**The reaction has an initial rate of 0.0530 M/s. What will the initial rate be if [A] is halved and [B] is tripled?**

**Answer:**

Solution

Given data

Reaction equation

A+ B ----> C+D

Rate = k[A][B]2

Initial rate = 0.0530 M/s

What will the initial rate be if [A] is halved and [B] is tripled

What will the initial rate be if [A] is tripled and [B ] is halved

Lets first calculate the rate constant for the reaction

Assume concentration of the [A] and [B] = 1

Now lets use these concentration to calculate the initial rate constant

Rate = k[A][B]2

0.0530 M/s = k [ 1] [1]2

K = 0.0530 M/s / 1

K= 0.0530 M/s

**Now lets use this rate constant and calculate the initial rates when the [A] is halved and [B] is tripled**

So new concentration of [A] = 1 /2 = 0.5 M

And [B] = 1 * 3 = 3 M

Rate = k[A][B]2

Rate = 0.0530 M/s * [0.5] [3]2

**Rate = 0.238 M/s**

**Now lets calculate rate when [A] is tripled and [B] is halved**

Then concentration of the [A] = 3 m and [B] = 0.5 M

Lets use these concentrations and calculate the rate

Rate = k[A][B]2

Rate = 0.0530 M/s*[3][0.5]2

**Rate = 0.0398 M/s**