Write the rate law for the following reaction and identify whi...

Question

Write the rate law for the following reaction and identify which molecules are present in the rate-determining step. Draw a possible transition state and propose a mechanism.

Chemical reaction showing bromide and hydroxide ions converting to alcohol and bromide ion.

Table displaying concentrations of [RBr] and [HO-] with corresponding reaction rates for control and trials.

Accepted Answer

Welcome back, everyone consider the following reaction. Benzyl chloride reacts with metox to produce an ether and chloride anion. Write the rate law for the reaction and identify the species involved in the rate determining step. We're also giving, we're also given the table with the rates measured for different concentrations of our starting materials. First of all, what we notice is that this is a substitution reaction because the chloral group is replaced with the metox group. We are given a primary alkyl hide which reacts with a strong nuclei because meth oxide has a negative charge. And therefore, we can conclude that this must be an essence to reaction. A bi molecular substitution reaction will have a rate which will depend on the concentration of the alkali and the nuclei. So we can say that the rate is equal to some rate constant K multiplied by the concentration of benzyl chloride, raise the power of a which we don't know yet and then multiplied by the concentration of the th side race to some power. Why our goal is to identify the value of K, the weight constant as well as the orders X and Y. We are always going to start with identifying the values of X and Y. First, generally, we can use the law of initial rates with varying concentrations, but we can simplify it really by just observing the data in the table. So let's suppose that we are simply interested in finding the value of X. So let's say that first of all, we are finding X, if we want to find the value of X, we want to make sure that we are choosing two trials where the concentration of meth oxide is exactly the same. So let's suppose that we are taking control in trial one because in those trials, the concentration of met oxide is exactly the same. Now, the concentration of benzyl chloride doubles if we go from control to trial one. So we can say that the concentration of benzyl doubles. And now if we observe the rate, we also notice that our rate doubles because we can get rid of our exponents. And notice that if we divide the rate from trial one by the rate of control, we get two. So the rate that is what meaning they are directly proportional and the order X must be equal to one. Now, similarly, we can find the value of why we want to choose two trials in which the value of the concentration of benzyl chloride is exactly the same. So we can go with control and trial to what we notice is that the concentration of meth oxide doubles once again. And now what happens to our rate? It essentially doubles again, right? We go from 2.02 to 4.04. So the rate doubles meaning they are directly proportional and there should be an exponent of one or basically speaking, why should be equal to one as a counter example, let's say if you use this method, by analogy, you can say that if you double your concentration, but let's say your weight quadruples, then the exponent would be two, right? Because when you're doubling it two, raised to the power of two would be equal to four. But whenever you're doubling your concentration and you get a double in your rate, that means your order will always be equal to one well done. So now that we have our values of X and Y, we essentially want to find the rate constant. And if we rearrange our expression for the rate constant K, we will essentially divide our rate by the concentration of benzoyl chloride, raise the power of X, which is one. We don't need to write that one. It's implicit multiplied by the concentration of meth oxide raised to the power of one. Once again. Now generally, we would need to calculate the rate constant for each trial. So we would say that the overall weight constant would be the average of the four data points that were given, we would take 1/4 multiplied by the sum of K one K two K three NK four. In other words, we would calculate the weight constant for each data set or data point. That would be more accurate to say we would add them together and divide by four to get the average. But in this case, we are lucky to see that our numbers are pretty ideal because whenever we are doubling the concentration, the rate doubles exactly in each case, meaning our data set is perfect. It's not experimental in measure, but it's theoretical. So it's sufficient for us to take any data point that we want. And we will get the same rate constant as the average itself. So let's suppose that we take control, meaning our rate constant K will be the rate of control. That would be 2.02 multiplied by 10 of the power of negative sixth molar per second. And on the bottom, we would essentially take the product of our concentrations. So let's see that the concentration of benzyl fluoride is 0.0250 molar. And we're going to multiply it by the concentration of meth oxide, which is also 0.0250 molar. Now, our goal is to essentially calculate the result to three significant figures and we end up with 3.23 multiplied by 10 raised to the power of negative third. Because this is a second order rate expression because one plus one equals two, we can state that the units would be molar to the negative first multiplied by second to the negative. First, we can also notice this from our calculation. If we take one of our molar, we can cancel it out with the molar term on the bottom. And we end up with one over M which is molar to the negative first. And of course, we have that one divided by second, which is second to the power of negative first, well done. So now that we have everything that's required, we can substitute everything into the right expression. So we can say that the rate low would be R equals 3.23 multiplied by 10 to the power of negative third molar raised to the negative. 1st, 2nd, raises to the negative first multiplied by the concentration of benzyl chloride and the concentration of meth oxide. And as we see from the rate expression, as well as knowing that this is an essence to reaction in which the way determining step depends on both substances. We can state that both species are involved in the way, determining step or NDR DS, which is the way determining step. And that would be our final answer. Thank you for watching.

Write the rate law for the following reaction and identify which molecules are present in the rate-determining step. Draw a possible transition state and propose a mechanism.

Key Concepts

Rate Law

Rate-Determining Step

Transition State Theory

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