To whom it may concern. please answer all my questions. This will be like 4times

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   To whom it may concern. please answer all my questions. This will be like 4times have posted multiple questions and only one or two questions will be answered out of the questions. or I'm I doing something wrong for posting multiple questions? Thank you very much

Determine the rate law from the data and calculate the rate constant k at 25.0 2. At 500 °C, cyclopropane (C3H6) rearranges to propene (CH3CH-CH2) C3H6 (g)CH3CH-CH2 (g) The reaction is first order and the rate constant is 6.7 x 10-4 s 1, (a) What is the rate law for the reaction, and; (b) if the initial concentration of C3H6 is 0.100 M, what is the concentration of cyclopropane of after 20 min? 3. Butadiene (C4H6) reacts with itself at 250 °C to form a dimer with the formula CsH12 2 C4H6 (g) CsHi2() The reaction is second order in C4H6- (a) What is the rate law for the reaction, and; (b) if the rate constant is 4.0 x 10-2 M-1.s-1, and the initial concentration of CaHe is 0.200 M, how long will t take for the concentration of C4H6 to reach 0.04 M? 4. a) In Q. 2 above what is the half life for the reaction and what is concentration of CH after 3 half-lives b) In Q. 3 above how long are the first and second half-lives?

Explanation / Answer

Rate = k[C3H6]

The first order rate law is : Ln (At/A0) = -kt

ln (At/0.1) = 6.7810^-4 /s * (20*60) sec

At = 0.04 M

Concentration after 20min will be 0.04 M

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rate = k[C4H6]^2

Second order rate law ; 1/[At]-1/[A0] = kt

or, (1/0.04)- (1/0.2) = 4*10^-2 M^-1 sec^-1* t

or t = 500 sec

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half life = 0.693/ 6.7810^-4 /s = 0.102 *10^2 sec

concentration after 1 half life =0.1/2

concentration after 2 half life = 0.1/4

concentration after 3 half lives = (0.1/4)/2 = 0.1/8 =0.0125 M

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half life for a second order reaction = 1/k[A0] =1/(4*10^-2 M^-1)*0.2 = 125 s

first half life is 125sec long. Second one is 250 sec long.

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