The following reaction is a single-step, bimolecular reaction:
CH3Br + NaOH --> CH3OH + NaBr
When the concentrations of CH3Br and NaOH are both 0.135 M, the rate of the reaction is 0.0070 M/s.
(a) What is the rate of the reaction if the concentration of CH3Br is doubled?
(b) What is the rate of the reaction if the concentration of NaOH is halved?
(c) What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of six?
(a) If the concentration of CH3Br is doubled, then the rate of the reaction will shout "Two times the fun!" and also double. So, the rate would be 0.0140 M/s.
(b) If the concentration of NaOH is halved, it might feel a bit low and less enthusiastic, but the rate of the reaction will also be halved. So, the rate would be 0.0035 M/s, taking it easy like a half-filled glass.
(c) If the concentrations of both CH3Br and NaOH are increased by a factor of six, then we would have a fantastic party going on! The rate of the reaction would also increase by a factor of six, making it a whopping 0.0420 M/s. Time to put on some party music!
To determine the rates of the reaction under different conditions, we can use the rate law expression for this bimolecular reaction. The rate law for this reaction can be represented as:
Rate = k[CH3Br][NaOH]
where k is the rate constant.
(a) If the concentration of CH3Br is doubled, the new concentration becomes 2 * 0.135 M = 0.27 M.
Using the rate law expression:
Rate1 = k[CH3Br]1[NaOH]
Rate2 = k[CH3Br]2[NaOH]
We can set these two rates equal to each other:
Rate1 = Rate2
k[CH3Br]1[NaOH] = k[CH3Br]2[NaOH]
k[CH3Br]1 = k[CH3Br]2 / [NaOH]
Simplifying the equation:
Rate2 = Rate1 * [CH3Br]2 / [CH3Br]1
Plugging in the values:
Rate2 = 0.0070 M/s * 0.27 M / 0.135 M = 0.0140 M/s
Therefore, if the concentration of CH3Br is doubled, the rate of the reaction becomes 0.0140 M/s.
(b) If the concentration of NaOH is halved, the new concentration becomes 0.135 M / 2 = 0.0675 M.
Using the rate law expression and following the same steps as in part (a):
Rate2 = Rate1 * [NaOH]2 / [NaOH]1
Rate2 = 0.0070 M/s * 0.135 M / (0.135 M / 2) = 0.0070 M/s * 2 = 0.0140 M/s
Therefore, if the concentration of NaOH is halved, the rate of the reaction becomes 0.0140 M/s.
(c) If the concentrations of CH3Br and NaOH are both increased by a factor of six, the new concentrations become 6 * 0.135 M = 0.81 M.
Using the rate law expression and following the same steps as before:
Rate2 = Rate1 * [CH3Br]2 / [CH3Br]1
Rate2 = 0.0070 M/s * 0.81 M / 0.135 M = 0.0420 M/s
Therefore, if the concentrations of CH3Br and NaOH are both increased by a factor of six, the rate of the reaction becomes 0.0420 M/s.
To determine the rate of the reaction under different conditions, we can use the rate law expression. The rate law expression describes the relationship between the rate of a reaction and the concentrations of the reactants. In this case, the rate law can be written as:
Rate = k[CH3Br][NaOH]
Where [CH3Br] is the concentration of CH3Br, [NaOH] is the concentration of NaOH, and k is the rate constant.
(a) To find the rate of the reaction if the concentration of CH3Br is doubled, we can assume that the concentration of NaOH remains constant. The rate law expression indicates that the rate is directly proportional to the concentration of CH3Br. Therefore, if we double the concentration of CH3Br, the rate will also double.
So, the rate of the reaction if the concentration of CH3Br is doubled is 0.0070 M/s * 2 = 0.014 M/s.
(b) To find the rate of the reaction if the concentration of NaOH is halved, we can assume that the concentration of CH3Br remains constant. The rate law expression shows that the rate is directly proportional to the concentration of NaOH. Therefore, if we halve the concentration of NaOH, the rate will also be halved.
So, the rate of the reaction if the concentration of NaOH is halved is 0.0070 M/s * (1/2) = 0.0035 M/s.
(c) To find the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of six, we can multiply the initial rate by 6 for both concentrations.
So, the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of six is 0.0070 M/s * 6 = 0.042 M/s.
Remember that these calculations are based on the assumption that the reaction is first-order with respect to both CH3Br and NaOH. If the reaction is not first-order, the rate may not change linearly with the concentration changes.