the average decomposition rate of the organic acid

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A monoprotic organic acid decomposes after an enzyme is added to the solution. The initial concentration of the acid is 1.55 mol/L. After 354 sec, you take a 15.00 mL sample, titrate it with 0.120 mol/L NaOH solution, and reach the equivalence point after adding 52.36 mL. Compute the average decomposition rate of the organic acid.
(in mol L^-1 s^-1)

1 mole of monoprotic organic acid can donate 1 mole of protons.
Therefore, 1 mole of the acid reacts with 1 mole of NaOH.

The NaOH used in the reaction is
m = (0.120 M) * (52.36 mL) = 6.2832 mmol.
Then 6.2832 mmol of NaOH reacts with 6.2832 mmol of acid.
Therefore, the concentration of the acid in the 15.00 mL samples is
c * (15 mL) = (0.120 M) * (52.36 mL)

c = 0.41888 mol/L.

That is, the concentration of the organic acid is 0.41888 mol/L after 354 seconds.

The average decomposition rate is

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