Question

A solution was prepared by adding 12.00 ml 3.00 mL of 0.130 M KNO_3, 5.00 mL of 0.0100 M Na_2S_2O_3, 1.00 mL of starch solution, and 15.00 mL of 0.130 M (NH_4)_2 S_2O_8. Assuming the volumes of the components are additive, calculate the concentration (M) of I^- in the final mixture before the chemical reaction. Show the calculation. Calculate the concentration (M) of S_2O_8^2- in the mixture prior to the reaction. Show the calculation. Calculate the concentration (M) of S_2O_3^2- in the mixture prior to the reaction. Show the calculation. Calculate ln(I^-) from the concentration obtained in 1a. Calculate In [S_2O_8^2-) from the concentration obtained in 1b. Use equation 18-4 to calculate the rate of reaction for the above mixture if it took 109 seconds for the blue starch-iodine complex to appear. Remember that the concentration of S_2O_3^2- in eq. 18-4 is the diluted concentration in the reaction mixture, calculated in part (c) above, not the concentration in the stock solution. A reaction has the rate law expression rate = k[A]^2[B]. Determine the rate constant for the reaction when the initial rate of the reaction is 5.08 times 10^-7 mol/(L middot s). the concentration of A is 8.72 times 10^-3 M and the concentration of B is 4.93 times 10^-2 M. Be certain to write the proper units for the rate constant.

Answer 1

1. From the given data

a. concentration [I-] = 0.13 m x 12 ml/36 ml = 0.043 M

b. concentration [S2O8^2-] = 0.13 m x 15 ml/36 ml = 0.054 M

c. concentration [S2O3^2-] = 0.13 M x 5 ml/36 ml = 0.018 M

d. from the above calculations,

ln[I-] = -3.15

e. from the above calculations,

ln[S2O8^2-] = -2.92

f. rate = -(-2.92/109) = 0.027 M/s

2. with rate = k[A]^2[B]

with given concentration of [A] and [B] and initial rate value we get,

rate constant k,

k = rate/[A]^2[B]

   = 5.08 x 10^-7/(8.72 x 10^-3)^2.(4.93 x 10^-2)

   = 0.1355 M-2.s-1