Question

Question #07304

Answer 1

Quadruple.

Explanation:

You know that the reaction

`A + 3B -> 2C`

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

`"rate"_1 = k * ["A"]^2" "`, where

`k` - the rate constant

So, if the concentration of `A` goes from `["A"]` to `2 * ["A"]`, 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,

`color(green)("rate"_2 = 4 xx "rate"_1) ->` the rate of the reaction will quadruple

Answer 2

It will be 4 times as fast.

Explanation:

For concentration of `c`:

`r=k*c^2`

For double concentration:

`r'=k*(2c)^2`

`r'=k*2^2*c^2`

`r'=4*(k*c^2)`

`r'=4*r`