The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.) C 2 H 5 Br(aq) + OH - (aq) → C 2 H 5 OH(l) + Br - (aq) Temperature (°C) k (L,mol •s) 25 8.81⨉10 -5 35 0.000285 45 0.000854 55 0.00239 65 0.00633 c. If a reaction mixture is 0.155 M in C 2 H 5 Brand 0.250 M in OH - , what is the initial rate of the reaction at 75 °C?

Hey everyone welcome back. So let's get started with this video. So here we're giving a reaction and we're told that at 25°C the rate of the reaction was measured using three different sets of initials concentrations. And then we're given a table with measurements. So for a they want us to determine the rate law for the reaction. Okay and recall that for rate law, it always follows the format. Oh really? Is equal to the rate constant K. In terms of concentration of reacting. So here we have these two reactant so then we have the concentration of X. And the concentration of the other reactant. Which is why but don't forget that they're also raised to a power. So here this is what we're going to be solving for which is an order. So we can put a. And be okay so we don't know the order. So we're going to have to figure that out. So let's start by let me move this screen down a bit so we can do some math. So let's start by finding a. So this would be the order for X. So then the way we would do this is they will use the table. Thank you. So this is for ex so what we're going to want to do is so the one we're looking for which is X. We're going to look at the table And find two values that are different. And then for the other reactant which is why we're going to find values that are the same. So remember for the value we're looking for we want different numbers. And for the other one we're going to want the same. So then we're looking for X. So the different numbers would be leased to Because if you see for the Y .2 point to it will be the same. So let's go ahead and use those numbers. So then it's going to be trial one in trial three and we want the larger number to be in the numerator. So then we're going to have concentration of trial three over the concentration of trial one. And this is going to be raised to the power of A. And it's going to equal the rate of the 3rd 1 Trout three and the rate of trial one. So with that being said we can go ahead and plug a number. So for trial three The concentration was 0.6. For trial one it was 0.2. And because these are both raced to a. We can rewrite it to look like this And then rate three Is 3.6 times 10 to the -2. And then for trial one Is 1.2 times to the -2. So then we do some math and we get 3 to the a. Is equal to three. So then what value can A. Be sorry For this equation to be true. Well 3 to the power of one will give us three. So then here we solved for A. Which is going to be one. Okay. So then we just sell for a. So now what we have to do is sulfur B. So let's go ahead and do the same for when we get a different color to find B. Okay. So remember this time we're finding B. So then we want we're going to be looking for why? So then we want different values for why? Same values for X. So then here we have same values for the X. And here we have different values. So let's go ahead and start plugging in numbers. We want the larger one in the numerator, Larger one being .4. So then it's going to be trial two, which is going to be to the power bi over trial one. R. B. Go to great two right one. Let me move this screen down. Okay. So then let's go ahead and plug in numbers. Um trial too is going to be 0.4 over 0.2. Once again, these are both raced to the B. So that we can rewrite it to look like this. And the rate to is 4.8 times 10 to the -2 over 1. times 10 to the -2. We do some math and we get to to the B. is equal to four. So then one number When we plug in for B will give us four. Well that is going to be two because two squared will give us four. So then be Is equal to two. So then we just got the orders. So then we can go ahead and actually fill in this equation to sulfur rate. So let me get my red color. Okay now let's plug in values so we're going to have three is equal to the constant K. The concentration of X. And then we solve for a. Which we said was one. We don't have to plug it in because we can assume one. But just for the sake we can put a one here and then why we solved for B. And we said it was too. Okay. So then this is going to be the answer for question A. So now let's go ahead and see what question be is asking. So B is telling us if the concentration of X. Is 0.35 polarity and the concentration of Y. Is 0.85 molar. Itty determine the initial rate of the reaction. Okay, so then here we have X. And Y. But we do not have K. The constant. To be able to solve for the initial rape. So what we can so what we can do is get one of the trials plug in the numbers because we have right there and then solve for the K. And then once we have K we can go ahead and use the values that they give us with the K. We solved for to solve for the rape. So let's go ahead and do that. So let's see let's use trial one. Just we could use any but let's just use try one because it's one so then let me go down. So then here remember we have rate is equal to K. Times reactant. So then here let's go back up our A. Is going to be 1.2 times 10 to the -2. So then let me go down. So we're going to have the rate and it's going to be 1.2 Times 10 to the -2 is equal to K. Which is what we're going to be solving for times. Let me go back up The concentration of X. is .2, concentration of Y. is also point to So then 0. and then we also have 0.2 for Y. But don't forget we sell for the order and why has the order to so then we're going to put a two here and then we're going to do the math and we're going to get Kay Is equal to 1.5 And the unit is second to the -1 because it's one over seconds. Okay so now we have K. Which is good because we need it to solve for the initial rate. So Okay this is not the answer because we still have to do one more step. So then let's go ahead and use the concentrations we were given. So then we're going to have rape, which is what we're looking for is equal to K because remember K. So The KV software, we said it was 1.5 and then we're actually given the concentrations here. So let's go ahead and plug them in. So X. He said it was zero 35. Why We're told is 0.0 85. But don't forget we said that the order is to so then we're going to have to square it. So then we do the math and we get the rate to equal 3.79 times To the negative three polarity over a second. So then this is going to be the answer for question. Be okay. Thank you for watching. I hope this helps. And I'll see you in the next video.

Ch.14 - Chemical Kinetics

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Ch.14 - Chemical Kinetics

Problem 99c

Chapter 14, Problem 99c

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)
C₂H₅Br(aq) + OH^- (aq) → C₂H₅OH(l) + Br^- (aq)
Temperature (°C) k (L,mol •s)
25 8.81⨉10^-5
35 0.000285
45 0.000854
55 0.00239
65 0.00633
c. If a reaction mixture is 0.155 M in C₂H₅Brand 0.250 M in OH^-, what is the initial rate of the reaction at 75 °C?

Verified step by step guidance

Identify the rate law for the reaction. Since the reaction is first order in each reactant and second order overall, the rate law is: \( \text{Rate} = k [\text{C}_2\text{H}_5\text{Br}] [\text{OH}^-] \).

Determine the rate constant \( k \) at 75 °C. Use the Arrhenius equation \( k = A e^{-E_a/(RT)} \) to extrapolate the rate constant at 75 °C from the given data.

Calculate the initial rate of the reaction using the rate law. Substitute the concentrations \([\text{C}_2\text{H}_5\text{Br}] = 0.155 \text{ M}\) and \([\text{OH}^-] = 0.250 \text{ M}\) into the rate law equation.

Substitute the extrapolated rate constant \( k \) at 75 °C into the rate law equation.

Solve the equation to find the initial rate of the reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Order

The order of a reaction indicates how the rate is affected by the concentration of reactants. In this case, the reaction is first order in each reactant, meaning that the rate is directly proportional to the concentration of each reactant. The overall order of the reaction is the sum of the individual orders, which is second order here, indicating that the rate depends on the concentrations of both reactants.

Rate Constant (k)

The rate constant (k) is a proportionality factor in the rate law that relates the rate of a reaction to the concentrations of the reactants. It varies with temperature, which is crucial for calculating the initial rate of the reaction at different temperatures. The provided k values at various temperatures allow for the determination of how the reaction rate changes with temperature, following the Arrhenius equation.

Initial Rate of Reaction

The initial rate of a reaction is the rate measured at the very beginning of the reaction when the concentrations of reactants are at their highest. It can be calculated using the rate law, which incorporates the rate constant and the concentrations of the reactants. For this reaction, the initial rate at 75 °C can be determined by substituting the appropriate k value and the concentrations of C2H5Br and OH- into the rate equation.