No idea if I did this correctly–I’m working on a worksheet, but I missed this da

Question

No idea if I did this correctly–I’m working on a worksheet, but I missed this day of class so I’m just trying to figure out how to solve this using google. I would appreciate if someone could check my work. The questions are in bold and my answers are not. Thank you.

a. Determine the rate law of the reaction.
Rate = K[A]x[B]y
Reaction 3/reaction 2 = (0.060 = k[.30]x[.20]y)/(0.040 = k[.20]x[.20]y)
1.00 = 1y
1 = y

Reaction 2/reaction 1 = (0.040 = k[.20]x[.20]y)/0.080 = k[.10]x[.40]y)
1.00 = 1x
1 = x

Rate = k[A]1[B]1

b. What is the overall rate of the reaction?
x + y => 1 + 1 = 2
2ndorder

c. What is the numerical value of the rate constant and what are its units?
k = Rate/[A][B] => 0.060/(.30)(.20) = 1
k = 1 m-1s-1
(I’m not sure why it would be m-1s-1though, this is just what I see on other rate laws)

d. Calculate the rate of this reaction when the initial concentrations of A and B are 0.10 M and 0.20 M respectively.
Rate = (1 m-1s-1)(.10)(2.0) = .02 m-1s-1
(I don’t think I did this correctly)

Explanation / Answer

a) Rate of Reaction

Its ok, you used initial method rates! Fine

b)

since x = 1 and y = 2 the overall is the addition of all coefficients, that is 2, second order, that is also correct!

c)

NOTE that for all rate constants, this values re variable, you cant assume its always L/s or mol^2/s^2 and many others since this will depend on the concentration powers.

Since we have:

Rate = k*Ca*Cb

Expect K have units of 1/(Concentration^2) that is => (mol/L)^-2

d)

Actually you are correct...

The rate for this case is:

R = K*Ca*Cb

Since k = 1

Ca = 0.1 and Cb = 0.2

R = 1*0.1*0.2 = 0.02

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