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For the reactionA(aq)B(aq)C(aq)Assume that the initial concentrations are [AJinit 0 M, [BJinit 0 M, and K equaled [_ [CJinit 10 M. Calculate the value of the reacti...

Question

For the reactionA(aq)B(aq)C(aq)Assume that the initial concentrations are [AJinit 0 M, [BJinit 0 M, and K equaled [_ [CJinit 10 M. Calculate the value of the reaction quotient Q (0.5 mark)(b) Describe what net reaction would occur and how the concentrations of A, B, and would . change (ie. increase; decrease or stay the same) t0 reach equilibrium mark)Assume you have an equilibrium mixture of [A]; [B]; and [C] at 298K and that the reaction is endothermic. What would happen t0 the concentration o

For the reaction A(aq) B(aq) C(aq) Assume that the initial concentrations are [AJinit 0 M, [BJinit 0 M, and K equaled [_ [CJinit 10 M. Calculate the value of the reaction quotient Q (0.5 mark) (b) Describe what net reaction would occur and how the concentrations of A, B, and would . change (ie. increase; decrease or stay the same) t0 reach equilibrium mark) Assume you have an equilibrium mixture of [A]; [B]; and [C] at 298K and that the reaction is endothermic. What would happen t0 the concentration of € in this mixture if the temperature is increased: (0.5 mark) And with this change in [C] what would happen to the value of K? (0.5 mark)


Answers

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.) $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH (l)+\mathrm{Br}^{-}(a q)$$ $$ \begin{array}{cl} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \boldsymbol{k}(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}) \\ 25 & 8.81 \times 10^{-5} \\ \hline 35 & 0.000285 \\ \hline 45 & 0.000854 \\ \hline 55 & 0.00239 \\ \hline 65 & 0.00633 \\ \hline \end{array} $$ a. Determine the activation energy and frequency factor for the reaction. b. Determine the rate constant at $15^{\circ} \mathrm{C}$. c. If a reaction mixture is $0.155 \mathrm{M}$ in $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}$ and $0.250 \mathrm{M}$ in $\mathrm{OH}^{-}$, what is the initial rate of the reaction at $75^{\circ} \mathrm{C}$ ?

To appropriately answer this question, you need to prepare an Iranian. It's plot. The reaction is C two H five br with hydroxide producing C two H 50 H liquid. Nbr minus. The problem tells you that it is first order in each of the reactant and second order overall. So to determine the activation energy and frequency factor of the reaction, you prepare an Iranian plot. And Iranians plot is a plot of the natural log of the rate constant as a function of one over the kelvin temperature. So we will take the temperatures converted to Kelvin. They gave us 25 through 65 which are the temperatures of 2 98 through 2 38. And then they give us the rate constants 8.81 times 10 to the -5, down to 6.33 times 10 to the -3. But we don't just plot temperature and the rate constant. We need to take one over the kelvin temperature and the natural log of the rate constant. The natural log that ends up being R. Y values and one over the temperature kelvin temperature are our X values. When this is the case, then we have an equation that gives us a slope equal to negative E. A. Over our. So if the slope is equal to negative ea over R then E. A. Is going to be equal to negative slope, multiplied by our. After we had plotted natural log of K as a function of one over the Calvin temperature and determine the slope of the line. Then we'll take the negative of that negative slope, which will give us a positive value here and multiply it by are so that we get an activation energy that is positive, activation energy is always positive. When we use this equation, the activation energy comes out in units of jewels per mole, 89475. Which we can convert to kill the jewels per mole by dividing by 1,089.5 killer jewels per mole. When we plot the natural log of the rate constant as a function of one over the kelvin temperature than the Y intercept ends up being equal to the natural log of a, the frequency factor. So the natural log of the frequency factor is 26.783 and we'll take the anti natural audible sides and a Is 4.3 times 10 to the 11. Yeah, Then it asks you to determine the rate constant at 15°C.. To do this, we need to use a another form of the Iranians equation. That is the natural log of the rate constant at a requested temperature will be equal to ea over r multiplied by one over the Kelvin temperature For which we know a rate constant -1 over the Kelvin temperature for which we are calculating a rate constant plus the natural log of the rate constant that we do know. So you can choose any temperature in any corresponding rate constant found in the table of data that is provided. I'm choosing the temperature of 25°C and the rate constant of 8.8, 1 times 10 to the -5. So this entire right side will be our activation energy, which we just calculated. We'll keep it in units of jewels per mole, divided by our Multiplied by one over. I'm choosing the 25°C or 2 98 Kelvin minus the temperature for which we want to calculate the rate constant -1 over that Kelvin temperature, 15°C or 288 kelvin Plus the log then of the rate constant at 298 or 25°C.. And we get the natural log of K. That we are calculating being equal to negative 10.59. So we'll take the anti natural log of both sides. And the rate constant is 25 times 10 to the negative five one over polarity seconds because its second order overall. And make sure this makes sense. If we go lower, should the rate constant get smaller? It was 8.81 times 10. The negative five. Now it's 2.5 times 10 the negative five. And it should the rate constant should always get smaller when temperature decreases. Then for part C, it asks if the reaction mixtures .155 Mueller in the C two H five B R and 50.250 moller in hydroxide. What is the initial rate of the reaction? At 75 degrees Celsius. So this is a loaded question in that first, you have to calculate the rate constant at 75°C.. So we're going to do what we just did up here, but instead of using 15°C or 2 88 Kelvin, We're going to use 75°C or 348 Kelvin. But everything else is the same as we did in part B. Then we will get a rate constant at 75°C or 348 Kelvin of .01581 over Mueller seconds. Now that we have the rate constant. Then we can use the differential rate law with the rate constant and the concentrations. To calculate the rate we're rate is equal to the rate constant, multiplied by each concentration raised to their order and the order is first order. So the rate, then the initial rate at these concentrations of the reaction at 75°C Is 6.1 times 10 to the negative, four Mueller per second.

