Rate Law & Half Life: Solve NOBr Reaction

[schu9600]

Homework Statement

Consider the following reaction
NOBr (g) → NO (g) + ½ Br2 (g)
The table below gives the time-dependent concentration of NOBr. Using this data,
determine, if the reaction is first- or second-order, with respect to NOBr. Give the rate
equation and determine the rate constant. To do this you’ll need to generate two plots,
and be sure to explain your reasoning. Then determine the half-live and lifetime for
NOBr. Watch your units! (7 marks)
t/s [NOBr]/mol dm-3
0.0 0.0250
6.2 0.0191
10.8 0.0162
14.7 0.0144
20.0 0.0125
24.6 0.011

Homework Equations

t_1/2 = ln2/k_r

The Attempt at a Solution

So I've been trying to work through this question for about an hour and a half and can't for the life of me figure out how to even start it off. I know I need to determine the order of the reaction first but can't even figure that out. If someone could give me two or three steps to get me going that would be great.

(I've made a graph of the time vs. concentration)

 

[epenguin]

You see we are supposed to help with confusions and difficulties and misunderstandings students run into when trying to carry out solutions. I am having a problem increasingly with students saying they have no idea where to start with perfectly standard problems that are in any and every book in the subject. We cannot be writing a book.

But I can come here now and then for say an hour now. (It would help if you posted your graph.)

Firstly what is a first or second order reaction? Find it in your book. If you are a desert island try and guess what it is from the question, which at least tells you what it roughly is about. Write down any equations that define them.

 

[schu9600]

Well the order of the reaction is the power by which the concentration of the substance is raised to in the rate law. I was thinking the NOBr would be a first order reaction because there is no coefficient in front of it in the equation so it would not be raised to any power (or it would be a power of 1). I know the half life part of the question requires the rate constant K_r to be able to solve it. I know that basic rate laws are in the form of
v=k_r [A]^a [C]^c...

This makes me want to believe the rate law for this question would be:
v=k_r [NOBr]^a

My graph I made is now attached.
By the way I legitimately was trying this question as I stated before. I'm not very good at this stuff and was looking for some assistance. This perfectly standard question does not come easily for me and it difficult for me to understand.

 

[schu9600]

The graph didnt attach trying again.

Says invalid file when I try to upload. Not sure how to attach the file =/

 

[epenguin]

The graph didnt attach trying again.

Says invalid file when I try to upload. Not sure how to attach the file =/

Maybe there are instructions or you can glean from other people's posts. I have only ever used tinypic here.

 

[schu9600]

Here is my graph? Can you please help me out? I want to figure out how to do this.

 

[epenguin]

I was thinking the NOBr would be a first order reaction because there is no coefficient in front of it in the equation so it would not be raised to any power (or it would be a power of 1).

I'm guessing second sentence means it is not something like 2NOBr -> ... Your conclusion does not actually follow. We may get on to connection kinetics-mechanism later, for now you are only asked to find out what are the kinetics that the data is telling you, a descriptive not yet mechanistic question.

"Well the order of the reaction is the power by which the concentration of the substance is raised to in the rate law."

Hence write out the equation rate = for a first order reaction and for a second order reaction with your substance. Using the d/dt notation.

I know that basic rate laws are in the form of
v=k_r [A]^a [C]^c...

So what do you think a would be for first order and for second order if

the order of the reaction is the power by which the concentration of the substance is raised to in the rate law.

?

 

[schu9600]

Well I would say that for a first order:

v=k_r [NOBr]¹

and a second order would be:

v=k_r [NOBr]²

Not really sure how to do the d/dt part. Never fully learned/understood derivatives.

 

[epenguin]

Well I would say that for a first order:

v=k_r [NOBr]¹

and a second order would be:

v=k_r [NOBr]²

Not really sure how to do the d/dt part. Never fully learned/understood derivatives.

OK that's right. Conventionally we'd drop the superscript 1 in 1st eq. and write

v=k_r [NOBr]¹ but it's not wrong as you wrote it.

 

[epenguin]

Now you say you never fully understood derivatives. We're not yet into their maths, just into what they mean. Derivatives in science basically rates of change of one thing with another, more often than not - and in our case - rates of change with time. So d[NOBr]/dt is the rate of change of [NOBr] with time.

On your fig. you've got [NOBr] and you've got time. How are you going to get this rate of change at any point from the fig? What are its units?

And by the way you can already quickly get a fair estimate of a half-life looking at that fig. About what would you say?

 

[schu9600]

So based on the graph the half life would be approximately 20 seconds.
Umm for the rate of change of the concentration would you take the slope of the tangent line at a spot on the graph? Or is there a way to get the change of concentration for the whole thing not just an instantaneous point.

Thanks for your help by the way.

 

[epenguin]

So based on the graph the half life would be approximately 20 seconds.
Umm for the rate of change of the concentration would you take the slope of the tangent line at a spot on the graph? Or is there a way to get the change of concentration for the whole thing not just an instantaneous point.

Thanks for your help by the way.

OK around 20 sec according to your fig. By the way I am wondering how the measurements in this experiment were done. How reaction was started. (Stuff was probably started somehow a time before the t=0 point.) Also why it wasn't measured for longer. Or wasn't there an experiment, were you just given numbers?

Yes I would take the rate at say half a dozen different times by slopes, after drawing a curve freehand. Then you'd get values for rates at different concentrations. You can plot those against what? , well concentration, and then try concentration squared so you can see whether it looks to be fitting one or the other of those two equations you gave, or at least one better than the other. And if it does fit either approximately or well you can obtain a rate constant according to equations you already gave.

Yes it is true there are better ways to assess the fit and the rate constants, ("integrated rate equations"). But you will get reasonably close this way and the point is you will understand all of what you are doing. You can do the procedure for the other ways, but understanding why they work needs a bit of math you say you don't have.

Look it's 4 a.m where I am, I got to go! Maybe SOMEONE ELSE CAN TAKE OVER?

I just leave you with this thought. For a first-order reaction the half-life is constant whatever point you start from, say at 0 sec or 10 sec or 20 sec. For a 2nd order one it isn't. Maybe you can think about how or why that is.

 

[epenguin]

I just leave you with this thought. For a first-order reaction the half-life is constant whatever point you start from, say at 0 sec or 10 sec or 20 sec. For a 2nd order one it isn't. Maybe you can think about how or why that is.

Mind you there may be not much scope for seeing that when they have only followed the reaction not much more than half way down. But the principle is more general. The time it takes for the concentration to fall from any point in time to e.g. 75% or any other given % of what it was at that time is the same for all starting times with a first order reaction.

Also quick early indication of the type of kinetics could be obtained by this reasoning: at t = t_1/2 the rate is what fraction of that at t = 0 for a first-order reaction, and for a second order reaction?

We'll wait to see the rate plots.

It occurred to me today that this reasoning:

I was thinking the NOBr would be a first order reaction because there is no coefficient in front of it in the equation so it would not be raised to any power (or it would be a power of 1).

is very dubious apart from what I said before about it. You have not taken into account the way they wrote the equation giving a half-molecule product! NOBr (g) → NO (g) + ½ Br₂ (g). I don't say a Br atom or radical couldn't be formed as a stage of the mechanism but a half-molecule surely isn't final product! They do not mean you to understand that it is even, it is just a manner of representing the stoichiometry.