On the basis of given data we can observe that the reactant concentration becomes half on doubling the time, which means that time is inversely proportional to the concentration here, therefore it is a first order reaction. The question is solved as follows:
![Soli For (CH₂) CI + Bri are Time (S) (CH3)₂ CBr t I above reaction, we [(CH3) C Br] (m) 0.600 0424 0300 O: 212 o 150 10.0 20.](http://img.homeworklib.com/questions/044c2540-e9d0-11ea-b123-6f65f91743d3.png?x-oss-process=image/resize,w_560)


4. For the reaction: (CH3),CBr +1 → (CH3),CI + Br* the following data were collected: (CH3)3Br)...
4. For the reaction: (CH3),CBr +1 → (CH3),CI + Br* the following data were collected: (CH3)3Br) (M) Time (s) 0.600 0 0.424 10.0 0.300 20.0 0.212 30.0 0.150 40.0 Determine the rate law for this reaction and calculate the rate constant. (20 points)
4. For the reaction: (CH3),CBr +1 → (CH3),CI + Br* the following data were collected: (CH3)3Br) (M) Time (s) 0.600 0 0.424 10.0 0.300 20.0 0.212 30.0 0.150 40.0 Determine the rate law for this reaction and calculate the rate constant. (20 points)
4. For the reaction: (CH3),CBr +1 → (CH3),CI + Br* the following data were collected: (CH3)3Br) (M) Time (s) 0.600 0 0.424 10.0 0.300 20.0 0.212 30.0 0.150 40.0 Determine the rate law for this reaction and calculate the rate constant. (20 points)
help!!
(CH3),CBr + 1 → (CH3),CI + Br the following data were collected: [(CH3)3Br] (M) Time (s) 0.600 0 0.424 10.0 0.300 20.0 0.212 30.0 0.150 40.0 Determine the rate law for this reaction and calculate the rate constant. (20 points) E (V) Thermodynamics: AGº - AH-TAS Nernst Equation: 8 = 8°-(RT/nF)InQ AGⓇ--RTINK At 25°C: 8-6° -0.0591/n)logQ AG RT K-e AG° = -nF8° Units/Constants: Volt: 1 V-1 JC Faraday: 1 F -96,485 C/mole Ea/RT Arrhenius Equation: k = Ae R-8.314...
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(A). For the hypothetical reaction A+B → C the following rate data (with and without catalyst) were obtained: Rate, M/s Rate, M/s (A), M [B, M (uncatalyzed) (catalyzed) 0.100 0.200 3.51 x 10 4 7.14 x 10-2 0.100 0.100 1.75 x 104 7.14 x 10-2 0.0500 0.200 1.76 x 104 3.57 x 10-2 0.0500 0.100 8.80 x 10-5 3.57 x 10-2 Based on these data: a. Determine the uncatalyzed and catalyzed rate laws for this reaction,...
For A - products, time and concentration data were collected and plotted as shown. 1/[A] [A] (M) t(s) 0.600 00.0 0.484 30.0 In[A] 0.405 60.0 0.349 90.0 Determine the reaction order, the rate constant, and the units of the rate constant. order units
For the reaction of (CH3)3CBr with OH-, (CH3)3CBr + OH- ÷ (CH3)3COH + Br- The following data were obtained in the laboratory. TIME (s) [(CH3)3CBr] 0 0.100 30 0.074 60 0.055 90 0.041 Plot these data as ln [(CH3)3CBr] versus time. Sketch your graph. Is the reaction first order or second order? What is the value of the rate constant?
Hi, it is one question with parts (a,b,c,d)
also complete the table
2. A flask is charged with 1.000 mol of A and allowed to react to form B according to the reaction A (g) 2B (g). The following data is obtained for [A] as the reaction proceeds. (10) Complete the table. 0,0 100 200 300 400 Time 0.00 20.0 30.0 40.0 min mol mol 0.835 0.714 0.625 0.555 1.000 [Bl Disappearance of A Appearance of B mol min Avg.rate...
41 The following data was experimentally obtained for the reaction: 2 of 2 (CH3)3CBr + H2O → (CH3)3COH + HBr t(h) (CH3)2CBr (mol/L) 0 10.39 x 102 3.15 8.96 x 102 6.20 7.76 x 102 10 6.39 x 102 18.30 3.53 x 102 30.80 2.07 x 102 a) Make the appropriate graphs and from them determine the order of reaction in relation to (CH3),CBr. In other words, the rate law expression will be like: Rate = k [(CH3)3CBr]" By concluding...
The following data were collected in two studies of the reaction below. A + 2B → C + D Time(s) Experiment #1 [A] (M) Experiment #2 [A] (M) 0 10.0 x 10-2 10.0 x 10-2 20 6.67 x 10-2 5.00 x 10-2 40 5.00 x 10-2 3.33 x 10-2 60 4.00 x 10-2 2.50 x 10-2 80 3.33 x 10-2 2.00 x 10-2 100 2.86 x 10-2 1.67 x 10-2 120 2.50 x 10-2 1.43 x 10-2 In Experiment #1,...