Question #07304

2 Answers
Jan 13, 2016

Answer:

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

Jan 13, 2016

Answer:

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#