition state to a rearkable degree and thus greatly increase the reaction rate. b) Enzyes are extreely specific and have the ability to change shape to adopt a perfect fit with the substrate. 6.9 A + B Products. Assue the reaction is first order with respect to A and first order with respect to B. Rate k[a][b] Mixture I: 0.00 ol A ( 6 spheres A) sphere Concentration of A 0. M 0.50 L Concentration of B ( 5 spheres B) 0.00 ol B sphere 0.0 M 0.50 L Use the rate law to find the value of k, the rate constant: Rate k[a][b] k Rate -4 [A][B] 8.3 x 0 ol/l i in 0.06967 L/ol in [0. ol/l][0.0 ol/l] Use this value of k to find the initial rate in Mixture II: 0.00 ol A ( 7 spheres A) sphere Concentration of A 0.4 M 0.50 L Concentration of B ( 8 spheres B) 0.00 ol B sphere 0.6 M 0.50 L Rate k[a][b] Rate 0.06967 L/ol in[0.4 ol/l][0.6 ol/l] Rate.5493 x 0 3.5 x 0 3 ol/l in 6.9 The overall order is equal to the su of the individual orders, i.e., the su of the exponents equals : + 4 + + +, so. The reaction is second order with respect to NAD. 6.93 k Ae E a /RT and k Ae E a /RT k Ae k Ae ln k E k Ea RT Ea RT a E RT ( E a - Ea ) e RT a 4.3 x 0 (8.34 J/ol K) ((73 + 37)K) ln E a E a 8.530 x 0 4 8.5 x 0 4 J/ol RT ln k k E a E a E a E a 6-
6.94 Rate is proportional to the rate constant, so if the rate constant increases by a certain factor, the rate increases by the sae factor. Thus, to calculate the change in rate the Arrhenius equation can be used and substitute rate cat /rate uncat k cat /k uncat. k Ae E a /RT Ea RT k Ae ( E a - Ea ) e E k a RT Ae RT ln k 3 Ea Ea 5kJ/ol 0 J k RT J.93998 (unrounded) kj 8.34 (( 73 + 37) K) ol K i k 6.9586 7 k The rate of the enzye-catalyzed reaction occurs at a rate 7 ties faster than the rate of the uncatalyzed reaction at 37 C. 6.95 Initially, the slow step in the echanis gives: Rate k [CHCl 3 ][Cl] However, Cl is an interediate, and should not be in the final answer. Step () shows the relationship between Cl and Cl. [Cl] (k /k ) [Cl ] [Cl] (k /k ) / [Cl ] / Substituting: Rate k (k /k ) / [CHCl 3 ][Cl ] / Cobining k s: Rate k[chcl 3 ][Cl ] / 6.96 First, find the rate constant, k, for the reaction by solving the first order half-life equation for k. Then use the firstorder integrated rate law expression to find t, the tie for decay. ln ln Rearrange t / to k k t / ln k yr 5.776 x 0 yr (unrounded) Use the first-order integrated rate law: ln [A] t ln [A] 0 kt ln[a] t ln[a] 0 t k ln[0. ppb] ln[75 ppb] - t 5.776 x 0 yr t 57.3765798 57 yr 6.97 Rearrange t / ln / k to k ln / t / k ln / 3.5 in 0.9804 in (unrounded) The proble states that the interval t /k: t / (0.9804 in ) 5.04948 5.0 in 6-
6.98 The rate law is: Rate k[a] [B] k is needed to finish the proble: 0.0 ol/l i s k Rate / [A] [B] [.0 ol/l ] [.0 ol/l ] 0.0 L /ol s Using the k just calculated with the rate law: Rate (0.0 L /ol s) [.0ol/L] [3.0ol/L].4 ol/l s 6.99 a) The rate constant can be deterined fro the slope of the integrated rate law plot. To find the correct order, the data should be plotted as ) [sucrose] versus tie linear for zero order, ) ln[sucrose] versus tie linear for first order, and 3) /[sucrose] versus tie linear for second order. All three graphs are linear, so picking the correct order is difficult. One way to select the order is to copare correlation coefficients (R ) you ay or ay not have experience with this. The best correlation