Winter 2012 question 4
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Winter 2012 question 4
for the second part of 4A, could someone explain how to derive the rate laws? What are k1, k1', k2? Why does rate=k2[C]?
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Re: Winter 2012 question 4
There is a similar question from quiz 2, where they ask you to solve for the rate constant k in the overall reaction. This is done similarly to that.
We know that rate = [rate constant] * [concentration of reactants]
We also know that rate constant k = [concentration of products] / [concentration of reactants]
Since the first reaction is the fast reaction, then A+B and C are in equilibrium with each other. In other words, the rate of formation of C is the same as the rate of dissociation to A+B.
These three facts gives us the following equations:
Rate (for slow equation) = k2*[C]
Rate (for fast equation) = k1*[A]*[B]=k-1*[C]
where k2 is the rate constant for the slow reaction, k1 is the rate constant of the fast reaction going forward, and k-1 is the rate constant of the fast reaction going reverse.
You use the second equation to solve for [C] in terms of everything else, and that gives you [C] = k1*[A]*[B]/k-1
You can plug this into the slow equation, and that gives you Rate = k2*k1*[A]*[B]/k-1
In that final rate equation, the constant k2*k1/k-1 can be set equal to the overall rate constant, which the answer key called "K".
We know that rate = [rate constant] * [concentration of reactants]
We also know that rate constant k = [concentration of products] / [concentration of reactants]
Since the first reaction is the fast reaction, then A+B and C are in equilibrium with each other. In other words, the rate of formation of C is the same as the rate of dissociation to A+B.
These three facts gives us the following equations:
Rate (for slow equation) = k2*[C]
Rate (for fast equation) = k1*[A]*[B]=k-1*[C]
where k2 is the rate constant for the slow reaction, k1 is the rate constant of the fast reaction going forward, and k-1 is the rate constant of the fast reaction going reverse.
You use the second equation to solve for [C] in terms of everything else, and that gives you [C] = k1*[A]*[B]/k-1
You can plug this into the slow equation, and that gives you Rate = k2*k1*[A]*[B]/k-1
In that final rate equation, the constant k2*k1/k-1 can be set equal to the overall rate constant, which the answer key called "K".
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