contestada

) 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.

Respuesta :

Answer:

The correct answer is option C.

Explanation:

The given rate law of reaction :

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

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

[tex]R'=k[X][2Y]^2[/tex]

[tex]R'=k[X]4[Y]^2[/tex]

[tex]R'=4k[X][Y]^2[/tex]

[tex]R'=4\times R[/tex]

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

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