coefficient is the one closest to a value of.00. Based on this selection criterion, the plot of ln[sucrose] vs. tie for the first order reaction is the best. Another ethod when linearity is not obvious fro the graphs is to exaine the reaction and decide which order fits the reaction. For the reaction of one olecule of sucrose with one olecule of liquid water, the rate law would ost likely include sucrose with an order of one and would not include water. The plot for a first order reaction is described by the equation ln[a] t kt + ln[a] 0. The slope of the plot of ln[sucrose] versus t equals k. The equation for the straight line in the first order plot is y 0.x 0.6936. So, k ( 0. h ) 0. h. Half-life for a first-order reaction equals ln /k, so t / ln / 0. h 3.3007 3.3 h. Integrated rate law plots 4 3 y 0.5897x +.93 R 0.9935 (see legend) 0 - - y -0.077x + 0.4896 R 0.9875 0 0.5.5.5 3 3.5 y -0.x - 0.6936 R 0.9998 tie, h Legend: y axis is [sucrose] axis is ln[sucrose] y axis is /[sucrose] b) If 75% of the sucrose has been reacted, 5% of the sucrose reains. The tie to hydrolyze 75% of the sucrose can be calculated by substituting.0 M for [A] 0 and 0.5 M for [A] t in the integrated rate law equation: ln[a] t ( 0. h )t + ln[a] 0. ln[a] t ln[a] 0 ( 0.h )t ln[0.5] ln[.0] ( 0. h )t t 6.604 6.6 h 6-3
c) The reaction ight be second-order overall with first-order in sucrose and first-order in water. If the concentration of sucrose is relatively low, the concentration of water reains constant even with sall changes in the aount of water. This gives an apparent zero-order reaction with respect to water. Thus, the reaction appears to be first order overall because the rate does not change with changes in the aount of water. 6.00 a) False, at any particular teperature, olecules have a range of kinetic energies. b) False, at reduced pressure, the nuber of collisions per unit tie is reduced, as is the reaction rate. c) True d) False, the increase in rate depends on the activation energy for the reaction. Also, biological catalysts (enzyes) ay decopose on heating, reducing their effectiveness. e) False, they also ust have the correct orientation. f) False, the activation energy is unique to the echanis of a particular reaction. g) False, since ost reaction rates depend to soe extent on the reactant concentrations, as these decrease during the course of the reaction, the reaction rate also decreases. h) False, see part f. i) False, a catalyst speeds up the reaction by lowering the activation energy. j) False, the speed of a reaction (kinetics) is separate fro the stability of the products (therodynaics). k) False, the frequency factor, A, is the product of the collision frequency which is affected by teperature and an orientation probability factor. l) True ) False, the catalyst changes the activation energy, not ΔH of reaction. n) True o) True p) False, biolecular and uniolecular refer to the olecularity or the nuber of reactant particles involved in the reaction step. There is no direct relationship to the speed of the reaction. q) False, olecularity and olecular coplexity are not related. 