Respuesta :
Answer:
The rate of reaction triples up if the concentration of B is tripled.
Explanation:
The rate law is given as
r = k [A]² [B]
The rate is 2nd order with respect to [A] and first order with respect to [B]
If the concentration of A is held constant, the rate law can be written as
r = K [B]
where K = k [A]²
If initially the concentration of B is x
r = K [x]
If it is then changed to 3x, the rate becomes
r (new) = K [3x] = 3 K [x]
r (new) = 3 × r
Hence, the rate of reaction triples up if the concentration of B is tripled.
The new rate of the reaction has been increased with the factor of 3 for the increase in the concentration of the B.
The rate of reaction has been given as the amount of reactant disappeared or the rate of formation of product in unit time.
Increase in rate of reaction
For the given reaction, the rate law has been given as:
[tex]\rm Rate=k[A]^2[B][/tex]
The concentration of A has been constant. Thus, the rate law with respect to B has been:
[tex]\rm Rate=k[B]\\Rate=\textit a[/tex]
The increase in the concentration of B has been 3 times. Thus, the new concentration of B has been 3B.
The new rate law (Rate') of the reaction has been:
[tex]\rm Rate'=k[3B]\\Rate'=3\;k[B]\\Rate'=3\;Rate[/tex]
The new rate of the reaction has been increased with the factor of 3 for the increase in the concentration of the B.
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