when the concentration of CHBr3 and NaOH are both .145, the rate of the reaction is .0040 M/s. What is the rate of the eaction if the concentration of CHBr3 is doubled?if the conc. is halved?if the two conc. are both increased by a factor of five?
CH3Br+NaOH-->CH3OH+NaBr
To determine the effect of changing the concentration of CHBr3 on the rate of reaction, we need to use the rate law for the reaction. The rate of a reaction is typically expressed as:
rate = k[CHBr3]^m[NaOH]^n
Where:
- rate is the rate of the reaction
- k is the rate constant
- [CHBr3] is the concentration of CHBr3
- [NaOH] is the concentration of NaOH
- m and n are the reaction orders with respect to CHBr3 and NaOH, respectively
Based on the given information, we know that when the concentration of both CHBr3 and NaOH is 0.145 M, the rate of the reaction is 0.0040 M/s. In this case, we can substitute the values into the rate law equation:
0.0040 M/s = k(0.145)^m(0.145)^n
To determine the rate of the reaction under different conditions, we can analyze the effect of changing the concentration of CHBr3:
1. If the concentration of CHBr3 is doubled:
- [CHBr3] becomes 2 * 0.145 = 0.29 M
- [NaOH] remains the same at 0.145 M
- The rate law equation becomes: 0.0040 M/s = k(0.29)^m(0.145)^n
- Since the concentration of CHBr3 is doubled, we have 0.29/0.145 = 2, so m = 1
- The rate of the reaction is directly proportional to the concentration of CHBr3, so the rate will also double:
= 2 * 0.0040 M/s = 0.0080 M/s
2. If the concentration of CHBr3 is halved:
- [CHBr3] becomes 0.5 * 0.145 = 0.0725 M
- [NaOH] remains the same at 0.145 M
- The rate law equation becomes: 0.0040 M/s = k(0.0725)^m(0.145)^n
- Since the concentration of CHBr3 is halved, we have 0.0725/0.145 = 0.5, so m = 0
- The rate of the reaction does not depend on the concentration of CHBr3, so the rate will remain the same:
= 0.0040 M/s
3. If both the concentration of CHBr3 and NaOH are increased by a factor of five:
- [CHBr3] and [NaOH] both become 5 * 0.145 = 0.725 M
- The rate law equation becomes: 0.0040 M/s = k(0.725)^m(0.725)^n
- Since both concentrations are increased by the same factor, we have 0.725/0.145 = 5, so m = n = 1/2
- The rate of the reaction is proportional to the square root of the concentration of CHBr3 and NaOH, so the rate will increase by a factor of sqrt(5) when both concentrations are multiplied by five:
= sqrt(5) * 0.0040 M/s ≈ 0.0089 M/s
Therefore, the rate of the reaction if the concentration of CHBr3 is doubled will be 0.0080 M/s, if the concentration is halved it will be 0.0040 M/s, and if both concentrations are increased by a factor of five, it will be approximately 0.0089 M/s.
To determine the effect of changes in concentration on the rate of a reaction, we can rely on the concept of reaction orders. The reaction order of each reactant gives us an indication of how the rate changes with respect to the concentration of that reactant.
In the given reaction, CHBr3 and NaOH are reactants, and their concentrations are initially 0.145 M. The rate of the reaction is 0.0040 M/s.
To determine the effect of doubling the concentration of CHBr3, we need to understand its reaction order. Let's assume it has a reaction order of "a." Now, we can set up a ratio to compare the rates before and after doubling the concentration of CHBr3:
Rate1 / Rate2 = ([CHBr3]1 / [CHBr3]2)^a
Since the initial concentration of CHBr3 is 0.145 M, and after doubling it becomes 2 * 0.145 M = 0.29 M:
Rate1 / Rate2 = (0.145 M / 0.29 M)^a
Since the reaction order of CHBr3 is not given, we cannot determine the exact rate without that information.
Similarly, to determine the effect of halving the concentration of CHBr3, we would use the ratio:
Rate1 / Rate2 = ([CHBr3]1 / [CHBr3]2)^a
Assuming the halved concentration is 0.145 M / 2 = 0.0725 M:
Rate1 / Rate2 = (0.145 M / 0.0725 M)^a
Again, without the reaction order of CHBr3, we cannot determine the exact rate change.
Finally, let’s determine the effect of both concentrations – CHBr3 and NaOH – being increased by a factor of five. Using the same ratio approach:
Rate1 / Rate2 = ([CHBr3]1 / [CHBr3]2)^a * ([NaOH]1 / [NaOH]2)^b
Since the concentrations are each increased by a factor of five:
Rate1 / Rate2 = (0.145 M / 5 * 0.145 M)^a * (0.145 M / 5 * 0.145 M)^b
Without the reaction orders of both CHBr3 and NaOH, we cannot determine the exact rate change for this scenario either.