A solution is Prepared by adding 15.00 mL of 0.110 M KI, 5.00 mL of 0.0100 M Na_

Question

A solution is Prepared by adding 15.00 mL of 0.110 M KI, 5.00 mL of 0.0100 M Na_2S_2O_3, 1.00 mL of starch solution, and 15.00 mL of 0.110 M (NH_4)_2S_2O_3 to a beaker. The volumes are additive. Calculate the concentration of iodide, I^-, and of peroxydisulfate, S_2O_8^2-, in the final solution. Calculate the concentration of thiosulfate, S_2O_3^2-, in the final solution. If the reaction described in problem 1 required 103 s for the blue-black color of the starch-iodine complex to appear, calculate the rate of the reaction by using the thiosulfate concentration from 1b and Equation (18-4). The rate law for the reaction performed in the laboratory exercise was given in Equation (18-7). Substitute the whole number orders of the reaction with respect to the two reactants that were determined by you in Experiment 18 into the rate law and rearrange the resulting equation to solve for the rate constant k. Use the initial concentrations calculated in problem 1, the rate calculated in problem 2 and the equation derived in problem 3 to determine the rate constant, k, of the reaction. Data from the Kinetics II experiment will be used to prepare a graph of ln(k) vs 1/T. If the reaction described in questions 1-2 of this pre-lab took place at 28 degree C, what are the x and y values of the data point for this reaction mixture that would be plotted on a graph of ln(k) vs 1/T? Remember that temperature in the Arrhenius equation is absolute temperature (Kelvin).

Explanation / Answer

1: (a) Total volume of the solution, Vt = 15.00 mL + 5.00 mL + 1.00 mL + 15.00 mL = 36.00 mL = 0.036 L

Moles of I-(aq) in the final slution = MxV = 0.110 mol/L x 15.00 mL x (1L / 1000 mL) = 1.65x10-3 mol

Hence I-(aq) in the fianal solution, [I-(aq)] = 1.65x10-3 mol / 0.036 L = 0.04583 M (answer)

Moles of S2O82-(aq) in the final slution = MxV = 0.110 mol/L x 15.00 mL x (1L / 1000 mL) = 1.65x10-3 mol

Hence S2O82-(aq) in the fianal solution, [S2O82-(aq)] = 1.65x10-3 mol / 0.036 L = 0.04583 M (answer)

(b): Moles of S2O32-(aq) in the final slution = MxV = 0.0100 mol/L x 5.00 mL x (1L / 1000 mL) = 5.0x10-5 mol

Hence S2O32-(aq) in the fianal solution, [S2O32-(aq)] =  5.0x10-5 mol / 0.036 L = 0.00139 M (answer)

Q.2: rate = (1/2) x 0.00139 M / 103 s = 6.75x10-6 M/s (answer)

Q.3: Lab data info required.

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