Question

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)

Answer 1

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 20.0 30.0 40.0 we alkyl incoming nucleophile; theus so Jablaydi Now, as can see in this reaction, we have alkyl bromide and nacle bromide bromide prefers mechanism of reaction; in colech is rate detere and rate the concen -tration (CH₃) C Br. prefers SNI termining first be dependent only (slow) stef of + B RDS + I ST Step I - (CH3)₂ CBR (CH₃)₂ C Step II (CH3)₂ C (CH₂), CI for this reaction, rate law will be written as- x [(CH₃)₂ C Br K [(CH₃)₂ C Br where K is rate constant. The it is first order reaction e r Therefore, it is a

Now, coe the basis the concentration initial 20 seconds. i es see that the remains half of after of (Crs). C Br. 1 m t which indicates the Rate law first order reaction again. Thus, for given reaction de & [(CH₃)₂ C Br K [CH₃ C Br dt for a first order reaction; [CH, C. Br + I- (CH₂CI+ Bri a t=0 tut, (a-x x da a Ca-x) dt dx dt dx kat (a-x) integrating both sides x=2 dx S a See (a-x) X=0 A