Question

Given the following timed data, convert mL 0.04 M NaOH to concentration of HCl, [HCl]_t, in 10 mL aliquots of the reaction of t-butyl chloride in isopropanol/water. Then convert [HCl]_t to concentration of t-butyl chloride, [RX]_t. Using a program such as Microsoft Excel, plot (Scatter Chart) time versus natural log of concentration of t-butyl chloride at time zero over concentration of t-butyl chloride at each time, ln([RX]₀/[RX]_t). Determine the slope of your plot as instructed in this video. The slope is the rate constant of first order kinetics as given by equation 20.2 of your text. This is equivalent to Figure 20.1 of your text but we don't have to use the factor 2.303, ln(10), because we are using natural versus base 10 logs. Before the advent of calculators, people actually used tables to look up log₁₀ values! Hint: [RX]₀ = ? so that [RX]_t = [RX]₀ - [HCl]_t which is the equation at the bottom of page 322 of your text?

run		time (min)		mL 0.04 M NaOH		[HCl]t		[RX]t		ln([RX]0/[RX]t)
1		9		3
2		22		4
3		34		8
4		49		13
5		74		14
6		104		17
7		infinity		27

Give only 2 significant digits without units or any other symbol. Do not use scientific notation!

Answer:

Answer 1

For run 1,

For run 2,

For run 3,

For run 4,

For run 5,

For run 6,

For run 7,

For run 7, at time = infinity,

For run 1,

For run 2,

For run 3,

For run 4,

For run 5,

For run 6,

The slope is the reciprocal of the rate constant. It is