[greentea11]

Don't really understand what this question means?

Under what conditions can a reaction with a rate law (attached)
be said to have a definite classification by order and  molecularity?

I've found that its a rate law for a linear chain reaction, but I don't really know what condition the question is hinting, any pointers would be much appreciated

thanks


Edit:
Next part of question gives time (t/s) and concentration P ( [P] molL-1)

Asks to determine (integer) order of reaction and its rate constant
I've plotted [P] vs time and its not linear so not 0th order, ln[P] vs time and its not linear, 1/[P] vs time and its not linear

what should I be plotting?

[mjc123]

(Second part) How does [P] vary with time for a first order reaction? Does it increase exponentially?

[Big-Daddy]

(First part) For the reaction to have a specific order (wrt each reactant) and (overall) molecularity, you need the rate law to be expressible in the form rate = c*[A]^n(A)[B]^n(B)[C]^n(C) etc. (where c is any constant independent of concentrations).

The question is asking under what approximations the given rate law will do as necessary to be rewritten in this form. So, what's the answer? I'll give you a hint: look for k/1=k.

As for the second part, you could try 1/([P]^n-1) against time, n≠1 and see if any value of n gives you a linear fit. In all likelihood the only one I can realistically think of is (t,[P]^-2) which would be 3rd-order.

[greentea11]

for the second part the data is

t/s 10   20     40     60     80  infinity
[P] 0.9  1.13 1.29  1.35  1.39 1.50

Have plotted [P] against time, its not linear so not zeroth order
 ln[P] against time, its not a linear line, so not first order

1/[P] is not a linear line so not second order

nor is 1/[P]^-2 linear, so not 3rd order

Although not sure how to plot the point at infinity?

Is that the right way to go about finding this order of reaction

EDIT: do I need to use any information from the first part?

[greentea11]

(First part) For the reaction to have a specific order (wrt each reactant) and (overall) molecularity, you need the rate law to be expressible in the form rate = c*[A]^n(A)[B]^n(B)[C]^n(C) etc. (where c is any constant independent of concentrations).

The question is asking under what approximations the given rate law will do as necessary to be rewritten in this form. So, what's the answer? I'll give you a hint: look for k/1=k.

As for the second part, you could try 1/([P]^n-1) against time, n≠1 and see if any value of n gives you a linear fit. In all likelihood the only one I can realistically think of is (t,[P]^-2) which would be 3rd-order.

Ah I see I need to get it into that form

Is it as simple as follows?:

when k'[A]  is negligible compared to k, then that term can be ignored, and it becomes just
 d[P]/dt =  k[A] [B] ^1/2

[Big-Daddy]

Is it as simple as follows?:

when k'[A]  is negligible compared to k, then that term can be ignored, and it becomes just
 d[P]/dt =  k[A] [B] ^1/2

I would have thought it would be when k'[A]<<1 or k'[A]>>1. In the former case we get d[P]/dt =  k[A][B]^1/2, in the latter case we get d[P]/dt = k/k' [B]^1/2 which also fits (since here we have a constant c, c=k/k', independent of the concentration of any species in the system).

[mjc123]

Have plotted [P] against time, its not linear so not zeroth order
 ln[P] against time, its not a linear line, so not first order

1/[P] is not a linear line so not second order

nor is 1/[P]^-2 linear, so not 3rd order

Although not sure how to plot the point at infinity?

Is that the right way to go about finding this order of reaction

As I was trying to hint in my earlier post, these expressions are those appropriate for the consumption of reactant, not for the appearance of product.
For example, a first order reaction A P; d[A]/dt = - k[A]; [A] = [A]₀e^-kt; ln[A] vs t is linear.
But d[P]/dt = k[A], [P] = [P]_∞(1-e^-kt) where [P]_∞ = [A]₀; ln[P] vs t is not linear, but ln([P]_∞ - [P]) vs t is.
For higher-order reactions the expression may vary depending on the exact stoichiometry, e.g. whether it's 2A  P, or A + B  2P, etc., but try plotting
{[P]_∞/([P]_∞ - [P])}^(n-1) vs t (for order n).

[greentea11]

Have plotted [P] against time, its not linear so not zeroth order
 ln[P] against time, its not a linear line, so not first order

1/[P] is not a linear line so not second order

nor is 1/[P]^-2 linear, so not 3rd order

Although not sure how to plot the point at infinity?

Is that the right way to go about finding this order of reaction

As I was trying to hint in my earlier post, these expressions are those appropriate for the consumption of reactant, not for the appearance of product.
For example, a first order reaction A P; d[A]/dt = - k[A]; [A] = [A]₀e^-kt; ln[A] vs t is linear.
But d[P]/dt = k[A], [P] = [P]_∞(1-e^-kt) where [P]_∞ = [A]₀; ln[P] vs t is not linear, but ln([P]_∞ - [P]) vs t is.
For higher-order reactions the expression may vary depending on the exact stoichiometry, e.g. whether it's 2A  P, or A + B  2P, etc., but try plotting
{[P]_∞/([P]_∞ - [P])}^(n-1) vs t (for order n).

Thank you, looks like plotting that with n = 2 is linear so that must be it
To get the rate constant, is that now the gradient of the line as opposed to negative gradient of line for consumption of product?

[mjc123]

In that case I think slope = (n-1)[P]_∞^(n-1)k. Try doing the maths.

[Big-Daddy]

In that case I think slope = (n-1)[P]_∞^(n-1)k. Try doing the maths.

Hmm what if the stoichiometric coefficients are not both 1? I got the slope to be

(n-1)[P]_∞^(n-1)(v(R)/v(P))^n-1v(R)k

where v(R) and v(P) are stoichiometric coefficients wrt the reactant and product each, e.g. such that d[R]/dt = -kv(R)[R]ⁿ.

I think without knowing anything about the stoichiometry all you can really calculate from the slope is (v(R)/v(P))^n-1v(R)k ?