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Rate Law from Concentration Vs. Time Data?

No2(g) + CO(g) -> NO(g) + CO2(g)

The rate of the above reaction depends only on the concentration of nitrogen dioxide at temperatures below 225°C.

At a temperature below 225°C the following data were obtained:

Time (s) [NO2]

0 0.580

1.45*10^3 0.502

3.62*10^3 0.418

5.44*10^3 0.367

1.09*10^4 0.268

2.18*10^4 0.175

Which of the following expressions for the rate law (either differential or integrated) are completely consistent with the above experimental data:

A) 1/[NO2]-1/[N02]0= kt

B) ln[N02]-ln[N02]0= -kt

C) k[CO]^2= d[CO2]/dt

D) [NO2] = [NO2]0e^-kt

E) d[CO2]/dt= k[NO2]^2

I plotted it in excel and found it to be a second order reaction with slope y = 0.0002x + 1.7283

R² = 0.99999

But I can't get this right! It's my last try please help

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so.. let's back up a bit and make sure you understand the basics here

*******

for the reaction

.. A --> B + C

rate can be defined as change in amount of A present per change in time. And if we assume volume is constant, rate = change in concentration of A / change in time. In math, "change in" is represented with "d" and in chemistry concentration is represented with [ ].. so that

.. rate = - d[A] / dt... . . ..( the - sign is because [A] is decreasing over time)

*********

in addition, we know that for a reaction to be successful, we need to have a collision between the reactants so it makes sense to say, the higher the concentration of A, the more collisions per unit time... and the faster the rate. So that rate is proportional to [A]. in math terms

.. rate α [A]

and in math we solve proportionalities by inserting a constant to make the equation an equality so that

.. rate = k x [A]

and because we have different mechanisms (sometimes 1 A can decompose on it's own, sometimes 2 must collide, etc) we place an exponent on that [A]

.. rate = k x [A]^n

where "n" is the order of the reaction with respect to A

.. n = 0 is zero order

.. n = 1 is first order

.. n = 2 is 2nd order

etc

**********

combining

.. rate = - d[A] / dt = k x [A]^n

and we have ourselves a fairly simply differential equation.

*** zero order ***

if n=0

.. d[A] / dt = -k x [A]°

and because Z° = 1

.. d[A] / dt = -k

now we can easily solve this if we assume these 2 data points

.. data point 1.... .at time = 0... concentration of A = [Ao]

.. data point 2.. .. .at time = t.. . concentration of A = [At]

so that

.. d[A] / dt = -k

rearranges to

.. d[A] = -k x dt

integrating between those 2 data points gives

.. ∫d[A] = -k x ∫dt

which becomes

.. .. .. . ..[At].. .. . ... ....t = t

.. [A].. . |.. .. . = -k x t.. |

.. .. . . .. [Ao].. ... ... .. .t = 0

or simply

.. [At] - [Ao] = -k x (t - 0)

which rearranges to

.. [At] = -kt + [Ao]

***********

likewise... if n=1

we integrate this

.. ∫1/[A] d[A] = -k x ∫dt

to get

.. ln[At] = -kt + ln[Ao]

likewise if n=2 we integrate this

.. ∫1/[A]² d[A] = -k x ∫dt

to get

.. 1/[At] = +kt + 1/[Ao]

*******

*******

got all that?

.. .. .order.. .. .. . non-integrated.. .. .. . ..integrated

.. .. . .. 0.. . .. . ...rate = k x [A]°... .. .. . . ..[At] = -kt + [Ao]

.. . .. .. 1.. . .. . ...rate = k x [A]¹... .. .. . . ln[At] = -kt + ln[Ao]

.. . .. . .2... . .. .. .rate = k x [A]²... .. .. . . 1/[At] = +kt + 1/[Ao]

***********

anyway.. if you look at those integrated rate equations, you'll see they are all of the form

.. y = mx + b

if we let

.. y = [At]... y = ln[At].. .. y = 1/[At]

and

.. x = time

because k and [Ao] are constants... (we start with an initial concentration of A and it decreases over time)

or if this helps

.. .. [A] = -kt + [Ao].. .. . . zero order

.. ln[A] = -kt + ln[Ao].. .. .first order

.. 1/[A] = +kt + 1/[Ao].. . 2nd order

*******

anyway.. since they are of the form y = mx + b.. which is a line

IF... IF...

.. we plot [A] vs t and we get a straight line, the rxn must be zero order in A

IF.. that [A] vs t does not give a straight line but rather a curver

... and IF ln[A] vs t gives a straight line, the reaction must be 1st order in A

and IF

.. neither [A] vs t.. nor ln[A] vs t is a straight line

but

.. 1/[A] vs t is a straight line... then the rxn is 2nd order in A

**********

I plotted the data here

http://oi67.tinypic.com/2r6zb81.jpg

**********

**********

now.. it looks like we're in agreement so far

.. the rxn is 2nd order in [A]

.. the line has equation y = 0.00018304x + 1.72830129

now.. because we're plotting the equation

.. 1/[A] = + kt + 1/[Ao]

where

.. y = 1/[A]

.. x = t

therefore

.. k = 0.00018304 / (M x sec)

.. 1/[Ao] = 1.72830129 /M... .. .(which was our t = zero data point.. right?)

******

so.. this is a 2nd order rate equation...

and that means it is of this form

.. 1/[NO2] = +kt + 1/[NO2]0

(A) is consistent with that equation

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