# Question 07304

Jan 13, 2016

#### Explanation:

You know that the reaction

$A + 3 B \to 2 C$

is second order with respect to $A$. Keep in mind, no information was provided about the overall order of the reaction, so you can't say that the reaction is also second order overall.

So, a general form for the rate would be

"rate" = -(d["A"])/(dt) = - 1/3(d["B"])/dt = 1/2(d["C"])/dt

Now, you don't really need the overall order of the reaction, since you are told that everything is kept constant with the exception of the concentration of $A$, which is said to double.

Keeping everything constant implies that the rate of the reaction will change exclusively with the change of the concentration of $A$.

You can thus say that the rate can be expressed, in this particular case, as

$\text{rate"_1 = k * ["A"]^2" }$, where

$k$ - the rate constant

So, if the concentration of $A$ goes from $\left[\text{A}\right]$ to $2 \cdot \left[\text{A}\right]$, you can say that

"rate"_2 = k * (2 * ["A"])^2

"rate"_2 = k * 4 * ["A"]^2 = 4 * overbrace(k * ["A"]^2)^(color(red)("rate"_1))#

Therefore,

$\textcolor{g r e e n}{{\text{rate"_2 = 4 xx "rate}}_{1}} \to$ the rate of the reaction will quadruple

Jan 13, 2016

It will be 4 times as fast.

#### Explanation:

For concentration of $c$:

$r = k \cdot {c}^{2}$

For double concentration:

$r ' = k \cdot {\left(2 c\right)}^{2}$

$r ' = k \cdot {2}^{2} \cdot {c}^{2}$

$r ' = 4 \cdot \left(k \cdot {c}^{2}\right)$

$r ' = 4 \cdot r$