For the reaction,3A + 2B to C + D,the differe | vedclass

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(A) For a general reaction $aA + bB 	o cC + dD$,the rate of reaction is expressed as:Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{

Vedclass · Accepted Answer

$-\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k[A]^n[B]^m$

Vedclass · Answer

(A) For a general reaction $aA + bB 	o cC + dD$,the rate of reaction is expressed as:Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} = k[A]^n[B]^m$Given the reaction $3A + 2B 	o C + D$,the rate expression is:Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt} = k[A]^n[B]^m$Comparing this with the given options,the correct representation is $-\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k[A]^n[B]^m$.

For the reaction,$3A + 2B \to C + D$,the differential rate law can be written as:

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