Asked by Violet
A chemist does a reaction rate analysis on the following reaction:
2CO (g) + O2 (g) → 2CO2 (g)
She collects the following data:
Trial Initial Concentration of
CO (M) Initial Concentration
of O2 (M) Instantaneous
Reaction Rate (M/s)
1 0.150 0.150 0.113
2 0.300 0.150 0.226
3 0.300 0.300 0.904
What is the rate equation for this reaction?
2CO (g) + O2 (g) → 2CO2 (g)
She collects the following data:
Trial Initial Concentration of
CO (M) Initial Concentration
of O2 (M) Instantaneous
Reaction Rate (M/s)
1 0.150 0.150 0.113
2 0.300 0.150 0.226
3 0.300 0.300 0.904
What is the rate equation for this reaction?
Answers
Answered by
DrBob222
The rate equation is rate = k(CO)^x(O2)^y.
Divide rate 2 by rate 1.
0.226/0.113 = 2 and
2 = (0.300)<sup>x</sup>(0.150)<sup>y</sup>/(0.150)<sup>x</sup>(0.150)<sup>y</sup>
Note that 0.150<sup>y</sup> cancels with 0.150<sup>y</sup> and leaves just
2= (2)<sup>x</sup> and x must be 1.
Do the same thing for rate 3 divided by rate 2 and determine y. You should get 2. Then substitute x = 1 and y = 2 into any of the three rate equations to determine k, then write the equation. If you get stuck, post your work in a new post at the top of the page (this post is close to the bottom when I read it) and I can help you through it.
Divide rate 2 by rate 1.
0.226/0.113 = 2 and
2 = (0.300)<sup>x</sup>(0.150)<sup>y</sup>/(0.150)<sup>x</sup>(0.150)<sup>y</sup>
Note that 0.150<sup>y</sup> cancels with 0.150<sup>y</sup> and leaves just
2= (2)<sup>x</sup> and x must be 1.
Do the same thing for rate 3 divided by rate 2 and determine y. You should get 2. Then substitute x = 1 and y = 2 into any of the three rate equations to determine k, then write the equation. If you get stuck, post your work in a new post at the top of the page (this post is close to the bottom when I read it) and I can help you through it.
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