RATES: If you double the concentration of reactant A in the reaction A + 2B �¨ C, what happens to the rate based on the rate law, rate = [A]2 [B]3? The rate would ___ by a factor of ___.
increase, 2
increase, 4
decrease, 2
decrease, 4
I can't read some of the signs; however, I THINK you probably wrote
A + 2B ==> C
then rate= k[A]^2[B]^3.
Doubling A will square that so rate would increase by 4.
Actually, as the coefficents determine the rate powers, the rate equation would actually be:
rate=k[A]^1[B]^2
To determine the effect on the rate based on the rate law, rate = [A]^2 [B]^3, you need to consider how doubling the concentration of reactant A would change the rate.
In the rate law, the exponent of [A] is 2. This means that the rate is proportional to the square of the concentration of A. Doubling the concentration of A would result in the concentration being multiplied by 2, and squaring that value would give you 2^2 = 4.
Therefore, doubling the concentration of A would result in the rate increasing by a factor of 4. So, the correct answer would be: increase, 4.