A second reaction mixture was made up in the following? 5 mL 4.0 M acetone + 20

Question

A second reaction mixture was made up in the following? 5 mL 4.0 M acetone + 20 mL 1.0 M HCL + 10 mL 0.0050 M I_2 + 15mL H_2O a. what were the initial concentrations of acetone, H+ ion1 and I2 in the reaction mixture? [acetone] M;[H+] M;[I2]0 M b. It took 250 seconds for the I2 color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in problem 2. what was the rate of the reaction? rate= M/sec rate = c. Divide the equation in part (b) by the equation in problem 2(b). The resulting equation should have the ratio of two rates on the left side and a ratio of H+ concentrations raised to the p power on the right. write the resulting equation and solve for the value of p, the order of the reaction with respect to H+. (Round off the value of p to the nearest integer.) 4. A third reaction mixture was made up in the following way: 5ml 4.0M acetate + 20 ml, 1.0 M HCI + 5 ml, 0.0050 MI2 + 20 ml, H2o If the reaction is zero order in l2, how long would in take for the l2 color to disappear at the temperature of the reaction mixture in problem 3?

Explanation / Answer

5 ml 4.0 M acetone + 20 mL 1.0 M HCL + 10 mL 0.0050 M I2 + 15mL H2O

a). Initial conc of acetone= 5ml*(4mole/1000ml)=0.02 moles

H+= 20ml*(1mole/1000ml)= 0.02 moles

I2= 10ml*(0.0050mole/1000ml)= 0.00005 moles

b. Rate= k[CH3COCH3][HCl]

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