Okay, this is a question in several parts that really stumps me:
Given the following reaction determine the rate of expression
(CH3)3CBr + OH -----> (CH3)3COH + r
[(CH3)3CBr] OH Initial reaction rate (mol L-1 sec-1)
.10 .10 1.0 x 10 ^ -3
.20 .10 2.0 x 10 ^ -3
.30 .10 3.0 x 10^ -3
.10 .20 1.0 x 10 ^-3
.10 .30 1.0 x 10 ^ -3
Rate expression (be sure to identify the unit of k):
If you have an initial concentration [(CH3)3CBr] of .45 and [OH] = .23, what is the initial rate of reaction?
Could you also add some explanation of how you got your answer so this could help me on tests that would be amazing! thank you!
To determine the rate expression for the given reaction, we need to analyze the data provided in the table. The rate of expression is typically represented by the letter k, which stands for the rate constant.
Based on the data provided, we can see that the reaction rate is dependent on the concentration of both (CH3)3CBr and OH. However, the concentration of OH remains constant throughout all the data sets, indicating that it is not involved in the rate expression. On the other hand, the concentration of (CH3)3CBr varies while keeping the concentration of OH constant.
To determine the rate expression, we need to compare the initial reaction rates for different concentrations of (CH3)3CBr while keeping OH constant. Looking at the data, we can see that when [(CH3)3CBr] is increased from 0.10 M to 0.20 M while OH remains constant at 0.10 M, the reaction rate doubles. Similarly, when [(CH3)3CBr] is increased from 0.10 M to 0.30 M while OH remains constant at 0.10 M, the reaction rate triples.
From this information, we can conclude that the rate expression for this reaction is as follows:
Rate = k[(CH3)3CBr]
Now, to find the unit of k, we examine the units given in the table. The rate is measured in mol L^-1 sec^-1, and the concentrations are in mol L^-1. Since the concentration of (CH3)3CBr is given in mol L^-1, the unit of k would be sec^-1.
To determine the initial rate of reaction, we can use the given concentrations of [(CH3)3CBr] = 0.45 M and [OH] = 0.23 M. Using the rate expression, Rate = k[(CH3)3CBr], we can substitute the given values to find the initial rate:
Rate = k * 0.45 M = (1.0 x 10^-3 mol L^-1 sec^-1) * 0.45 M = 4.5 x 10^-4 mol L^-1 sec^-1
Therefore, the initial rate of reaction is 4.5 x 10^-4 mol L^-1 sec^-1.
I hope this explanation helps you understand how to determine the rate expression and calculate the initial rate of reaction. Let me know if you have any further questions!