# 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?

Dec 15, 2015

By a factor of $8$.

#### Explanation:

$\text{A" + 2"B" -> "products}$

the rate law takes the form

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

As you can see, the rate is proportional to the concentration of $\text{A}$ and the square of the concentration of $\text{B}$.

Let's say that initially, you have $\left[\text{A}\right] = x$ and $\left[\text{B}\right] = y$. The rate law in this case is equal to

$\text{rate 1} = k \cdot x \cdot {y}^{2}$

Now you double the concentrations of $\text{A}$ and $\text{B}$ to new values of $2 x$ and $2 y$. The new rate law will be

$\text{rate 2} = k \cdot \left(2 x\right) \cdot {\left(2 y\right)}^{2}$

$\text{rate 2} = k \cdot 2 x \cdot 4 {y}^{2} = 8 \cdot x \cdot {y}^{2}$

As you can see, you have

$\text{rate 2"/"rate 1} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} \cdot 8 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x \cdot {y}^{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x \cdot {y}^{2}}}}} = \textcolor{g r e e n}{8}$

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