Question

For a reaction, the expression for rate of reaction is given by: R=K[X]^2[y]. If the concentration of X and Y is increased by 3 times, then what will be the increase in the of reaction? Answer

Answer 1

Answer:

New rate = 27 × Older rate

here's the answer and hope it helps you.

Answer 2

Answer:

The rate of reaction increases by 27 times.

Explanation:

We are given that the expression for the rate of reaction to be $R = K[x^2][y]$ for the reaction $X + Y \rightarrow P$

When broken down, we can figure this out to be the rate of reaction is directly proportional to the square of the concentration of reactant X, and directly proportional to the concentration of reactant Y.

This means, if the concentration of X increases by two times, the rate of reaction increases by 4 times. Similarly, if the concentration of Y increases by two times, the rate of reaction increases by 2 times.

Now, we are given that the concentration of X and Y increases by 3 times.

Thus, X increases the rate of reaction by 9 times, and Y increases the rate of reaction by 3 times.

So, the reaction, overall increases by 27 times.

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