At 500 °C, cyclopropane (C3H6) rearranges to propene (CH3CH=CH2): C3H6 ( g ) ---

Question

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 C8H12.

2 C4H6 (g) ----> C8H12 (l)

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 C4H6 is 0.200 M, how long will it 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 C3H6 after 3 half-lives b) In Q. 3 above how long are the first and second half-lives?

help with question 2, 3 and 4. i already did them, i just want to compared answers so please show steps.

2. AL 500 C C3He reamasers to propene (CHyCH CH2: The reaction is fine order and the rate comuni 57x 104 l a What is the ate law for une reaction, and (b) if the initial cyclopropane of after 20 min? 3. Butadiene (C4Hol reacts with itself at 250 CID form a dimes with the formula ost12. The reaction is second order in C4H6 (a) What is the rale law for the reaction, and,o) ir me rave 40 x io 2 M-1 s-1 and the initial concentration of cHisaznoM, how long will take for the concentration of C4Ho reach 00 4 a) In Q.2 above what is the halflife for the reaction and what is concentrarion of ciHe after 3 half lives b) In Q. 3 above how long are the first and second half lives?

Explanation / Answer

2) a) Rate = -d [C3H6] / dt = 6.7 *10-4[C3H6 ]
This is the rate law for the first order reaction
   b) ln [A] - ln[Ao] = -Kt
   A = ? [Ao] = 0.100M ; t=20 min = 20*60 = 1200 sec
   ln[A] - ln[0.100] = - 6.7*10-4 s-1 * 1200 sec
ln[A] = -0.804 +ln [0.100] = -0.804 - 2.30258 = -3.10658
[A] = e-3.10658 = 0.044753 M = 0.045 M approx

3) a) rate = d[C4H6] /dt = - K[C4H6]2 is the rate law
   b) 1 / [A] = 1 / [Ao] + Kt ----------------(2)
   [A] = 0.04 M ; [Ao] = 0.2M ; K = 4*10-2M-1s-1 ; t =?
(1 / 0.04) = (1 / 0.2)+ 4*10-2 *t
25 = 5 + 4*10-2 *t
   t = 20 / 4*10-2 = 500 sec = 500/60 = 8.34 min

4. a) t1/2 = ln 2 / K = 0.693147 / 6.7*10-4 = 1034.548 sec = 17.24 min
   b) t1/2 = 1 / K[Ao] = 1 / (4*10-2 * 0.2) = 125 sec = 2.0834 min ( Put [A] = Ao/2 in equation 1)
Ao ----- t1/2------> Ao/2 -------t1/2 -------------> Ao/4
3 / K [Ao] = t1/2   ( put [A] =Ao/4 in equation 1)
   second t1/2 = 3 / (4*10-2 * 0.2) = 375 sec = 6.25 min

  
  

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