For the reaction, A + 2B C + 2D, some measurements of the initial rate of reacti

Question

For the reaction, A + 2B C + 2D, some measurements of the initial rate of reaction at varying concentration gave the following data.

Experiment [A] [B] Rate (mol L-1 s-1 ) 1 0.100 0.200 0.000360 2 0.150 0.200 0.000540 3 0.150 0.250 0.001055

the rate law is therefore: rate = k[A][B]2 the rate law is therefore: rate = k[A]2[B] the rate law is therefore: rate = k[A]2[B]2 the rate law is therefore: rate = k[A][B] the rate law is therefore: rate = k[A][B]3 I feel confident that the first rate is 1, but I don't know if I should be confident. But I am for sure confused on B

Explanation / Answer

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

choose point 1 and 2, so B cancels out

0.000360/0.000540 = (0.1/0.15)^a

a = ln(0.000360/0.000540) / ln( (0.1/0.15)) = 1

1st order with respect to [A]

for B; choose 2 and 3

0.001055/0.000540 = (0.25/0.200)^b

ln(0.001055/0.000540) / ln((0.25/0.200)) = b

b = 3

3rd order with respect to [B]

then

Rate = k[A][B]^3

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