I\'m having trouble with this T jump equation, I doubt know where to start i tri

Question

I'm having trouble with this T jump equation, I doubt know where to start i tried deriving t and couldn't even get that to match up with the given equation. can someone fully explain how I would do this

2) Consider the equilibrium reaction below If the reaction mechanism occurs in a single step, then the reaction will be first order in A, B, and C. In this case it can be shown that the relaxsation time is givern by where the subscript "eq" denotes the equilibrium concentrations after the T-jump (Note: This is different than what we derived because here there are two reactants on the left and we only had a single reactant.) Acid-base reactions typically follow the kinetics described above. Consider the H'(aq) + C,H,COO (aq-C,HsC00H(aq) 1.6 x 104 (at 298 K) a) Show that equilibrium constant for this reaction is the ratio of rate constants in the forward and reverse directions. What are the units of the rate constants in the forward and reverse directions? Use these units to determine the units of K. Does this give you the same results as plugging concentrations into the equilibrium expression? b) c) Suppose a 0.015 M benzoic acid solution is prepared. The solution is subjected to a T-jump experiment where the final temperature is 298 K. If the relaxation time is 15x 10 s, then what are the rate constants in the forward and reverse directions?

Explanation / Answer

            k1[H+] [C6H5COO-] = k -1[C6H5COOH]     

Rearrange as

           [C6H5COOH] / ([H+] [C6H5COO-] ) = k1/k-1

By definition    [C6H5COOH] / ([H+] [C6H5COO-] ) = K

Thus K =   k1/k-1

b.

- dc/dt = k1[H+] [C6H5COO-]             

The units for k1 are

<k1> = L.mol-1s-1

- dc/dt = k-1[C6H5COOH]     

      The units for k1 are

        <k1> = s-1

       

The units for K are L.mol-1s-1 / s-1 = L.mol-1 , the same as you can obtain directly from the expression [C6H5COOH] / ([H+] [C6H5COO-] ) = K

Note: In fact K is used as an adimensional quantity (see question text).

c.

C6H5COO-         +   H+ = C6H5COOH   

          0…………….0…………0.015 M           initial state

           x……………..x…………0.015-x           at equilibrium

K = (0.015-x)/x2

Assume x << 0.015 and solve for x

(I can’t solve, because I can read the power of 10 in the value of K).

Use the equation of (I can’t read the power of 10 in value) and replace:

[A]eq = [B]eq = x

       k1 = K· k -1

Calculate k-1 , then k1 .

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