A second reaction mixture him made up in the following way: 20.0 ml. 4.0 M aceto

Question

A second reaction mixture him made up in the following way: 20.0 ml. 4.0 M acetone + 10.0 ml. 1.0 M HCI + 10,0 ml. 0.0050 M l2 + 10.0 ml. H2O What were the initial concentrations of acetone. II* ion. and lj in the reaction mixture? Show your work! It took 140 seconds for the h color to disappear from hick reaction mixture when it occurred at the same temperature as the reaction in Problem 2. What was the rate of the reaction? Write the rate law as it would apply to the second reaction mixture: Rate Divide the equation in Part b by the equation in Problem 2b. The resulting equation should have the ratio of the two rates on the left side and a ratio of acetone concentrations raised to the m power on the right. Write the resulting equation and solve for the value of m, the order of the reaction with respect to acetone. (Round off the value of m to the nearest integer.)

Explanation / Answer

(a) Total volume of the reaction mixture, V = 20 mL + 10 mL + 10 mL + 10 mL = 50 mL
[I2]o = (0.005 M)*(10 mL) / 50 mL = 0.001 M

Similarly calculate initial concentrations of H+ and acetone.

[acetone]o = 4 M * 20 mL / 50 mL = 1.6 M

[H+]o = 1 M * 10 mL / 50 mL = 0.2 M

(b) To first order, the reaction rate is 0.001 M / 140 s = 7.14e-6 M/s

Rate law: Rate = k [CO(CH3)2]^a [I2]^b [HCl]^c
7.14e-6 = k (1.6)^a (0.001)^b (0.2)^c

But if you know that H+ is a catalyst and that this reaction is 1st order in acetone and 1st order in iodine according to the most likely reaction mechanism, then
Rate = k [CO(CH3)2] [I2]
7.14e-6 = k (1.6) (0.001)

(c) Problem 2b is not given, so this question can't be answered.

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