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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: hiro on May 13, 2005, 09:53:27 AM

Title: rate equation
Post by: hiro on May 13, 2005, 09:53:27 AM

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

Title: Re:rate equation
Post by: Donaldson Tan on May 13, 2005, 11:52:42 AM

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.

Title: Re:rate equation
Post by: GCT on May 13, 2005, 01:02:46 PM

Quote from: hiro on May 13, 2005, 09:53:27 AM

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.

Title: Re:rate equation
Post by: Donaldson Tan on May 13, 2005, 07:05:44 PM

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.

Title: Re:rate equation
Post by: chemical on May 14, 2005, 08:44:00 AM

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

Title: Re:rate equation
Post by: hiro on May 14, 2005, 10:48:38 AM

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?

Title: Re:rate equation
Post by: Donaldson Tan on May 14, 2005, 02:20:04 PM

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.

Title: Re:rate equation
Post by: chemical on May 17, 2005, 07:34:12 AM

Quote from: geodome on May 14, 2005, 02:20:04 PM

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?

Title: Re:rate equation
Post by: Donaldson Tan on May 18, 2005, 04:34:40 AM

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]