Question

(4) 14 points) The table below pertains to the hypothetical reaction shown. A+ B2 → C+D Initial rate data: Experiment (A), M [B2), M 0.37 0.49 2 L 0.74 0.49 3 0.37 0.98 Initial Rate, M/S 1.60 x 105 3.20 x 105 6.40 x 105 (a) Use the method of initial rates to determine the reaction orders of A and B2, and write the overall rate law. You must explicitly show how you used the data in this table to obtain the orders. Stating what the orders are without evidence will earn you little in the way of points!

Answer 1

Let the rate law for this reaction be

Rate=k[A]a[B​​​​​​2]b

(a) Substitute the values for trial 1 in the given rate law

1.60 x 10-5 M/s=k(.37 M)a(0.49 M)b....(1)

Substitute the values for trial 2 in the given rate law

3.20 x 10 M/s=k(0.74 M)a(0.49 M)b.....(2)

Divide (2) by (1) we get

$\frac{3.20\times 10^{-5}\;M/s}{1.60 \times 10^{-5}\;M/s}=\frac{k(0.74\;M)^a(0.49 \;M)^b}{k(0.37\;M)^a(0.49\;M)^b}$i.e. 2=(0.74 M)a/(0.37 M)a

2=(0.74 M/0.37 M)a

21=2a

a=1

So order of reaction with respect to A=a=1

Substitute the values for trial 3 in the rate law

6.40 x 10-5 M/s=k(0.37 M)a(0.98 M)b....(3)

Divide (3) by (1) we get

$\frac{6.40\times 10^{-5}\;M/s}{1.60 \times 10^{-5}\;M/s}=\frac{k(0.37\;M)^a(0.98 \;M)^b}{k(0.37\;M)^a(0.49\;M)^b}$4=(0.98 M)b/(0.49 M)b

22=(0.98 M/0.49 M)b

22=2b

b=2

So order of reaction with respect to B​​​​​​2=b=2

So the rate law can be written as

Rate=k[A]1[B]2

i.e. Rate=k[A][B]2

b) Now substitute the values from trial 1 in this rate law

1.60 x 10-5 M/s=k(0.37 M)(0.49)2

k=(1.60 x 10-5 M/s)/(0.37 M)(0.49 M)2

=1.80 x 10-4 M-2s​​​-1

So the rate constant with units is 1.80 x 10-4 M-2s-1.