A certain reaction has the following general form.

aA → bB

At a particular temperature and [A]0 = 2.80 ✕ 10-3M, concentration versus time data were collected for this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of + 3.00 ✕ 10-2 L mol-1 s-1.

(a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. (Rate expressions take the general form: rate = k. [A]a. [B]b.)

rate law = k [A]

Integrated rate law: ln [A] = - kt + ln[A]₀

m = - 3.00 x 10⁻² Lmol⁻¹ s⁻¹

Explanation:

To determine the rate law we need to know the order of the reaction respect to the concentration of A.

In general, the rate law for a given equation is:

r = K [A ]^n

where rate, r, is the change in time of the concentration of A, and n is the order of the reaction.

So what we need to solve this question is find out which order of reaction conforms with the fact that a plot of 1/[A] versus time resulted in a straight line.

If zero order:

ΔA/Δt = - k [A]º = - k ⇒ ΔA = - k Δt

From calculus:

∫ [A] d[A] = - ∫ kdt ⇒ [A] = - kt + [A]₀

A graph of this this equation will result in a straight line only graphing [A] versus time, and not 1/[A] vs time as stated in the question.

If first order.

r = - k[A] ⇒ ΔA/Δt = - k[A]

Δ[A]/[A] = - kΔt

A plot of 1/[A] vs t will result in a straight, so we now know the reaction is first order.

from calculus we know that this integrated gives us:

∫d[A]/[A] = - ∫k dt

and the integral is:

ln [A] = - kt + ln[A]₀ where [A]₀ is the intial concentration of A.

you can see this equation has the form y = mx + b

So our reaction is first order, the integrated rate law is ln [A] = - kt + ln[A]₀, and the values of the rate constant is the negative of the slope:

m = - k ⇒ k = - m = - 3.00 x 10⁻² Lmol⁻¹ s⁻¹

In case you are wondering about the units for k : we are plotting 1/[A] vs time so it follows k will have the units of L/mol per s.