Question

Four solutions listed below are made by mixing the reagents as indicated at 25 degree C. In each case the total volume of the solutions is 100 00 mL. Solution 1 25.00 ml. of 0.600 M H_2SeO_3 25.00 mL of 1.00 M KI 25.00 mL of 0.800 M HCI 25.00 mL KCI Solution 2 50.00 mL of 0.600 M H_2SeO_3 25.00 mL of 1.00 M KI 25. 00 mL of 0.800 M HCI Solution 3 25.00 mL of 0.600 M H_2 ScO_3 50.00 mL of 1.00 M KI 25.00 mL of 0.800 M HCI Solution 4 25.00 ml. of 0.600 M H_2SeO_3 25.00 mL of 1.00 M KI 50.00 mL of 0.800 M HCI What is the order of the reaction with respect to H_2SeO_3? What is the order of the reaction with respect to Gamma? What is the order of the reaction with respect to H6+? What is the overall order of reaction? Write the expression for the rate law Calculate the k values for each run. Write the rate law with the average k value and units

Answer 1

Run-1:

Total volume, Vt = 100 mL = 0.100 L

moles of H2SeO3 taken = MxV = 0.600 mol/L x 25 mL x (1L / 1000 mL) = 0.015 mol

Hence [H2SeO3]₀ = 0.015 mol / 0.100L = 0.15 M

moles of KI(or I^-) taken = MxV = 1.00 mol/L x 25 mL x (1L / 1000 mL) = 0.025 mol

Hence [I^-]₀ = 0.025 mol / 0.100L = 0.25 M

moles of HI(or H⁺) taken = MxV = 0.800 mol/L x 25 mL x (1L / 1000 mL) = 0.020mol

Hence [H⁺]₀ = 0.020 mol / 0.100L = 0.20 M

$\Delta$ Se = [H2SeO3]₀ = 0.15 M  

Rate = $\Delta$ Se / $\Delta$ T = 0.15 M / 300 s = 5.0x10^-4 M.s-1

Run-2:

Total volume, Vt = 100 mL = 0.100 L

moles of H2SeO3 taken = MxV = 0.600 mol/L x 50 mL x (1L / 1000 mL) = 0.030 mol

Hence [H2SeO3]₀ = 0.030 mol / 0.100L = 0.30 M

moles of KI(or I^-) taken = MxV = 1.00 mol/L x 25 mL x (1L / 1000 mL) = 0.025 mol

Hence [I^-]₀ = 0.025 mol / 0.100L = 0.25 M

moles of HI(or H⁺) taken = MxV = 0.800 mol/L x 25 mL x (1L / 1000 mL) = 0.020mol

Hence [H⁺]₀ = 0.020 mol / 0.100L = 0.20 M

$\Delta$ Se = [H2SeO3]₀ = 0.30 M  

Rate = $\Delta$ Se / $\Delta$ T = 0.30 M / 300 s = 1.0x10^-3 M.s-1

Run-3:

Total volume, Vt = 100 mL = 0.100 L

moles of H2SeO3 taken = MxV = 0.600 mol/L x 25 mL x (1L / 1000 mL) = 0.015 mol

Hence [H2SeO3]₀ = 0.015 mol / 0.100L = 0.15 M

moles of KI(or I^-) taken = MxV = 1.00 mol/L x 50 mL x (1L / 1000 mL) = 0.050 mol

Hence [I^-]₀ = 0.050 mol / 0.100L = 0.50 M

moles of HI(or H⁺) taken = MxV = 0.800 mol/L x 25 mL x (1L / 1000 mL) = 0.020mol

Hence [H⁺]₀ = 0.020 mol / 0.100L = 0.20 M

$\Delta$ Se = [H2SeO3]₀ = 0.15 M  

Rate = $\Delta$ Se / $\Delta$ T = 0.15 M / 37.5 s = 4.0x10^-3 M.s-1

Run-4:

Total volume, Vt = 100 mL = 0.100 L

moles of H2SeO3 taken = MxV = 0.600 mol/L x 25 mL x (1L / 1000 mL) = 0.015 mol

Hence [H2SeO3]₀ = 0.015 mol / 0.100L = 0.15 M

moles of KI(or I^-) taken = MxV = 1.00 mol/L x 25 mL x (1L / 1000 mL) = 0.025 mol

Hence [I^-]₀ = 0.025 mol / 0.100L = 0.25 M

moles of HI(or H⁺) taken = MxV = 0.800 mol/L x 50 mL x (1L / 1000 mL) = 0.040mol

Hence [H⁺]₀ = 0.040 mol / 0.100L = 0.40 M

$\Delta$ Se = [H2SeO3]₀ = 0.15 M  

Rate = $\Delta$ Se / $\Delta$ T = 0.15 M / 75 s = 2.0x10^-3 M.s-1

Q.1: In Run-2, the rate of reaction doubles on doubling the concentration of H2SeO3, by keeping other concentrations constant. Hene the rate of reaction with respect to H2SeO3 is 1

Q.2: In Run-3, on doubling the concentration of I^-, by keeping other concentrations constant, the rate of reaction increases 8 times. Hence the order of reaction with respect to I^- is 3.

Q.3: In Run-4, on doubling the concentration of H⁺, by keeping other concentrations constant, the rate of reaction increases 4 times. Hence the order of reaction with respect to H⁺ is 2.

Q.4: Overall order = 1+3+2 = 6

Q.5: Rate law is

Rate = k x [H2SeO3]x[I^-]³ x [H⁺]²

Q.6: For run-1,

Rate = 5.0x10^-4 M.s-1 = k x (0.15M) x (0.25M)³ x (0.20M)²

=> k = 5.0x10^-4 M.s-1 / (0.15M) x (0.25M)³ x (0.20M)² = 5.33 M^-5.s-1