Question

1. The solutions below were mixed together to carry out the iodination of acetone at 250 °C. 5.0 mL of 3.4 M acetone 5.0 mL of 12 M HCl solution 100 ml. of 0.0074 M l solution 5.0 mL of DI HO If the reaction took 516 seconds to go to completion, then what is the value for the rate constant, k, for the reaction at 250 °C? The initial reaction rate is equal to the average reaction rate for this reaction include the units for k (7) rate law rate = k[acetone] [H] rate constant, k units for 2. What is the overall reaction order for iodination of acetone (see the rate law in question 1)? (1) 3. Use the provided experimental data to calculate the equilibrium constant, K, for the reaction given below. Note the stoichiometry is not all I to 1. (7) Include the units for K Co(aq) + 2SCN-(a) = CO(SCN)2(aq) 5.0 ml of 0.10 M CO(NO) was mixed with 50 ml of 0.050 MKSCN The absorbance of the equilibrium mixture at 491 nm was 0.157 The linear regression equation for the Co(SCN)") calibration curve was y = (4.3 x 10 x + 0.0074

Answer 1

Question 1.

[Acetone] = 5 mL * 3.4 M/(5+10+5+5) mL = 0.68 M

[I2] = 5 mL * 1.2 M/25 mL = 0.24 M

Here, the limiting reagent is I2, whose concentration is 10 mL * 0.0074 M/25 mL = 0.00296 M

i.e. Rate = 0.00296 M/516 s = 5.74*10^-6 M/s

From the rate law: 5.74*10^-6 M/s = k * 0.68 M * 0.24 M

i.e. The rate constant (k) = 3.515*10^-5 M^-1.s^-1