But this year is a very fun derivation to do to create something that we call the Van Hoff equation. So we're going to start with two equations that I'm sure you're very familiar with. That all stem around our standard for energy here. So we know that our standard free energy is equal to our standard entropy of reaction minus the temperature times, the standard entropy of reaction. We also know that our standard for energy of reaction is equal to negative universal gas, constant times, temperature times, the natural log of RK. So if we do some algebra, we can actually set these two equations equal to each other and try to come up with a new equation. So let's go ahead and do that. We have Delta H standard here minus T Delta US Standard is equal to negative Are times t times a natural log of K. We're going to divide both sides by negative are times teeth and see what we get. So here we get. The natural log of K is equal to our standard entropy divided by a negative or standard, and will be divided by our times, teeth minus. We're going to cancel out our teas here. A negative delta US standard over art. So we can go ahead and do some cleaning up here. Two negatives make a positive and we're going to go ahead and separate out are one over teeth and I will show you why. So we get a final derivation for a van. Half equation equal to negative. Are entropy over our times, one over tea waas are Delta us over our. So this equation might not look super familiar yet, but let's make one minor adjustment. When we think about algebra, we think about linear rising numbers so we can easily solve equations. So this equation may look awfully familiar to a why Equals and X plus b our slope form. So we have a Y variable and an X variable. We have our slope and we have our intercept. So to answer the question on how different experimental values of Delta, H, Delta S and K can lead us, they're at several different temperatures can lead us to the simplified equation. Let's think about a graph. So here is an arbitrary graph, and I'm going to go ahead and just draw a line. Okay, so using our slope intercept form here. We know that the Y axis is going to be the natural log of K and that our X axis is going to be negative. Delta h overall. So this may look kind of strange. Excuse me? Are ex access is actually one of her t mess it up. Our access access is one over. Chief, you have 1/2. So this may look strange, but now that we have different data points here, so we can pick an arbitrary data point are here. The slope of this line is equal to negative h over our and we know our because the constant. So by one data point, we can figure out the k, the temperature and the entropy. And once we have a line, we can figure out our intercept. So when you have multiple temperature data points in multiple different K's, you confined values for entropy and entropy very easily. By utilizing this Van Hoff equation

