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Novay_Z [31]
2 years ago
11

Given the following rate law, how does the rate of reaction change if the concentration of x is doubled? rate = k [x]2[y]3

Mathematics
1 answer:
tiny-mole [99]2 years ago
8 0

The rate of reaction will increase by a factor of 4. The correct option is D.

<h3>What is the rate of reaction?</h3>

In a reaction mechanism, a chemical reaction is depicted step-by-step. It is composed of separate chemical processes known as elementary stages. An elementary reaction is one that only requires one step to complete.

In an elementary reaction, the orders relative to each reactant are equal to the stoichiometric coefficient of that reactant. A provided rate law must be consistent with both the experimental rate law and the overall rate law that is being suggested.

The rate of reaction will be calculated as below:-

Rate (R) = k[x][y]²

R' = k[x][2y]²

R'=4k[x][y]²

R'=4R

Hence, the rate of reaction will increase by a factor of 4.

The complete question is given below:-

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.

Learn more about the rate of reaction here:

brainly.com/question/8592296

#SPJ4

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