Friday

April 18, 2014

April 18, 2014

Posted by **elvis** on Saturday, February 18, 2012 at 9:18pm.

Initial Rate,

RUN #(A, M) (B, M) (C, M) (mol L-1 s-1

1 0.200 0.100 0.600 5.00

2 0.200 0.400 0.400 80.0

3 0.600 0.100 0.200 15.0

4 0.200 0.100 0.200 5.00

5 0.200 0.200 0.400 20.0

The rate constant for this reaction (all in the same units) is

a. 6667 b. 208 c. 2083 d. 139 e. 2500

help me!

chemistry - DrBob222, Friday, February 17, 2012 at 10:29pm

The secret here is to determine the order of the reaction. For example, it is zero order in C. Do you know how to do the others? After you determine the orders, then k is determined as in your post above.

Saw the other post didnt help at all but I need the answer and simple step by step solution

- chemistry -
**DrBob222**, Saturday, February 18, 2012 at 9:46pmI don't remember all the posts you had yesterday. Did you read how to determine the rate law expression. Can you do that from these data? I seem to remember that it was zero with respect to C, 2nd order with respect to B and 1st order with respect to A. so the rate law expression is

rate = k(A)^1(B)^2(C)^0 which becomes simply rate = k(A)(B^2) since we don't need to show exponents of 1 (and remember anything to the zero power is 1)

Pick any trial in your post and plug the values into the rate law expression above. The only unknown is k. Solve for that.

For example, pick the first one.

rate = k*(A)(B)^2

5.00 = k*(0.200)(0.1)^2 and solve for k.

- chemistry -
**Frank A**, Tuesday, February 12, 2013 at 4:29pm2500

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