The following reaction is a single-step, bimolecular reaction:
CH3Br + NaOH ---> CH3OH + NaBr
When the concentrations of CH3Br and NaOH are both 0.150 M, the rate of the reaction is 0.0090 M/s.
a) what is the rate of the reaction if the concentration of CH3Br is doubled?
______ M/s
b) What is the rate of the reaction if the concentration of NaOH is halved?
______ M/s
c) What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of five?
______ M/s
Please help! I have no idea how to start this problem off!
jjh
bro, not cool
To solve this problem, we can use the concept of reaction rates and the rate law expression. The rate law expression for a bimolecular reaction is given by:
rate = k[A][B]
Where:
- rate is the rate of the reaction,
- k is the rate constant (specific to the reaction), and
- [A] and [B] are the concentrations of the reactants A and B, respectively.
Let's break down the problem into three parts:
a) When the concentration of CH3Br is doubled:
We need to find the rate of the reaction. Since the reaction is a single-step bimolecular reaction, the rate law expression will be:
rate = k[CH3Br][NaOH]
Now, we know that the concentration of CH3Br is doubled. So, the new concentration of CH3Br will be 2 * 0.150 M = 0.300 M.
Substituting the given values into the rate law expression, we have:
rate = k * (0.300 M) * (0.150 M) = 0.0090 M/s
Therefore, the rate of the reaction, when the concentration of CH3Br is doubled, is 0.0090 M/s.
b) When the concentration of NaOH is halved:
Similar to part a, we need to find the rate of the reaction. The rate law expression remains the same:
rate = k[CH3Br][NaOH]
Now, the concentration of NaOH is halved. So, the new concentration of NaOH will be 0.150 M/2 = 0.075 M.
Substituting the given values into the rate law expression, we have:
rate = k * (0.150 M) * (0.075 M) = 0.0090 M/s
Therefore, the rate of the reaction, when the concentration of NaOH is halved, is 0.0090 M/s.
c) When the concentrations of CH3Br and NaOH are both increased by a factor of five:
Again, we need to find the rate of the reaction using the rate law expression:
rate = k[CH3Br][NaOH]
Now, both the concentrations of CH3Br and NaOH are increased by a factor of five. So, the new concentrations will be 5 * 0.150 M = 0.750 M.
Substituting the given values into the rate law expression, we have:
rate = k * (0.750 M) * (0.750 M) = 0.0090 M/s
Therefore, the rate of the reaction, when the concentrations of CH3Br and NaOH are both increased by a factor of five, is 0.0090 M/s.
In summary:
a) The rate of the reaction, when the concentration of CH3Br is doubled, is 0.0090 M/s.
b) The rate of the reaction, when the concentration of NaOH is halved, is 0.0090 M/s.
c) The rate of the reaction, when the concentrations of CH3Br and NaOH are both increased by a factor of five, is 0.0090 M/s.
According to the problem, the rate constant, k, has units of M/s which is moles/L*s and that is for a zero order equation. The equation then is r = k and the concns don't enter into the equation at all. You can read more about it here.
http://www.chm.davidson.edu/vce/kinetics/DifferentialRateLaws.html