For a given reaction 2A + B --> C, if the concentration of A doubles then the reaction rate doubles. If the concentration of both A and B doubles then the reaction rate increases by factor of 8. What is the order of the reaction with respect to B?
To determine the order of the reaction with respect to B, we need to analyze the effect of changing the concentration of B on the reaction rate.
We are given that when the concentration of A doubles, the reaction rate also doubles. This implies that the reaction is first order with respect to A.
When both the concentrations of A and B are doubled, we are told that the reaction rate increases by a factor of 8. This means that the rate of the reaction becomes 8 times the original rate. We can express this mathematically as:
(rate with doubled concentrations of A and B) = 8 * (rate with original concentrations of A and B)
If we let the original rate of the reaction be denoted as r, and the concentrations of A and B be denoted as [A] and [B], respectively, the equation becomes:
(k*[A]^[order of A]*[B]^[order of B]) = 8 * (k*[A]^[order of A]*[B]^[order of B])
Here, k is the rate constant and the orders of A and B are the exponents that determine how the reaction rate depends on their concentrations.
Since we know that the order of A is 1 (because doubling [A] doubles the rate), we can rewrite the equation as:
(k*[A]*[B]^[order of B]) = 8 * (k*[A]*[B]^[order of B])
Canceling the k and [A] terms on both sides of the equation, we have:
[B]^[order of B] = 8
To find the order of B, we need to determine what power we need to raise 2 to in order to obtain 8. In this case, we find that the order of B is 3, because 2^3 = 8.
Therefore, the order of the reaction with respect to B is 3.