Question #97778

The following reaction is second order in A and first order in B: 2A + B → 3C.
Determine the rate law for the reaction 4A + 2B→6C.

Expert's answer

The rate law for the first reaction is


2A+B3C2A + B\rightarrow 3C

V=[A]2[B]V = [A]^2[B]

We can see that in the second reaction the concentration of all species is doubled, then the rate law is:


4A+2B6C4A + 2B \rightarrow 6C

V=[2A]2[2B]=22[A]22[B]=8[A]2[B]V = [2A]^2[2B] = 2^2[A]^22[B]= 8[A]^2[B]

we can see then when we double the concentration of all species the rate of the reaction increases eight times.


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