Question

A) What is the rate constant for the reaction with an average time of 48.6 seconds, between 3.00 ml of 0.1 M t-butyl chloride, 3 ml 0.01 M NaOH, 4 ml water with 3 drops bromophenyl blue solution.

B) What is the rate constant for the reaction with an average time of 48.6 seconds, between 3.00 ml of 0.1 M t-butyl chloride, 6 ml 0.01 M NaOH, 1 ml water with 3 drops bromophenyl blue solution.

Answer 1

This is a 1st order reaction and the arte of the reaction depends only on the concentration of the substarte (t-butylchloride).

Initial molarity of the substrate = 0.1M. Final volume of the mixture = 3+3 + 4= 10mL

molarity of the substrate in the mixture = 3mL*0.1M/10mL = 0.03 M

rate = -d[t-butylchloride]/dT = - (0-0.03)/48.6-0 = 6.17*10^-4 M/s

rate = k[t-bytyl chloride]

k = rate/[t-butylchloride] = 6.17*10^-4 M/s/0.1M = 6.17*10^-3 /s

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Initial molarity of the substrate = 0.1M. Final volume of the mixture = 3+6 + 1= 10mL

molarity of the substrate in the mixture = 3mL*0.1M/10mL = 0.03 M

rate = -d[t-butylchloride]/dT = - (0-0.03)/48.6-0 = 6.17*10^-4 M/s

rate = k[t-bytyl chloride]

k = rate/[t-butylchloride] = 6.17*10^-4 M/s/0.1M = 6.17*10^-3 /s

rate constant will be same. Because the rate does not depend on the concentration of NaOh . It depends ont he concentration of t-butyl chloride only.