6.0 a) To find concentration at a later tie, given starting concentration and the rate constant, use an integrated rate law expression. Since the units on k are s -, this is a first order reaction. Use the first-order integrated rate law: ln [N O 5 ] t ln [N O 5 ] 0 kt ln [N O 5 ] t ln[.58 ol/l] (.8 x 0 3 s 60 s ) ( 5.00 in) in ln [N O 5 ] t 0.38578 [N O 5 ] t 0.680 0.68 ol/l b) Fraction decoposed [(.58 0.680) M] / (.58 M) 0.5689 0.57 6.0 Fro first two steps: Fro step (): k [I ] k [I] ; [I] (k /k ) / [I ] / Fro step (): k [H ][I] k [H I]; [H I] k /k [H ][I] Rate law for slow step: Rate k 3 [H I][I] Substituting for [H I]: Rate k 3 [k /k [H ][I]][I] Rate k 3 k /k [H ][I] Substituting for [I]: Rate k 3 k /k [H ][(k /k ) / [I ] / ] Rate k 3 k /k (k /k ) [H ][I ] Cobining k s: Rate k[h ][I ] 6.03 a) Conductoetric ethod. The HBr that fors is a strong acid in water, so it dissociates copletely into ions. As tie passes, ore ions for, so the conductivity of the reaction ixture increases. b) Manoetric ethod. The reaction involves a reduction in oles of gas, so the rate can be deterined fro the change in pressure (at constant volue and teperature) over tie. 6-4
6.04 To solve this proble, a clear picture of what is happening is useful. Initially only N O 5 is present at a pressure of 5 kpa. Then a reaction takes place that consues the gas N O 5 and produces the gases NO and O. The balanced equation gives the change in the nuber of oles of gas as N O 5 decoposes. Since the nuber of oles of gas is proportional to the pressure, this change irrors the change in pressure. The total pressure at the end, 78 kpa, equals the su of the partial pressures of the three gases. Balanced equation: N O 5 (g) NO (g) + / O (g) Therefore, for each ole of dinitrogen pentaoxide.5 oles of gas are produced. N O 5 (g) NO (g) + / O (g) Initial P (kpa) 5 0 0 total P initial 5 kpa Final P (kpa) 5 x x / x total P final 78 kpa Solve for x: P + P + P (5 x) + x + / x 78 N O5 NO O x 35.3333 kpa (unrounded) Partial pressure of NO equals x (35.3333) 70.667 7 kpa. Check: Substitute values for all partial pressures to find total final pressure: (5 35.3333) + ( x 35.3333) + ((/) x 35.3333) 78 kpa The result agrees with the given total final pressure. 6.05 a) Rearrange t / ln / k to k ln / t / k ln / 90 in 7.706 x 0 3 in (unrounded) [ Aspirin] ln kt Aspirin [ ] t 0 [ Aspirin] t [ ] [ Aspirin] t [ g/00l] ln g/00l (7.706 x 0 3 in ) (.5 h) (60 in / h).554 0 0 0.3498 (unrounded) [Aspirin] t ( g / 00 L) (0.3498) 0.6996 g / 00 L 0.6 g / 00 L b) The antibiotic pill PILL. The pill is taken at the fever teperature, so use the fever k. [ PILL] ln kt PILL ln [ ] t 0 [ /3 PILL] t [ PILL] 0 (3.9 x 0 5 s ) t (3600 s / h) t.88798.9 h Pills should be taken at about three hour intervals. 6-5
c) Convert the teperatures to C then to K. C [T( F) 3] (5/9) K 73.5 + C 73.5 + [T( F) 3] (5/9) k 3. x 0 5 s T 73.5 + [98.6 F 3] (5/9) 30.5 K (unrounded) k 3.9 x 0 5 s T 73.5 + [0.9 F 3] (5/9) 3.98 K (unrounded) E a? ln k Ea k R T T ( ) 5 3.9 x 0 s k J 8.34 ln Rln ol K 5 E a k i ( 3.x0 s ) T T 3.98 K 30.5 K E a.009 x 0 5 J/ol x 0 5 J/ol The subtraction of the / T ters leaves only one significant figure. 6.06 No. The uncertainty in the pressure, P, is 5%. The reaction rate is proportional to [P] 4. The relative reaction rate with 5% error would be [.05] 4. or % in error. The rate easureent has an uncertainty of % so a 0% change in rate is not significant. 6.07 a) The iodide ion approaches fro the side opposite the relatively large chlorine. H _ I H C H b) The backside attack of the I inverts the geoetry at the carbon bearing the Cl, producing this product: CH CH 3 Cl I C CH 3 H c) The planar interediate can be attached fro either side, producing a raceic ixture (that is, an equal ixture of two optical isoers): CH CH 3 H 3 CH C I C CH 3 H and H 3 C H C I 6-6
