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Enter your answer in tbe box provided.Consider the following reaction, which takes place in a single elementary step:kf2A +B = AzB krIf the equilibrium constant Kci...

Question

Enter your answer in tbe box provided.Consider the following reaction, which takes place in a single elementary step:kf2A +B = AzB krIf the equilibrium constant Kcis 15.8 at a certain temperature and k = 5.10 X 10 ? s, calculate the value of kf:M-2s-1

Enter your answer in tbe box provided. Consider the following reaction, which takes place in a single elementary step: kf 2A +B = AzB kr If the equilibrium constant Kcis 15.8 at a certain temperature and k = 5.10 X 10 ? s, calculate the value of kf: M-2s-1



Answers

Consider the following reaction, which takes place in a single elementary step: If the equilibrium constant $K_{\mathrm{c}}$ is 12.6 at a certain temperature and if $k_{\mathrm{r}}=5.1 \times 10^{-2} \mathrm{s}^{-1},$ calculate the value of $k_{\mathrm{f}}$

In this question were given the value of the equilibrium constant K C, which is equal to 12. 6 and the value of K R, which is equal to 5.1 tons, 10 to the minus two seconds to the minus one. Where else to calculate the value of K F? So to do so, K C is equally K F over K R and rearranging here, K f is equal. The K C Times K are so this is going to be equal to 12.6 times 5.1 times 10 to the minus two seconds to the minus one and key. See here. If we look at Casey, it would be in units of K. C. Is the majority of aid to be over the mill Arat e of a squared be So this is equal to mill Arat E mill, Arat e squared morality. So this would counsel and it would be equal to m to the minus two. Phil, if we plug in units here, this should be em to the minus two, and this would be equal to six. Um, wait. So let's work this out here. Multiplying this out 12 6 times 5.1 exploded minus two 0.64 M minus two s minus one, and this would be the value of K F.

This question has been given an equilibrium waiting in which we have two moles of a that are creating would be from 82. The and we have an equilibrium of the equilibrium constant. What reactions is K. F. And what is. Okay, perfect. So what we want to determine here is body off K. F. And we know that the value of K. C. Which is the equilibrium constant. This entire reaction is given by the ratio of the equilibrium constant of rather than hate content, squad rotation that off the best way. So we know this to be equal 1-26 and this is going to be equal to our K. F, divided by 5.1 x 10 to the negative tour, which is the red consent of the backward creation. Now that we have this, we can just make our K. F. Subject of the formula R K F is going to be equal to one, Multiplied by 5.1 By 10 to the power. So I'm making this calculation, our K which is the red constant of the forward reaction. This is going to be equal to 0.6. Well now, just to look for the units, we know that the force protection. If this going to be an elemental reaction, we know that the rate is going to be given by they say the rate of the forward reaction. This is going to be given by concentration of which is the K. F. Multiplied by the concentration of a square, multiplied by the concentration of the. Now looking at this information means allocate going to pay the rate divided by concentration, which is now raised to the power of so looking at the rate, this is going to be mp second divided by the concentration which is um and this is raised to the power of prayer. So the units that we're going to have here, these are going to be one over M squared S. Therefore the units of KF 0.64 m squared here.

And this problem, we need to determine the value of the rate constant. For the reverse of this reaction, we're given the value for the rate constant case of one for the forward reaction as well as the value of the equilibrium constant k. C. We know that we can write out the equilibrium expression for Casey. Based on this reaction, be the equilibrium concentration of A to be invited by the equilibrium concentration of a squared as equilibrium, concentration of B. And we're told that this ratio comes out to 12.6 for the equilibrium, constant Casey, and we can rate out the rates of each each direction of this reaction for the forward rate. This is equal to the forward rate, constant times, the concentration of a squared times, a concentration of B and for the reverse rate or some negative one, this is equal to the rate constant of the reverse reaction case sub negative one times a concentration of a to be and we know that when we were at equilibrium, these concentrations are at their equilibrium. Concentrations and the forward and reverse rates are equal, so we can set those two expressions equal to one another. So K one, it's a concentration of a squared. I'm speed is equal to que negative one times the concentration of a to be. And now, if we rearranged this in order to get it in the form of the equilibrium expression for Casey, you see that if we divide over these two concentrations to this side and isolated by dividing over the K night of one to the other side, then we can see that the concentration of a to be divided by the concentration of a squared times The concentration of B is equal to U K from what we road out previously. And this is also equal to the forward rate constant divided by the reverse re constant. And so now, based on this relationship, we can rearrange that expression to solve for the river Serie Constant K negative one. And this will be equal to the forward rate. Constant divide about you. The equilibrium constant que sub one was given as 5.1 times 10 to the negative, too. In verse seconds we divide this by the unit list 12.6 for the equilibrium constant. And when we take that ratio, we find that the re constant for the reverse of this reaction comes out to about 4.0 times, 10 to the negative, third in verse seconds.

For this question we have the reaction of so two cl two going to S. 02 plus C. L. To destroy geometry for the entire process is 1 to 1. All coefficients of one. If we have an initial concentration of CO two for so two cl two and then we end up increasing the concentrations of co two and seal to to the same extent this occurs because the stock geometry is 1-1, so we know the final concentration of cl two is 12 times 10 to the negative too, because the stock geometry is 1 to 1. That also will be the final concentration of CO two. And because everything is 1-1, that will be a decrease in concentration of S. O. to seal too. So we'll subtract uh 1.2 times tend the negative two from two and we'll get the equilibrium concentration of so two cl two and then K C will be equal to these two concentrations multiplied by each other. For the products divided by the concentration of the reactant 8.8 times 10 the negative three, And we get AKC value of .18.


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