Solutions to: Kinetics Homework Problem Set S.E. Van Bramer

Transcription

1 Solutions to: Kinetics Homework Problem Set S.E. Van Bramer. The following experimenat data was collected for the reaction at 98 K. 2 N 2 O 5 --> 4 NO 2 + O 2 a. What is the average rate of N2O5 loss and the average rate of the reaction between.0 and. seconds? Assume the reaction is zero order and calculate the rate constant (k) using this value. Assume the reaction is first order and calculate the rate constant (k) using this value. Assume the reaction is second order and calculate the rate constant (k) using this value rate N2O5 :. sec.0 sec rate N2O sec : 2 rate N2O sec N2O5_avg : Assume zero order: k : N2O5_avg 0 k mol s liter Assume first order: k : N2O5_avg k s Assume second order: k : N2O5_avg 2 k mol s liter

2 b. What is the average rate of N2O5 loss and the average rate of the reaction between 5.0 and 5. seconds? Assume the reaction is zero order and calculate the rate constant (k) using this value. Assume the reaction is first order and calculate the rate constant (k) using this value. Assume the reaction is second order and calculate the rate constant (k) using this value rate N2O5 : 5. sec 5.0 sec rate N2O sec : 2 rate N2O sec N2O5_avg : Assume zero order: k : N2O5_avg 0 k mol s liter Assume first order: k : N2O5_avg k s Assume second order: k : N2O5_avg 2 k mol s liter Based upon the answers the reaction is first order. Because when we assumed that the reaction was first order the rate constant (k) was the same for both problems. When we assumed that the reaction was zero order or second order, the rate constant (k) was very different for the two problems.

3 d. Graphic analysis of rate consnstant. Graph the data to check if it is zero order, first order, and second order. Using these graphs determine the rate constant (k). i : 0,.. 7 Zero Order Plot t : C i N2O5i : 0 sec sec. sec 2 sec 5 sec 5. sec 0 sec 20 sec Concentration N2O5 mole/liter C N2O5i Zero Order Plot t i First Order Plot sec Time

4 6 First order kinetics plot 8 ln(concentration) ln C N2O5i t i sec time Second Order Plot Second Order Plot Concentration N2O5 mole/liter CN2O5 i Since the first order (ln vs t) plot is linear, the rate constant for N2O5 is the slope of the plot. The rate constant for the reaction is the /2 slope of the above plot. The /2 comes from the stoiciometry of the balanced equation: t i sec Time C N2O5i y : ln x : t i i i k : 2 slope( x, y) k s Note this is the same value as obtained above.

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6 2. The above experiment is repeated at several temperatures to obtain the following experimental data: Based upon this experimental data, determine the activation energy for this reaction and the preexponential factor. The activation energy and preexponential may be determined using any two of these data points. If you have access to a spreadsheet, a linear regression may be used to include all of the data points (this is ideal, but not practical for an exam) Recall the Arrhaneous equation: k A e This experiment is preformed at two temperatures (T) to find two rate constants (k): T : ( ) K k : sec T 2 : ( ) K k 2 : sec R : 8.34 joule mole K From the Arrhaneous equation k A e k 2 A e 2 These may be combined into one equation by dividing both sides so that : k k 2 A e A e 2 Which simplifies to: k ( T 2 T ) exp k 2 R ( T T 2 )

7 Substitute in known values and solve for Ea: : ln ( ) k k 2 ( ) joule mole Now substitute this back in to either equation and solve for A k A e A : exp k ( ) A s Alternatively using graphical methods the solution is found by graphing ln(k) vs /T. The slope of this is the activation energy and the intercept is the prexponential factor. This relationship is derived from the Arrhaneus equation as follows: k A e ln( k) ln A e ln( k) ln( A) + ln e ln( k) ln( A) ln( k) ln( A) R T This has the form ymx + b where x axis is /T y axis is ln(k) y intercept is ln(a) slope is -Ea/R

8 i : 0,.. 4 T : k : i i K K 33.5 K K K sec sec.2993 sec sec sec x : 0 y : 0 4 Arrhanius Plot 2 ln( k i sec) T i x : i T i R : 8.34 joule K mole y : i ln( k sec i ) : slope( x, y) R mol 2 joule mole A : e intercept( x, y) sec A s

9 3. Using the constants determined above if the initial concentration of N 2 O 5 is 3.0 x 0-4 M C initial : a. What is the concentration of NO2 and N2O5 after 5 seconds at 0 C? T : ( ) K t : 5 sec k : A e k_n2o5 : 2 k k_n2o s From the integrated rate equation find the concentration of N2O5 ln C initial C t k_n2o5 t C t : C initial exp( k_n2o5 t) C t The concentration of NO2: ( ) C NO2 : 2 C initial C t C NO

10 b. What is the concentration of NO2 and N2O5 after 5 seconds at 20 C? T : ( ) K t : 5 sec k : A e k_n2o5 : 2 k k_n2o s From the integrated rate equation find the concentration of N2O5 C t : C initial exp( k_n2o5 t) C t The concentration of NO2: ( ) C NO2 : 2 C initial C t C NO