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(b) Check the controllability of the following system:1 =Compute contTOI u(t) which takes the system state fTom the origin at time to (2.1)T at time Compute control...

Question

(b) Check the controllability of the following system:1 =Compute contTOI u(t) which takes the system state fTom the origin at time to (2.1)T at time Compute control u(t) which takes the system state from (2. 1)T at time to (3.3)" at time

(b) Check the controllability of the following system: 1 = Compute contTOI u(t) which takes the system state fTom the origin at time to (2.1)T at time Compute control u(t) which takes the system state from (2. 1)T at time to (3.3)" at time



Answers

The following data are for the system $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}}(\text { atm }) & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\text {B }} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline \end{array} $$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after $30 \mathrm{~s}$ ? After $90 \mathrm{~s}$ ?

If we look at the given graph, we see that the concentrations Or in this case pressures of A&B stop changing right here, so remains constant at .6 to 1.76. So, um it takes 80 seconds to reach equilibrium and that's our answer for part a. Once we get to 80 seconds we've reached the concentrations that the uh that are going to be unchanging for this system. Under these equilibrium conditions. Um for part B At time t equals 30 seconds. Actually, it's a lowercase T. Um Which would be right here between 20 and 40. We see that the pressure of A is going down and the pressure of B is going up. So if is going down and B is going up, that means the reaction is going in the forward direction to make B and use A. So the forward reaction is going to have a faster rate than the reverse reaction. And then when we're we get to 90 seconds, the system is equilibrium, which it reaches after 80 seconds. So by definition, equilibrium, the forward and reverse reactions are the same. That's why the concentrations do not experience any net change. So right forward equals rate reverse.

So we've got um we've got A and B. And you can see that the partial pressures of A. Are decreasing over time and the partial pressures of B are increasing. Okay, so the amounts of reactant are decreasing in, the amounts of products are increasing. And what we want to do is see when does this stop changing? One of these pressures? Stop changing because that's when equilibrium has been reached. Okay, So if you look at your chart, you see that they've stopped changing at about 75 seconds. Okay. So at 75 seconds is when you've reached equilibrium. Mhm. So we want to compare how fast the forward reaction and the reverse reaction are going We're talking about 45 seconds. Well, we have not reached equilibrium yet. Okay, so before we reach equilibrium, the rate of the forward reaction. All right, there's always going to be greater than the rate of the reverse reaction. So, the rate of the reverse. So at that point the rate of the forward reaction is definitely faster than the rate of the reverse reaction. Okay, now, much later, when we're after 90 seconds, okay, we are. We have reached equilibrium. And sort of by definition, that means that the rates are equal. Okay, the equilibrium means the rate of the forward reaction equals the rate of the reverse reaction.

So we can see for the given reaction that the concentrations stop changing after 75 seconds. Um So that's our answer for part of the system reaches equilibrium at 75 seconds when the net concentrations do not change any longer. Um If you're at 45 seconds, that's going to be right here. You can see that the pressure of A is going down and the pressure of B is going up, which corresponds to the reaction um happening more in the forward direction to use up A. And make be. So that means that the forward re is going to be faster than the reverse, right? And then at 90 seconds because the system reaches equilibrium after 75 seconds when it's in equilibrium. By definition, the forward and reverse reactions have equal rates. So we can say that at 90 seconds the rate of the forward reaction is equal to the rate of the reverse reaction.


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