KINETICS Practice Problems and Solutions
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1 Nae: AP Cheistry Period: Date: Dr. Mandes The following questions represent potential types of quiz questions. Please answer each question copletely and thoroughly. The solutions will be posted on-line on Monday. 6. Consider the reaction: P H 4 PH 3. A rate study of this reaction was conducted at 98 K. The data that were obtained are shown in the table. [P 4], ol/l [H ], ol/l Initial Rate, ol/(l. s) x x x 0-4 a. What is the order with respect to: P 4 0. H. b. Write the rate law for this reaction. rate = k[h ] c. Deterine the value and units of the rate constant, k. plug and chug using the rate law & data fro exp t and solving for k, we get k = s - 7. Consider the reaction: SO + O 3 SO 3 + O. A rate study of this reaction was conducted at 98 K. The data that were obtained are shown in the table. [SO ], ol/l [O 3], ol/l Initial Rate, ol/(l. s) a. What is the order with respect to: SO. O 3 0. b. Write the rate law for this reaction. rate = k[so ] [O 3] 0 c. Deterine the value and units of the rate constant, k. plug and chug using the rate law & data fro exp t and solving for k, we get k =.36 ol. L -. s - 8. Consider the following echanis. A + B R + C (slow) A + R C (fast) a. Write the overall balanced cheical equation. A + B C b. Identify any interediates within the echanis. R c. What is the order with respect to each reactant? A st ; B st d. Write the rate law for the overall reaction. rate = k [A ][B ] 9. Consider the following echanis. O 3 O + O (fast) O 3 + O O (slow) a. Write the overall balanced cheical equation. O 3 3 O
2 b. Identify any interediates within the echanis. O c. What is the order with respect to each reactant? O 3 nd (once in rds, then once when sub for interediate) d. Write the rate law for the overall reaction. rate = k [O 3] 0. Consider the reaction: B C + 3D. In one experient it was found that at 300 K the rate constant is 0.34 L/(ol. s). A second experient showed that at 450 K, the rate constant was L/(ol. s). Deterine the activation energy for the reaction. at 300 K: 300 at 450 K: 450 Ea k Ae Ea k Ae ln k450 A Ea ln( k ) ln( A) where ln( A) ln( k ) so that Ea Ea ln( k ) [ln( k ) ] k450 Ea ln( k ) ( ) 300 R T 300 T450 Ea plug and solve for Ea, Ea = 0.8 kj Ea MORE PROBLEMS>>>> Deterining rate law fro echaniss (use the rate-deterining step to get the orders).. One ethod for the destruction of ozone in the upper atosphere is: O 3 + NO NO + O (slow) NO + O NO + O (fast) overall rxn O 3 + O O a. Which species is an interediate? b. Which species is a catalyst? c. Which is the rate-deterining step (rds)? d. Nuber of ties each reactant is used in the rds? e. Write the rate law for the reaction.
3 Deterining rate law fro Initial Rates. (Use the ratio of initial rates to get the orders).. Consider the table of initial rates for the reaction: ClO + OH - - ClO ClO + H O. Experient [ClO ] o, ol/l [OH - ] o, ol/l Initial Rate, ol/(l. s) x x x 0 - a. Order with respect to ClO : b. Order with respect to OH - : c. Rate law for this reaction: d. Value and units for the rate constant: 3. Consider the table of initial rate for the reaction between heoglobin (Hb) and carbon onoxide. Experient [HB] o, ol/l [CO] o, ol/l Initial Rate, ol/(l. s) a. Order with respect to HB: b. Order with respect to CO: c. Rate law for this reaction: d. Value and units for the rate constant:
4 Deterining rate law fro tie and concentration data. (Use the integrated rate laws and graphing to get orders). 4. The rate of this rxn depends only on NO : NO + CO NO + CO. The following data were collected. Tie (s) 0 [NO] (ol/l) a. Order with respect to NO : b. Rate law for this reaction: c. [NO ] at.7 x 0 4 s after the start of the rxn. 5. The following data were obtained for the decoposition of N O 5 in CCl 4. The following data were collected. Tie (s) 0 [NO5] (ol/l) a. Order with respect to N O 5: b. Rate law for this reaction: c. [N O 5] at 3.5 x 0 3 s after the start of the rxn. SOLUTIONS!!!!!! TO MORE PROBLEMS >>>>>. a. Which species is an interediate? NO b. Which species is a catalyst? NO c. Which is the rate-deterining step (rds)? slow step d. Nuber of ties each reactant is used in the rds? O 3 is used once so order is O is used zero ties, so order is 0 e. Write the rate law for the reaction. rate = k[o 3]
5 Note that for a free-response question you ust show the work (ratio of rate laws), but not for ultiple choice n n rate k[ ClO ] [ OH ] rate k[ ClO ] [ OH ]. n n rate k[ ClO ] [ OH ] rate k[ ClO ] [ OH ] n =, so order is = n, so order is a. Order with respect to ClO : b. Order with respect to OH - : c. Rate law for this reaction: so, rate = k[clo ] [OH - ] d. Value and units for the rate constant: k = 30 L ol s get the value by subbing the data for exp t into the rate law and solving for k 3. rate k[ HB] [ CO] rate k HB CO n n s rate k[ HB] [ CO] rate k HB CO.6 n n n n =, so the order is = n, so the order is n a. Order with respect to HB: b. Order with respect to CO: c. Rate law for this reaction: so, rate = k[hb] [CO] d. Value and units for the rate constant: k = 0.8 L ol s get the value by subbing the data for exp t into the rate law and solving for k
6 4. Graph for zeroeth order: [NO ] vs. tie [y vs. x; y = ax +b] slope = -.7 x 0-5 y-intercept = 0.45 r = 0.90 General integrated rate law: [A] = -kt + [A]o This reaction's integrated rate law: [H O ] = (-.7 x 0-5 )t r = 0.90 Graph for first order: ln[no ] vs. tie [y vs. x; y = ax +b] slope = x 0-5 y-intercept = r = 0.97 General integrated rate law: ln[a] = -kt + ln[a] o This reaction's integrated rate law: ln [NO ] = (-5.78 x 0-5) t + (-0.770) r = 0.97 Graph for second order: [NO ] - vs. tie [y vs. x; y = ax +b] slope =.0 x 0-4 y-intercept =.0 r = best so order is General integrated rate law: [ A ] = kt + [ A ] This reaction's integrated rate law: [NO ] - =.0 x 0-4 t +.0 r = o Graph with the greatest r value: [NO ] - vs. tie, so the order is second order a. Order with respect to NO : b. Rate law for this reaction: rate = k[no ] c. [NO ] at.7 x 0 4 s after the start of the rxn. Subbing.7 x 0 4 s for tie in [NO ] - =.0 x 0-4 t +.0 [NO ] = 0.30 ol/l 5. Graph for zeroeth order: [N O 5] vs. tie [y vs. x; y = ax +b] slope = x 0-4 y-intercept =.3 r = General integrated rate law: [A] = -kt + [A]o This reaction's integrated rate law: [N O 5] = (-4.54 x 0-4 )t +.3 r = Graph for first order: ln[n O 5] vs. tie [y vs. x; y = ax +b] slope = x 0-4 y-intercept = r = General integrated rate law: ln[a] = -kt + ln[a] o This reaction's integrated rate law: ln[ N O 5] = (-6.05 x 0-4) t r = best so order is
7 Graph for second order: [N O 5] - vs. tie [y vs. x; y = ax +b] slope = 9.8 x 0-4 y-intercept = 0.57 r = 0.97s General integrated rate law: [ A ] = kt - + [ A ] This reaction's integrated rate law: [N O 5] - = 9.8 x 0-4 t r = 0.97 o Graph with the greatest r value: ln [N O 5] vs. tie, so the order is first order Order with respect to N O 5: Rate law for this reaction: a. Order with respect to N O 5: b. Rate law for this reaction: rate = k[n O 5] c. [N O 5] at 3.5 x 0 3 s after the start of the rxn. Subbing 3.5 x 0 3 s for tie in ln[ N O 5] = (-6.05 x 0-4) t +.3 [N O 5] = 0.7 ol/l
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