Question

The kinetics of the reaction A + 3B → C + 2D were studied and the following results obtained, where the rate law is: – = k[A]n[B]m For a run where [A]0 = 1.0 × 10–3 M and [B]0 = 5.0 M, a plot of ln [A] versus t was found to give a straight line with slope = –5.0 × 10–2 s–1. For a run where [A]0 = 1.0 × 10–3 M and [B]0 = 10.0 M, a plot of ln [A] versus t was found to give a straight line with slope = –7.1 × 10–2 s–1.Reference: Ref 12-9 What is the value of n?

Answer 1

The problem deals with pseudo-nth order kinetics.
Since B is present in large excess, its change in concentration is negligible small. That means you can take [B] to be constant at [B]₀ throughout each experiment. The rate law changes to
-d[A]/dt = k∙[A]^n∙[B]₀^m = k' ∙[A]^n
k' = k∙[B]₀^m is the pseudo-nth order rate constant.

Since a plot of ln[A] versus t yields a straight line, the reaction is first order with respect to A.

the n value is - 1

(or)

in the above problem the slope of the reaction is Negative. so generally slope is Negative for 1st order reactions only.

for Zero order reactions the slope is zero, straight line passes through origin

For 2nd ,3rd , 4th.....n th order reactons the slope is Possitive/

So, finally the order of n is 1.