CH 203 problem 15.12

15.12 Consider the reaction

We will assume that both the forward and reverse reactions are elementary processes and that the value of the equilibrium constant is very large.

(a) Which species predominate at equilibrium, reactants or products?

If both the forward and reverse reactions are elementary processes, then we can write the rate laws based upon the molecularity of the steps, right? So we have

forward reaction: rate = kf [A][B]

reverse reaction: rate = kr[C][D]

Chemical equilibrium occurs when opposing reactions are occuring at equal rates; therefore, if our system is at equilibrium, then the rate of the forward reaction must equal the rate of the reverse reaction (this is just what we did when solving rate laws which contained intermediates):

rate of forward reaction = rate of reverse reaction

kf[A][B] = kr[C][D]

now, we rearrange this expression a bit:

kf / kr = [C][D] / [A][B]

we know that kinetic rate constants depend only on temperature; therefore, we have

kf / kr = [C][D] / [A][B] = constant

we call this constant the equilibrium constant for a reaction, and we give it the symbol K:

K = [C][D] / [A][B]

thus, at equilibrium, the concentrations of products divided by the concentrations of reactants is a constant. If the equilibrium constant has a large value then the numerator of K must be larger than the denominator. If this is so, then the products must predominate.

(b) Which reaction has the larger rate constant, the forward or the reverse?

From what we did above, at equilibrium,

rate of forward reaction = rate of reverse reaction

kf[A][B] = kr[C][D]

kf / kr = [C][D] / [A][B] = K

If the equlibrium constant K is large then kf must be larger than kr, so the forward reaction has the larger kinetic rate constant.

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DAC 8/6/08