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) 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.

User PeteAUK
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1 Answer

4 votes

Answer:

The correct answer is option C.

Step-by-step 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* R

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

User Felypp Oliveira
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