[hiro]

experiment    [NO2]at start(M)    [F2]at start(M)     initial rate(molL-1s-1)
   1                     0.001                    0.005                  2x10^-4
   2                     0.002                    0.005                  4x10^-4
   3                     0.002                    0.01                    8x10^-4

the equation of this is 2NO2+F2-->2NO2F

but how to write the rate equation for this reaction?

thx

[Donaldson Tan]

i give you a clue.

rate = k[NO2]^a[F2]^b

find the ratio of experimental rate of (1) over (2) and that for concentration of reactants.

[GCT]

experiment    [NO2]at start(M)    [F2]at start(M)     initial rate(molL-1s-1)
   1                     0.001                    0.005                  2x10^-4
   2                     0.002                    0.005                  4x10^-4
   3                     0.002                    0.01                    8x10^-4

the equation of this is 2NO2+F2-->2NO2F

but how to write the rate equation for this reaction?

thx

how can you find the order of each reactant with respect to the rate?

what can you do to find k?

It doesn't get any simpler than this.

[Donaldson Tan]

comparing experiments (1) and (2), the rate doubles when [NO2] doubles.

comparing experiments (2) and (3), the rate doubles when [F2] doubles.

This should be more than sufficient for you to arrive at your answer.

[chemical]

hi,

if NO2 is increased the rate doubles this means it is a first order reaction
if F2 is increased the rate doubles also this means it is first order reaction thus the overal reaction is 2nd order

K[NO2]¹[F2]¹

[hiro]

o...icic thx...
^^''i weak at rate of reaction...and energy things...
so if the rate triples it means that one of the compund will also triples?

[Donaldson Tan]

rate = k[NO2]^ak[F2]^b

for experiment 1 & 2, [NO2] is constant, so

rate2/rate1 = 0.002^a/0.001^a = 2^a
rate2/rate1 = (4 x 10^-4) / 2 x 10^-4 = 2
=> 2 = 2^a => a = 1

you do the same for comparing experiments 2 & 3, where [F2] is constant, then you can find the value of b.

[chemical]

rate = k[NO2]^ak[F2]^b

for experiment 1 & 2, [NO2] is constant, so

rate2/rate1 = 0.002^a/0.001^a = 2^a
rate2/rate1 = (4 x 10^-4) / 2 x 10^-4 = 2
=> 2 = 2^a => a = 1

you do the same for comparing experiments 2 & 3, where [F2] is constant, then you can find the value of b.

was my answer wrong?

[Donaldson Tan]

chemical: your answer is right. i merely showed how it is done mathematically.

after you have work out the reaction order, just substitute the various values, rearrange the rate equation to find the rate constant, ie.

k = rate/[NO2][F2]