The reaction A+2B →products , was found to have the rate law , rate= k[A][B]². Predict by what factor the rate of reaction will increase when the concentration of A and B is doubled?

1 Answer
Dec 15, 2015

Answer:

By a factor of #8#.

Explanation:

For your reaction

#"A" + 2"B" -> "products"#

the rate law takes the form

#"rate" = k * ["A"] * ["B"]^2#

As you can see, the rate is proportional to the concentration of #"A"# and the square of the concentration of #"B"#.

Let's say that initially, you have #["A"] = x# and #["B"] = y#. The rate law in this case is equal to

#"rate 1" = k * x * y^2#

Now you double the concentrations of #"A"# and #"B"# to new values of #2x# and #2y#. The new rate law will be

#"rate 2" = k * (2x) * (2y)^2#

#"rate 2" = k * 2x * 4y^2 = 8 * x * y^2#

As you can see, you have

#"rate 2"/"rate 1" = (color(red)(cancel(color(black)(k))) * 8 * color(red)(cancel(color(black)(x * y^2))))/(color(red)(cancel(color(black)(k))) * color(red)(cancel(color(black)(x * y^2)))) = color(green)(8)#

Therefore, the rate of the reaction has increased by a factor of #8#.