A solution was prepared by adding 12.00 mL of 0.130 M KI, 3.00 mL of 0.130 M KNO

Question

A solution was prepared by adding 12.00 mL of 0.130 M KI, 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)_2S_2O_8. a. 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. b. Calculate the concentration (M) of S_2O_8^2- in the mixture prior to the reaction. Show the calculation. c. Calculate the concentration (M) of S_2O_3^2- in the mixture prior to the reaction. Show the calculation. d. calculate In[I^-] from the concentration obtained in 1a. e. Calculate In[S_2O_8^2-] from the concentration obtained in 1b. f. 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-s), the concentration of A if 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.

Explanation / Answer

Concentration of I- ion = (0.13 * 12/1000)*1000/(12+3+5+1+15)
Concentration of I- ion = 0.0433
ln[I-] = ln(0.0433) = -3.1396
Concentration of S2O82- = (0.13*15/1000) *1000/(12+3+5+1+15)
Concentration of S2O82- = 0.0542
ln[S2O82-] = ln(0.0542) = -2.915
Concentration of S2O32- = (0.01*5/1000) *1000/(12+3+5+1+15)
Concentration of S2O32- = 0.0014

Rate = K * [A]^2 [B]
5.08*10^-7 = K * (8.72*10^-3)^2*(4.93*10^-2)
K = 0.1355

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