Results and calculations (calculate the concentrations of iodine([I2]0) and iodi

Question

Results and calculations (calculate the concentrations of iodine([I2]0) and iodide ([I-]0) that would result from the dilutions if a reaction did not occur. Using the absorption measurement for triiodide (A) and its extinction coefficient, calculate the triiodide concentration ([I3-]) that results from the achievement of equilibrium, then calculate the final concentration of iodine([I2]0) and iodide ([I-]0) by subtraction. Finally, calculate the value of the equilibrium, constant.

wavelength of max. absorption = 370

Absorption = .382

extinction coefficient at absorption max = 1.14 X E^4

# mL I2 mL KI total vol (mL) [I2]0 (mol/L) [I-]0(mol/L) A temp (c) [I3-](mol/L [i2] [i-] Keq 1 2.00 1.00 4.00 .099 22 2 2.00 2.00 4.00 .115 14 Step 1. Stock solutions of of about vided for this experiment. 2.00x103 M KI and 1.00x104 M iodine should be prepared or the iodine solution into a cuvette, the 2.00x103 and wavelength ected in 1.00 mL water. 1.00 mLof in Part A above. After measuring the measure its absorption Step 2. the cuvette with a 110°C absorption, determine the temperature Prepare Record about 200 mL of an ice-water slush a 250 mL beaker. Also heat a 400 mL beaker with water on a hot plate to about 75oC Record Step 3. Pipet 2.00 mL ofthe KI solution and 2.00 ml of 1.00x10 Miodine solution into a dry cuvette, mix and measure its absorption and the temperature as in Step 1 above your results in the tables and til Step 4. Now immerse the cuvette from Step 3 in the ice bath and stir with the thermometer the the temperature dips below 10°C. Quickly reinsert the cuvette into the spectrometer, measure absorption value and read its temperature. Record your results in the table under C-2. Step 5. Now put the cuvette into the hot water and stir with the thermometer until the temperature reaches about 300C. Again determine the absorption value and the temperature. Record your results in the table under C-2 Step 6. Repeat the heating, absorption measurement and temperature measurements two more times increasing the temperature about 100C each time until you reach about 50°C. Record our results in the table under C-2 Step 7. Use the analysis in Experiment 24 with the data from Steps 3 6 to determine the thalpy, entropy and free energy for the reaction 1) Determine the equilibrium constant at room temnerature for additiona

Explanation / Answer

Let us consider the values for Run 1

we are using 2 mL of 1 x 10^-4 M I2 solution

Moles of I2 we have = 1 x 10^-4 mol/ L * 1 L/1000 mL * 2 mL = 2 x 10^-7 mol of I2

Similarly we can find moles of KI solution . We are using 1 mL KI solution

moles of KI = 2 x 10^-3 mol/ L * 1 L/1000 mL * 1 mL = 2 x 10^-6 mol KI

KI is a strong electrolyte and dissociates as

KI -----> K^+ + I^- ( mol ratio of KI and I^- is 1:1, therefore mol of KI = mol of I^-)

Therefore we have, moles of I^- = 2 x 10^-6 mol

mol of I2 = 2 x 10^-7 mol

Total volume of solution = 4 mL   

Let us convert this to L which is 4 mL * 1 L/1000 mL = 0.004 L

Molarity of I2 = mol of I2/total volume in L = 2 x 10^-7 mol / 0.004 L = 5 x 10^-5 M

Molarity of I^- = mol of I- /total volume in L = 2 x 10^-6 mol/ 0.004 L = 5 x10^-4 M

So we have for Run 1

[I2]o ( mol/L ) = 5 x 10^-5

[I-]o ( mol/L) = 5 x 10^-4

We know that the reaction mixture will obey Beer's Law. so If the reaction had taken place, we would have I3^- and its concentration can be found out using Beer's law which is

A = e * l* C

where A = absorbance = 0.099( for run 1 )

e = molar absorptivity coefficient = 1.14 x 10^4

l = path length of the cuvette ( This value is not given, so I am assuming standard value which is 1 cm)

C = concentration of the solution

Plugging in the values, we get

0.099 = 1.14 x 10^4 * 1 * C

C = 0.099/ 1.14 x 10^4

C = 8.68 x 10^-6 M

Therefore [I3^-] at equilibrium is 8.68 x ^10^-6 M

Now final concentration of I2 & I- can be calculated by subtracting this value from their initial concentrations

So we have [I2] = [I2]0 - [I3^-] = 5 x 10^-5 - 8.68 x 10^-6 = 4.132 x 10^-5

[I-] = [I-]0 - [I3^-] = 5 x 10^-4 - 8.68 x 10^-6 = 4.913 x 10^-4

[I2] - ( mol/L ) = 4.132 x 10^-5

[I-] ( mol/L) = 4.913 x 10^-4

Summarizing all the results for Run 1 we get,

mL

I2

mL

KI

total

vol

mL

[I2]0

mol/L

[I-]0

mol/L

Temp

(C)

[I3^-]

mol/L

#

mL

I2

mL

KI

total

vol

mL

[I2]0

mol/L

[I-]0

mol/L

A

Temp

(C)

[I3^-]

mol/L

[I2] [I-] Keq 1 2.00 1.00 4.00 5 x 10^-5 5 x 10^-4 0.099 22 8.68 x 10^-6 4.132 x 10^-5 4.913 x 10^-4

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