the rate limiting step is unimolecular with A as the sole reactant.
A+B-----C+D
If A and B are both 0.200M, then the rate reaction is 0.0050M/s
A. What is the rate of the reaction if A is doubled?
B. starting with original concentrations what is the rate reaction id B is halved?
C. Starting with original concentration what is the rate reaction if A and B are both increased by a factor of six?
To determine the rate of the reaction in different scenarios, we need to understand the concept of the rate-determining step and how it relates to the rate equation.
In this given reaction: A + B → C + D, it is mentioned that the rate-limiting step is unimolecular and involves only reactant A. This implies that the reaction rate is directly proportional to the concentration of A raised to a certain power, and the concentration of B has no effect on the rate. Let's denote the power of A as n. So, the rate equation can be written as follows:
Rate = k[A]^n
Given that the initial concentrations are both 0.200 M and the rate of reaction is 0.0050 M/s, we can substitute these values into the rate equation and solve for k and n. However, since the value of k is not explicitly given, we can use the ratio method to determine the effect of changing the concentrations of A and B on the rate of reaction.
A. What is the rate of the reaction if A is doubled?
If A is doubled to 0.400 M, we need to determine the new rate. Since the concentration of B remains unchanged, we can use the ratio of the new concentration to the original concentration to find the ratio of the new rate to the original rate. Let's calculate it:
Ratio = ([A]_new / [A]_original)^n
Ratio = (0.400 M / 0.200 M)^n
Assuming n is a constant, the ratio will be the same for any value of n. Therefore, we can write:
(0.400 M / 0.200 M)^n = (2)^n
So, if A is doubled, the rate of reaction will be 2^n times the original rate.
B. Starting with the original concentrations, what is the rate reaction if B is halved?
Since B has no effect on the rate of reaction, halving its concentration will not change the rate. Therefore, the rate will remain the same.
C. Starting with the original concentrations, what is the rate of reaction if A and B are both increased by a factor of six?
If A and B are increased by a factor of six, their new concentrations will be:
[A]_new = 6 * [A]_original
[B]_new = 6 * [B]_original
Again, using the ratio method, we can find the ratio of the new rate to the original rate:
Ratio = ([A]_new / [A]_original)^n * ([B]_new / [B]_original)^0
Since [B] does not affect the rate, its exponent is 0. Therefore, the ratio simplifies to:
Ratio = (6/1)^n = 6^n
So, if both A and B are increased by a factor of six, the rate of reaction will be 6^n times the original rate.
Please note that without the specific values of k and n, we cannot determine the exact numerical rate values. These calculations only provide the relationships between the different scenarios in terms of ratios.