please show the steps for every part, thank you. 20. The reaction of CV with OH
ID: 1033241 • Letter: P
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please show the steps for every part, thank you.
20. The reaction of CV with OH was studied by monitoring the concentration of CV+ over time in a manner similar to that encountered in Experiment M. The concentration of OH was much higher than that of CV (IOH- [CV ]) so that the order of reaction with respect to [CV'] could be determined. OH: (aq) CV+ (aq) CVOH (aq) + The following data was recorded (see the table below), and three plots were prepared: [Cv] versus time, In [CV'] versus time, and 1/[CV] versus time. Of the three plots, only the plot of In [CV-] versus time yielded a straight line, with a best-fit linear equation of y -7.439x103x - 11.47 (as reported by spreadsheet software). 1/[CV+] L/mol) 9.62× 104 1.02 × 105 1.10 x 10s 1.20 x 105 1.29 x 105 Time (s) [CV 1 10 20 30 40 mol/L 1.04 x 10-5 9.77 x 10-6 9,06 x 10-6 8.34 x 106 7.76 x 10-6 In [CV*] 11.47 11.54 11.61 -11.69 a. What is the order of reaction with respect to [CV 1 b. In the case of the plot of In [CV-] versus time, what are the units of slope? What are the units of the y-intercept? c. What is the value of the pseudo-rate constant k', including units? d. Assuming that the concentration of [OH] was 0.250 M and that the reaction is first- order with respect to [OH1 what is the value of the rate constant, k, including units? e. What would be the concentration of CV+ remaining when t 80 s? f. At what point in time will only 40% of the original amount of CV. remain?Explanation / Answer
a) It is given that that only the plot of ln [CV+] vs time is a straight line. A linear plot of ln [CV+] vs time indicates that the order of the reaction with respect to CV+ is 1, i.e, the reaction is 1st order with respect to [CV+].
b) The integrated rate law for the pseudo 1st order reaction in [CV+] is given as
ln [CV+] = k’*t + ln [CV+]0
where k’ = pseudo 1st order rate constant and [CV+]0 is the initial concentration of [CV+].
The left side is dimensionless (natural logarithmic quantities are unitless). Hence, the product k’*t on the right side must be dimensionless. t has unit s. In order to make k’*t unitless, k’ must have the unit of reciprocal time, i.e, s-1.
The y-intercept is equal to ln [CV+]0. Since this is a natural logarithmic quantity, hence, the y-intercept is dimensionless or unitless.
c) Compare the intergrated rate law with the linear equation
y = -7.439*10-3x – 11.47
Comparing the two equations we have,
k’ = -7.439*10-3 s-1 (since y = ln [CV+] and x = t).
d) The order of the reaction with respect to [OH-] is 1. Again, [OH-] is held constant at 0.250 M.
The second order rate law is given as
Rate = k*[CV+]*[OH-] ? k*[CV+]*[OH-]0 (the concentration of OH- remains constant throughout).
=====> Rate = k’*[CV+]
where k’ = pseudo 1st order rate constant = k*[OH-]0 where k = 2nd . order rate constant for the reaction.
Plug [OH-]0 = 0.250 M and k’ = -7.439*10-3 s-1, we have
-7.439*10-3 s-1 = k*(0.250 M)
=====> k = (-7.439*10-3 s-1)/(0.250 M) = -0.029756 M-1 s-1 (ans).
e) We have already noted that
y = ln [CV+], ln [CV+]0 = -11.47 and given t = 80 s.
Put the values in the integrated first order reaction and get
ln [CV+] = (-7.439*10-3 s-1)*(80 s) – 11.47
=====> ln [CV+] = -0.59512 – 11.47 = -12.06512
=====> [CV+] = exp(-12.06512) = 5.7568*10-6 ? 5.76*10-6
Offcourse, the concentration will have unit M and hence, the concentration of CV+ after 80 s will be 5.76*10-6 M (ans).
f) Again, use the integrated rate law as
ln [CV+] = (-7.439*10-3 s-1)*t + ln [CV+]0
======> ln [CV+]/[CV+]0 = (-7.439*10-3 s-1)*t
Given [CV+] = 40% of [CV+]0, we have [CV+]/[CV+]0 = 40% = 40/100 = 0.4, we have
ln (0.4) = (-7.439*10-3 s-1)*t
======> -0.91629 = (-7.439*10-3 s-1)*t
======> t = (-0.91629)/(-7.439*10-3 s-1) = 123.1738 s ? 123 s (ans).
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