For the reaction: 3A(. g .) arrow 2B(. g .) , the rate of formation of B at 298K is represented as ln((d [B]/dt))= -4.606+2ln [A] . The order of reaction is

Question

For the reaction: 3A(. g .) arrow 2B(. g .) , the rate of formation of B at 298K is represented as ln((d [B]/dt))= -4.606+2ln [A] . The order of reaction is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

ln((d [B]/d t))=-4.606+2ln [.A]. 2.303log((d [B]/dt))=-4.606+2× 2.303 log [A] log((d [B]/dt))=(- 4 .606/2 .303)+2log[A] log((d [B]/dt))=-2+2 log[A] log((d [B]/dt))=log (10)- 2+log([A])2 log((d [B]/dt))=log((10)- 2 ([A])2) ((d [B]/d t))=(1 0- 2 ([A])2) Compare with ((d [B]/d t))=k([A])m ((d [B]/d t))=(1 0- 2 ([A])2) Order = 2

Q. For the reaction: 3A(g)​→2B(g)​ , the rate of formation of B at 298K is represented as ln(dtd[B]​)=−4.606+2ln[A] . The order of reaction is

0

1

2

3

Q. For the reaction: $3 A_{(g)} \to 2 B_{(g)}$ , the rate of formation of $B$ at $298 K$ is represented as $l n (\frac{d [ B ]}{d t}) = - 4.606 + 2 l n [A]$ . The order of reaction is