Question

For the given reaction 2A + 3B C , If i) the rate of reaction doubles when conc. of A is doubled ii) the rate remains the same when the conc. of B is halved. Find the rate law for the given reaction. Also write the unit of rate constant for this reaction.​

Answer 1

Answer:

The expression for the rate of the reaction is r=k[A]n[B]m−(1)

When the concentration of A is doubled and the concentration of B remains same, the rate expression becomes 2r=k[2A]n[B]m−(2)

When the concentration of both A and B is doubled, the rate expression becomes 8r=k[2A]n[2B]m−(3)

Divide equation (2) with equation (1) to obtain n=1

Divide equation (3) with equation (1) to obtain n+m=3

Hence, m=2

The overall order of the reaction is =1+2=3.

Explanation:

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