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.