Of the reaction for the dissociation of Die Atomic A into two singular a molecules. And we're going to discuss some of the entropy and the P and energy properties of this reaction. So our first step here is to look at the signs our standard adults H and R standard. So the standard Delta US is a little bit easier to find. You can look right at this reaction equation and look at the number of moles on either side. Remember that if there's more moles of gas, the entropy of that reaction mixture will be higher. So let's look here. We have one more gas reacting on the reacting side, and then we have two moles of gas on the product side. Therefore, we have more Mel's on the product side, and the entropy of this forward direction reaction is going to be positive. To find Delta H. We want to look at these two reaction mixtures. Here we have initial state one with eight Die Atomic A's and two singular A's, and we have reaction. We have this reaction make sure equilibrium that has six Die Atomic A's and six singular A molecules so visually here you can see that majority of the die atomic A molecules are still in the reaction at equilibrium. This must mean that the reaction is endo thermic and requires an input of energy in order to fully dissociate die atomic A into singular A. And because this reaction is an author, Mick, that means our Delta H is greater than zero. It is positive. So for our next part here, we're going thio distinguish between what Delta s means versus Delta s standard. So our delta s standard so we can conceptually understand the difference between these two just from the idea of standards alone. So our Delta s standard is the amount of entropy in the entire reaction when associating a 21 mold A to completely converting it into two moles off singular A. So that is a standard for this reaction. Whereas this Delta H with the note or this Excuse me, this dealt, asked with the no standard is the entropy for the reaction at its current progress. So this number can change. This number cannot change. So for our third part of the reaction or third part of our question here we are talking about the sign of our Delta G standard. So when you think about Delta G standard, we can use our equation that Delta G standard is equal to Delta H Standard minus the temperature times the standard entropy of the reaction. So we ran into a little bit of a problem here. We have adult H. That's positive. We have adults, us that's positive. And that makes it almost impossible for us to determine the sign of our Delta G standard here so we can call up a few scenarios. So at points when our temperature is very high, that will cause our negative T Delta s term to increase. This in turn, will cause our Delta G standard to be negative at lower temperatures are negative. T Delta s term here will decrease. Therefore, our delta G standard will increase will be greater than zero. So now that we know are signs for Delta G, we can go ahead and look into our K p. So our KP is our equilibrium constant here in our reaction. So when we have our equation of Delta G standard is equal to the negative. Rt l n of Cape er Gibbs Free energy equation as we increase our temperature from up here. Remember, as we increase our temperature, our Delta G standard will become less than zero. It will become a negative number. Therefore, this entire term here, we'll need to be larger. So we can say that as temperature that was a Delta, uh, temperature increases R K p will also increase. So then, for our final part here, we're thinking about Delta G not standard. Just normal. Delta G. So far a delta Gee, at equilibrium when we're at equilibrium are Delta G is going to be equal to zero because there is no change in free energy at equilibrium. There's no change. This Delta G is going to be equal to zero, even though our adult that you standard or at different points of the reaction this Delta G value can vary when we're at equilibrium. Furthest association reaction are Delta G will be equal to zero

Okay, so the reaction was broader. As a function of time, A B reacts to produce A and B a lot of the one over a B concentration time. Use a straight line with the slope of 0.55 Um, m inverse inverse. That means that our slow about K is equal to zero point Bill by five. And Denver? No, we have, ah, one over a B concentration. We know that our, um reaction is second order. So we have a great that is equal to K times A B through power of two. And we also know the half life so half light is he would have is equal to one over two of a B initially. And we have all this data that's still point Girl. Fine time. There are 0.55 That means our half life is 33 seconds now for quite the us to find it's the initial initial concentration of a B is 0.25 polarity and the reaction mixture initially contained no product. What are the concentrations of A and B after 75 seconds? So you are integrated, great wall. And then we saw for a B concentration That's one over, Katie. That's one over. Maybe you Nicholas, which is equal to one over 0.55 time 75 plus one over 10.5, which is equal to 1.123 polarity. So now let's write out, and I stable a be reactive form A and B. So our initial concentration was a 0.25 for reactions and bill for a product negative and then partner. They were 0.25 old my neck. And to find the concentration of a B, this is equal to 0.250 minus X, which is equal to what we have to pay 0.123 So that means that X is equal to 0.127 polarity. And based on that, we can find both eh and B, which are products that will be good to, which is equal to 0.127 polarity


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