[kimi85]

In the reaction A --> products, 4.40 min after the reaction is started, [A] = 0.588 M.  The rate of reaction at this point is found to be rate = -delta [A]/ delta t = 2.2 x 10^-2 M/min.  Assume that this rate remains constant for a short period of time.
a. What is [A] 5.00 min after the reaction is started?
b. At what time after the reaction is started will [A] = 0.565 M

I really don't know how to approach this kind of problem.

Anybody can give me a hint.

Thanks!

[Dan]

This is how I'd do it, but it's been some time...

Start by integrating the rate equation:

-d[A]/dt = k                            (where k = 2.2exp(-2))

so,        (integral, limits: [A]₀ to [A]_t) -d[A] = (integral, limits: 0 to t) k dT

This gives an equation that, using the data provided, can be solved for [A]₀. Once you have [A]₀ you can solve for [A]_t for a given t or vice versa. I hope that makes sense to you, it's difficult for me to explain maths in text...

[kimi85]

thank you. But is there another way without doing integration?

[lavoisier]

I think this is a pretty tricky (and silly) problem, and I disagree with Dan's version.

First, you don't know what the reaction order is, so you can't integrate any rate law unless you make arbitrary assumptions (like r = k [A] in Dan's reply).

But then, you don't need to do that, because they ask you to consider the rate constant (r = k, pseudo zero order!), for a 'short period of time' (without specifying any actual time interval - very scientific!).

So the actual answer is that after t = 4.40 min, the reaction proceeds at the constant rate of 2.2x10^-2 M/min. The integrated rate law for zero order reactions is:

k(t1-t0) = A0-A1

So for the concentration at 5 min:

A1 = A0 - k(t1-t0) = 0.588 M - 2.2x10^-2 M/min * (5.00-4.40) min = 0.5748 M

And for the reverse question:

t1 = t0 + (A0-A1)/k = 4.40 min + (0.588-0.565) M / (2.2x10^-2 M/min) = 5.445 min

Anyway, I find this problem completely useless for learning kinetics, because there are hardly any actual reactions behaving like this, and it's more down to getting the trick in the text than to understanding what happens in reality.

[Dan]

I assumed r = k as was stated in the question, although I originally used C instead of k, I have edited that so that our posts are comparable.

The difference I made was to take t₀ = 0 and solve for [A]₀. The main problems with this are a) you don´t need to, I was overcomplicating the problem, and b) this assumes that "a short time" as stated in the question covers 5 minutes. Otherwise it's the same thing, and of course gives the same answers. We were essentailly measuring from different reference points.

Basically, my integrated rate equation was:

kt = [A]₀ - [A]_t

and so, kt = 0.6848 - [A]_t