Check out some similar questions!

western50 · Oct 16, 2010, 02:59 PM

At 620 K, butadiene dimerizes at a moderate rate.
The following data were obtained in an experiment involving this reaction.

T (s) [butadiene] (M)
0 0.01000
1000. 0.00629
2000. 0.00459
3000. 0.00361

1. What is the order of the reaction in butadiene?
2. how many seconds is the dimerization 2.0 % complete?
3. what is the half-life of the reaction is the initial concentration of butadiene is 0.0800 M?

Unknown008 · Oct 17, 2010, 12:30 AM

1. You will have to draw a graph for this problem and see if there is a constant half life. If there is a constant half life, then the reaction is of first order with respect to the concentration of butadiene.

When you did this, look for the half life. If there is a constant half life, then it's a first order reaction.

If the half life is such that it is the squared of the time (ie, the time for the substance to halve is the square of the initial time except zero), then it's of second order.

2. Use the rate equation.

$R = k[C_4H_6]^2$

$R \propto \frac{1}{t}$

$\frac{h}{t} = [C_4H_6]^2$

$h = t[C_4H_6]^2$

Find the value of the constant h.
Then, find the time it takes to go from 0.01 to 0.0098 (ie, 2% of 0.01 M reacted).

3. Make another graph and continue like the first part.

western50 · Oct 17, 2010, 02:04 AM

I still don't get it?

Unknown008 · Oct 17, 2010, 02:11 AM

Have you plot the graph of Concentration against time?

Look for the half life. What can you say about it?

western50 · Oct 17, 2010, 02:25 AM

How can we get the answer from the graph? What do you mean by looking for half life? How can you know it is first order or second order?

Unknown008 · Oct 17, 2010, 02:36 AM

If it's a straight line, it's of zero order.

If it's a curve, and if the half life (time for the concentration to halve) is constant, it's first order.

If it's a curve, and if the half life is squared each time (eg. At 3s, it's 800 and it is at 9 seconds that it becomes 400, or at 5 seconds it's 20 and it's only at 25 seconds that it becomes 10), then it's second order.

Etc.

western50 · Oct 17, 2010, 03:25 AM

I don't get the steps for the second problem

Unknown008 · Oct 17, 2010, 03:49 AM

Calculate the value of the constant h first.

h = 1000(0.00629)^2 = 0.0396

h = 2000(0.00459)^2 = 0.0421

h = 3000(0.00361)^2 = 0.0391

Average those and we get h = 0.0403

Then, use this:

0.0403 = t(0.0098)^2

0.0098 because 2% of 0.01 is 0.0002 and thus, (0.01 - 0.0002 =) 0.0098 M is left.

t = 419 s