Question

Question #706ce

Answer 1

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

`["NO"_ 2]_ "new" = color(red)(2) * ["NO"_2]`

`["Cl"_ 2]_ "new" = color(red)(2) * ["Cl"_2]`

The rate of the reaction will now be

`"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`.