# Question 706ce

May 1, 2017

The answer is (E) $8$.

#### Explanation:

You know that

color(blue)("rate" = k * ["NO"_2]^2 * ["Cl"_2])

You now double the concentrations of the two reactants

$\left[{\text{NO"_ 2]_ "new" = color(red)(2) * ["NO}}_{2}\right]$

$\left[{\text{Cl"_ 2]_ "new" = color(red)(2) * ["Cl}}_{2}\right]$

The rate of the reaction will now be

$\text{rate"_ "new" = k * ["NO"_ 2]_ "new" * ["Cl"_ 2]_ "new}$

This is equivalent to

"rate"_ "new" = k * (color(red)(2) * ["NO"_2])^2 * (color(red)(2) * ["Cl"_2])

"rate"_ "new" = k * color(red)(2)^2 * color(red)(2) * ["NO"_2] * ["Cl"_2]

"rate"_ "new" = 8 * color(blue)(k * ["NO"_2] * ["Cl"_2])

This means that you have

"rate"_ "new" = 8 * color(blue)("rate")#

The rate of the reaction will increase by a factor of $8$.