Question

Consider the rate below

R=K[A][B]^2
Which step would quadruple the rate?
doubling the concentration of A
doubling the concentration of B
doubling the concentration of both A and B
doubling the concentration of A but halving the concentration of B

Answer 1

Doubling the concentration of B.
if you put in random numbers you will see the effects of the equation.

Answer 2

B.Doubling the concentration of B.

Step-by-step explanation:

Consider the rate below

`R=k[A][B]^2`

We have to find the rate of reaction is quadruple means the rate of reaction is 4 times the rate of initial rate of reaction.

Let [A}=a,[B]=b then

`R_1=kab^2`

a. doubling the concentration of A

It means [A]=2a

Therefore , substituting the value

Then we get the rate of reaction

`R_2=k[2a][b]^2=2kab^2=2R_1`

Therefore , the rate of reaction is 2 times the initial rate of reaction.Hence, option A is false.

B.Doubling the concentration of B

It means [B]=2b

Therefore , substituting the value

Then we get the rate of reaction

`R_2=k*a*(2b)^2=4 kab^2=4R_1`

Hence, the rate of reaction is 4 times the initial rate of reaction .Therefore, the rate of reaction is quadruple.Therefore, option B is true.

C.Doubling the concentration of both A and B

[A]=2a, [B]=2b

Substitute the values then the rate of reaction

`R_2=k(2a)(2b)^2`

`R_2=8kab^2=8R_1`

Hence, the rate of reaction is 8 times the initial rate of reaction .Therefore, it is not quadruple.Therefore, option C is false.

D.Doubling the concentration of A but halving the concentration of B

[A]=2a, [B]=
`(1)/(2)b`

Substitute the values then the rate of reaction

`R_2=k(2a)((1)/(2)b)^2=(2kab^2)/(4)=(1)/(2)kab^2`

`R_2=(1)/(2)R_1`

Hence, the rate of reaction is
`(1)/(2)` times the rate of initial rate of reaction.Therefore, option D is false.