Question

) Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? Rate = k [X][Y]2 A) The rate of reaction will increase by a factor of 2. B) The rate of reaction will increase by a factor of 5. C) The rate of reaction will increase by a factor of 4. D) The rate of reaction will decrease by a factor of 2. E) The rate of reaction will remain unchanged.

Answer 1

Answer:

The correct answer is option C.

Explanation:

The given rate law of reaction :

$Rate(R) = k [X][Y]^2$

Rate of the reaction after doubling the concentration of Y : R'

$R'=k[X][2Y]^2$

$R'=k[X]4[Y]^2$

$R'=4k[X][Y]^2$

$R'=4\times R$

The rate of the reaction will increase by the factor of 4.