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		18		5
3		33		6
4		48		9
5		71		14
6		95		14
7		infinity		27

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

Answer 1

hSuJ- * * Page No - 0 Given timed data, Convert ML 0104 M Naoh to concentration of Hal. Hel], in lome aliquots of the reaction of t-butyl chloride in isopropanol , wortu. Then convert (Heidt to concentration of t-butyl - Chloride, (Rx]t. * Hel + Nach Nach +H₂O CHR = CNOOH X V NOCH VHU for run = , 0.04M X 3mL 10mL Chen 0.12 10 = 0.012 M for run 2, -0.04 MXUML Chel Tome - 10 0.016 M = for run 3, 0.04 MX 8ml - 0.32 - Chel Tome 10 for 0.032 M = run 4, 'OUM X13m CHO 1 = 0.52 10 = 0.052M COML

for nun 5, Page No - CHCl = 0.04 MXlUML — = 0.56 lome 10 for = 01056 M run 6, Chu = 0.04 X17ml = 0.68 10mL 10 -0.068 H for run 7, 0.oux 27mL Tome =1.08 To = 0.1084 R-ce +H20 R-OH THE for run 7, at time = infinity [HU] = {Rx] = 0.108M (RxIt = [RX]o-[+c]t for run 1, CRx= 0.108 M - CHU]t = 0.1084 - 0.012M = 0.09614 For run 2, (Rx), - OilOS M - [Heldt = 0.1084 -0.016M = 0.042 M for run 3. CRx) = 0.1084 - [HU]+= 0.108M - 0032M -0.07614 Co% Pานา 4, CRx) = 0.1084 - (tert = 0.1084 - 0:0S2M 0.056 For ren 5, PRx]. - 0.105M - (HU)t = 0.1084 - 0.058 M = 0.052M

for rund, Page No- CRx), - 0.1084 ftult = 0.1084 - 0.068 M = 0.04 M. Irun time (min) (Rx][ 0.096 12 0.092 0.076 0.056 74 -0.052 104 0.04 Infinity moyo {[Rx) ol (Rx]+) 0.117783 Lo.lb0343 0.351398 -0.65678 0.730888 0.993252

y=97.44x time (min) 0.2 1 1.2 0.4 0.6 0.8 In([RX]/[RX]t)

i The slope is the reciprocal of the constant. rate ie n It I = 0.010. is 97.44