Question

A certain reaction has the following general form.

Answer 1

Answer:

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.