6.08 k egg / 4.8 in T (73. + 90.0) K 363. K k egg / 4.5 in T (73. + 00.0) K 373. K E a? The nuber of eggs () is exact, and has no bearing on the significant figures. ln k Ea k R T T ( ) k J egg/4.5in Rln 8.34 ln E a k ol K i ( egg/4.8in) T T 373. K 363. K E a 7.730 x 0 3 J/ol 7.3 x 0 3 J/ol 6.09 () H SO 4 H 3 O + + HSO 4 + SO 3 [fast] () SO 3 + C 6 H 6 H(C 6 H + 5 )SO 3 [slow] (3) H(C 6 H + 5 )SO 3 + HSO 4 C 6 H 5 SO 3 + H SO 4 [fast] (4) C 6 H 5 SO 3 + H 3 O + C 6 H 5 SO 3 H + H O [fast] a) Add the steps together and cancel: C 6 H 6 + H SO 4 C 6 H 5 SO 3 H + H O b) Initially: Rate k [SO 3 ][C 6 H 6 ] (fro the slow step) [SO 3 ] k [H SO 4 ] Result: Rate k [k [H SO 4 ] ][C 6 H 6 ] Rate k[h SO 4 ] [C 6 H 6 ] 6.0 a) Starting with the fact that rate of foration of O (rate of step ) equals the rate of consuption of O (rate of step ), set up an equation to solve for [O] using the given values of k, k, [NO ], and [O ]. rate rate k [NO ] k [O][O ] 3 9 k[ NO] ( 6.0 x 0 s ) 4.0 x 0 M [O].4 x 0 k[ O] 6 (.0 x 0 L/ ol s) 5 M i.0 x 0 M b) Since the rate of the two steps is equal, either can be used to deterine rate of foration of ozone. rate k [O][O ] (.0 x 0 6 L/ol s) (.4 x 0 5 M) (.0 x 0 M).4 x 0 ol/l s 6. a) At tie 0.00 in assue [A] 0.00, thus at tie 3.00 in, [A] t 0.00 (fro the equation for % inactivation). Using this inforation: ln [A] t / [A] 0 k t [ A] [ 0.00] ln [ ] t ln A [ ] t 0.00 0 k.30585.3 in t 3.00 in b) Fro the equation for % inactivation, 95% inactivation results in [A] t 0.05: ln [A] t / [A] 0 k t [ A] [ 0.05] ln [ ] t ln A [ ] t 0.00 0 t k.30585 in.3003.3 in 6-7
6. I. Rate k [N O 5 ] No II. Rate k [N O 5 ] No III. Rate k [NO ][N O 5 ] k K eq [N O 5 ] [NO 3 ] k[n O 5 ] [NO 3 ] No IV. Rate k [N O 3 ][O] k K /3 eq [N O 5 ] /3 [NO ] /3 [N O 3 ] /3 No V. Rate k [N O 5 ] No 6.3 This proble involves the first-order integrated rate law (ln [A] t / [A] 0 kt). The teperature ust be part of the calculation of the rate constant. The concentration of the aoniu ion is directly related to the aonia concentration. a) [NH 3 ] 0 3.0 ol/ 3 [NH 3 ] t 0.35 ol/ 3 T 0 C k 0.47 e 0.095(T-5 C) ln[nh 3 ] t kt + ln[nh 3 ] 0 ln[nh 3] 0 ln[nh 3] t t k 3 3 ln[3.0 ol/ ] ln[0.35 ol/ ] t 0.095(0-5 C) 0.47e t.847.8 d b) Repeating the calculation at the different teperature: [NH 3 ] 0 3.0 ol/ 3 [NH 3 ] t 0.35 ol/ 3 T 0 C k 0.47 e 0.095(T-5 C) ln[nh 3] 0 ln[nh 3] t t k 3 3 ln[3.0 ol/ ] ln[0.35 ol/ ] t 0.095(0-5 C) 0.47e t 7.35045 7.4 d c) For NH + 4 the rate k [NH + 4 ] Fro the balanced cheical equation: + Δ NH 4 Δ[ O ] Δt Δt Thus, for O : rate k [NH + 4 ] 0.095 Rate ( 0 5) 3 ( ) 0.47e 3.0 ol/ 4.5346 4.5 ol/3 ( ) 6.4 a) Rate / Rate 4 [(.7 x 0 7 ol/l s) / (. x 0 7 ol/l s)] [k(0.00 ol/l) / k(0.044 ol/l) ].5.77 Rate k [CS ] b) First, calculate the individual k values; then average the values. k Rate / [CS ] k (.7 x 0 7 ol/l s) / (0.00 ol/l).7 x 0 6 s k (. x 0 7 ol/l s) / (0.080 ol/l).75 x 0 6.8 x 0 6 s k 3 (.5 x 0 7 ol/l s) / (0.055 ol/l).77 x 0 6.7 x 0 6 s k 4 (. x 0 7 ol/l s) / (0.044 ol/l).77 x 0 6.7 x 0 6 s k avg [(.7 x 0 6 s ) + (.75 x 0 6 ) + (.77 x 0 6 ) + (.77 x 0 6 )] / 4.76 0 6.7 x 0 6 s 6-8