# Rate of Reaction

## Homework Statement

[/B]
I don' t want to be too specific, but it's basically "A + B -> C has the rate law Rate=k[A][ B], How long does it take [A] to reach concentration X. K is given and the initial concentrations of A and B are given."

## Homework Equations

The integrated rate laws, whether I need first or second or is one of the things that's confusing me.

## The Attempt at a Solution

I know this is simple but I'm quite confused :/ The reaction is first order with A, so I'm tempted to use the given value of A and the integrated rate law for first order reactions. But then doesn't the concentration of B effect the rate? If I need to take this into account I'm confused as to how I should proceed. I don't want to much info but some intuition would be great. Thanks!

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It won't let me edit it!! I meant Rate=k*[A]*B. (For some reason whenever I put the B in square brackets it won't save...).

epenguin
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It is just a first order reaction as your first impulse. Because they told you. That is the experimental fact.

A reaction of two substances is not necessarily second order, and a reaction of one only not necessarily first order. The kinetics of a reaction depends on the reaction mechanism (and conversely contains information about this) you cannot tell it just by the number of molecules reacting.

One way of having first order kinetics in that reaction is when it is 'pseudo-first'. If the concentration of B is much more than of A, B will be at almost constant concentration during the reaction and you can treat the time course of the reaction just as a first order reaction even when it is really second-order or something else. Then you'd do the experiment at various different to find its effect. You say you know the concentrations so you can see if this is possible.

But even when this is not the case the reaction can still be first order in one only of the reactants. Try and think up a reaction mechanism in which this would happen.

• sciencegem
epenguin
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It won't let me edit it!! I meant Rate=k*[A]*B. (For some reason whenever I put the B in square brackets it won't save...).
It's reading the [ B as a typographical instruction for bold (does this with [ and a few other letters]). You just have to leave a space after the bracket [ B] .

It is just a first order reaction as your first impulse. Because they told you. That is the experimental fact.
Thank you so much, that's really helpful. It's not a pseudo-first reaction. I just want to double check how you know it's first order? was having trouble editing, so I typed Rate=k[A] first but I mean Rate=k [ A ] [ B ]. Isn't that a second order reaction? When you say "But even when this is not the case the reaction can still be first order in one only of the reactants", how do I know when it still is first order in one of the reactants? (Sorry I'm very slow).

epenguin
Homework Helper
Gold Member
Is the question saying the rate law is rate = k[A][ B] ?

If so you have to treat it as second order! Unless it gets simplified by very different concentrations.

• sciencegem
Is the question saying the rate law is rate = k[A][ B] ?
Yes.

epenguin
Homework Helper
Gold Member

Is the question saying the rate law is rate = k[A][ B] ?

If so you have to treat it as second order! Unless it gets simplified by very different concentrations.
I think I get it! Thank you! So it's a second order reaction, but I can still just punch in the concentration of A into the integrated rate law for second order reactions, because A is behaving second order. Is that correct?

epenguin
Homework Helper
Gold Member
Sounding a bit confusing /confused. Big excess of one and time course like first order. Different simplification if initial [A] = initial [ B] because then [A] = [ B] at all times. Otherwise have to treat according to general 2nd order equation which is more complicated (so experimentalists try to avoid that condition!).

• sciencegem
Sounding a bit confusing /confused. Big excess of one and time course like first order. Different simplification if initial [A] = initial [ B] because then [A] = [ B] at all times. Otherwise have to treat according to general 2nd order equation which is more complicated (so experimentalists try to avoid that condition!).
Ahh, I get it now. The concentrations are similar which makes a lot more sense. Thank you so much for all you help!

Ahh, I get it now. The concentrations are similar which makes a lot more sense. Thank you so much for all you help!
Nope actually, I was so close but I'm still a little confused . Is that different simplification you were referring to that the rate law can basically be written Rate=k[A]^2... or can it? How does that make physical sense?

Different simplification if initial [A] = initial [ B] because then [A] = [ B] at all times.
Is that different simplification you were referring to that the rate law can basically be written Rate=k[A]^2... or can it? How does that make physical sense?
My reasoning now is Rate=k[A][ B] when [A] halves, [ B] halves, and the rate is effected by a factor of 1/4, so I can determine different concentrations of [A] at different times using the second order integrated rate law 1/[A] = 1/[A_0] + kt. Where's the flaw in that reasoning?

epenguin
Homework Helper
Gold Member
Nope actually, I was so close but I'm still a little confused . Is that different simplification you were referring to that the rate law can basically be written Rate=k[A]^2... or can it? How does that make physical sense?
You can't change the rate law, but choose within various practical limits the initial concentrations so that it simplifies, nothing needing physical justification.

My reasoning now is Rate=k[A][ B] when [A] halves, [ B] halves, and the rate is effected by a factor of 1/4, so I can determine different concentrations of [A] at different times using the second order integrated rate law 1/[A] = 1/[A_0] + kt. Where's the flaw in that reasoning?
Yes in kinetics you follow with time (at different times, but these days if possible continuously) either [A] or [ B] or the appearance of product, and can analyse the results in different ways, e,g. fit the results to an equation like that (but for second order that one is wrong). Usually you'd want several different experiments at different starting concentrations.

Yes in kinetics you follow with time (at different times, but these days if possible continuously) either [A] or [ B] or the appearance of product, and can analyse the results in different ways, e,g. fit the results to an equation like that (but for second order that one is wrong). Usually you'd want several different experiments at different starting concentrations.
Awesome thanks! Confused by what you mean when you say that's wrong for second order though, you mean 1/[A]=1/[A_0] +kt? I'm pretty sure that's what my textbook says it is.

epenguin
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Sorry, that's right, I was confusing with another form.

Chestermiller
Mentor
For the rate law you wrote down,

$$\frac{d[ A]}{dt}=\frac{d[ B]}{dt}=-k[ A][ B]$$

If the initial concentrations of A and B are [A0] and [B0], and the amount of A and B that reacts up to time t is x, then

[A]=[A0]-x

[ B]=[B0]-x

If we substitute these into the differential equation, we obtain:

$$\frac{dx}{dt}=k([A_0]-x)([B_0]-x)$$

This is the differential equation you need to solve subject to the initial condition x = 0 at t = 0.

Chet

• sciencegem
For the rate law you wrote down,

$$\frac{d[ A]}{dt}=\frac{d[ B]}{dt}=-k[ A][ B]$$

If the initial concentrations of A and B are [A0] and [B0], and the amount of A and B that reacts up to time t is x, then

[A]=[A0]-x

[ B]=[B0]-x

If we substitute these into the differential equation, we obtain:

$$\frac{dx}{dt}=k([A_0]-x)([B_0]-x)$$

This is the differential equation you need to solve subject to the initial condition x = 0 at t = 0.

Chet
Just saw this. Awesome, thanks!