Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Experiment 4: Chemical Kinetics, Lab Reports of Chemistry

Main purpose of this lab is to determine the rate law for the reaction of the dye crystal violet with hydroxide.

Typology: Lab Reports

2020/2021

Uploaded on 05/11/2021

anshula
anshula 🇺🇸

4.4

(12)

244 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
Experiment 4: Chemical Kinetics, Part 2
Purpose: Determine the rate law for the reaction of the dye crystal violet with hydroxide.
Reading: Olmstead and Williams, Chemistry, sections 13.3 and 13.4.
Introduction
The determination of the rate law for the reaction of crystal violet with hydroxide is completed in
this experiment. The order of the reaction with respect to CV is determined by comparison with
the integrated rate laws for first and second order reactions. The rate constant is also determined.
rate of disappearance of CV = rate of appearance of CVOH
= k [CV]
x
[OH
-
]
y
(1)
Comparing the Time Course to the Integrated Rate Laws
An initial rate study is an excellent method for determining the order with respect to hydroxide,
since hydroxide is not a colored species. However, since crystal violet is the colored reactant, a
time course study is a better method for determining the reaction order with respect to crystal
violet. In a time-course study, the concentration of a reactant or product is followed as a function
of time and compared to the integrated rate laws for different order reactions.
First-Order Rate Expression
The rate law for a reaction that is first-order in the reactant is:
rate = – d[A]
dt = k [A] (2)
This rate law applies to the type of reaction with one reactant, for example A Products. How
can we apply this rate law for our reaction, which is in the form of A + B products? In
comparison with Eq. 1, the first order rate law in equation 5 results if x = 1 and y = 0 or if [OH
-
]
y
is held constant and combined with the rate constant k to give an effective rate constant:
rate = (k[OH
-
]
y
) [CV]
x
= k
eff
[CV] (3)
The effective rate constant is k
eff
= (k[OH
-
]
y
). In other words, the first order rate law works fine,
but we get an effective first order rate constant, k
eff
, that includes the effect of the concentration
of hydroxide. Keeping [OH
-
]
y
constant can be done experimentally by having hydroxide in large
excess, in which case, the overall concentration change of hydroxide during the course of the
reaction is negligible. The process is then said to be pseudo-first order in CV. The integrated rate
expression is:
ln [CV]
[CV]
o
= – k
eff
t (4)
or ln [CV] = – k
eff
t + ln[CV]
o
(5)
where [CV]
o
is the initial concentration of CV at time zero and [CV] is the concentration at any
time, t.
Second-Order Rate Expression
If x = 2 and y = 0 or [OH
-
]
y
is held constant in Eq. 1:
rate = – d[CV]
dt = k
eff
[CV]
2
(6)
pf3
pf4

Partial preview of the text

Download Experiment 4: Chemical Kinetics and more Lab Reports Chemistry in PDF only on Docsity!

Experiment 4: Chemical Kinetics, Part 2

Purpose: Determine the rate law for the reaction of the dye crystal violet with hydroxide. Reading: Olmstead and Williams, Chemistry , sections 13.3 and 13.4.

Introduction

The determination of the rate law for the reaction of crystal violet with hydroxide is completed in this experiment. The order of the reaction with respect to CV is determined by comparison with the integrated rate laws for first and second order reactions. The rate constant is also determined.

rate of disappearance of CV = rate of appearance of CVOH = k [CV]x^ [OH-]y^ (1)

Comparing the Time Course to the Integrated Rate Laws An initial rate study is an excellent method for determining the order with respect to hydroxide, since hydroxide is not a colored species. However, since crystal violet is the colored reactant, a time course study is a better method for determining the reaction order with respect to crystal violet. In a time-course study, the concentration of a reactant or product is followed as a function of time and compared to the integrated rate laws for different order reactions.

First-Order Rate Expression The rate law for a reaction that is first-order in the reactant is:

rate = –

d[A] dt = k [A]^ (2)

This rate law applies to the type of reaction with one reactant, for example A→ Products. How can we apply this rate law for our reaction, which is in the form of A + B→ products? In comparison with Eq. 1, the first order rate law in equation 5 results if x = 1 and y = 0 or if [OH-]y is held constant and combined with the rate constant k to give an effective rate constant:

rate = (k[OH-]y) [CV]x^ = keff [CV] (3)

The effective rate constant is keff = (k[OH-]y). In other words, the first order rate law works fine, but we get an effective first order rate constant, keff, that includes the effect of the concentration of hydroxide. Keeping [OH-]y^ constant can be done experimentally by having hydroxide in large excess, in which case, the overall concentration change of hydroxide during the course of the reaction is negligible. The process is then said to be pseudo-first order in CV. The integrated rate expression is:

ln

[CV]

[CV]o = – keff^ t^ (4) or ln [CV] = – keff t + ln[CV]o (5)

where [CV]o is the initial concentration of CV at time zero and [CV] is the concentration at any time, t.

Second-Order Rate Expression If x = 2 and y = 0 or [OH-]y^ is held constant in Eq. 1:

rate = –

d[CV] dt = keff^ [CV]

the reaction is pseudo-second-order in CV and integration of Eq. 6 gives:

1 [CV] –^

[CV]o = keff^ t^ (7)

Determining the Reaction Order Since absorbance is directly proportional to concentration, according to Beer’s Law A=εlc, the absorbance can be used for the curve fitting in place of the concentrations. For a first order reaction, the absorption coefficient cancels in the numerator and denominator of the ln term in Eq. 8, so either concentration or absorbance may be used to directly determine the rate constant. For a second order reaction, since [CV] = A/εl, 1 A –

Ao =

keff εl t (8)

where Ao is the initial absorbance. To determine the order of a reaction, absorbance versus time measurements are collected and the data are plotted. For a zero order reaction, the concentration or absorbance versus time plot is a straight line. According to Eqs. 4 and 5, for a first order reaction a plot of ln [CV] or ln A versus t should yield a straight line with a slope of keff. According to Eqs. 7 and 8, for a second-order reaction a plot of 1/[CV] or 1/A versus t is a straight line.

Procedure

Note that your reference cuvette should contain water for all calibration procedures below.

A. Time-Course Kinetics Experiments

  1. As for last week, the total volume of each trial should be 3.00 mL. Your first kinetics trial should contain the volume of crystal violet that gives an absorbance of about 1.0-1.3. Choose the amount of sodium hydroxide from last week’s experiment that gives a half-time of about 1.5 mins; the half-time or half-life is the time it takes for the concentration to drop by a factor of two. Using the SpectroVis Plus User Guide instructions provided in the laboratory, collect data every 4 seconds for maximum time of 900 seconds. Add the crystal violet to your cuvette, then the appropriate volume of water that will make your final volume 3.00 mL (be sure to account for the NaOH that you’re about to add). Stir rapidly with the stir bar. Finally, add the NaOH, mix with the pipette tip, and immediately start collecting kinetic data with the spectrophotometer. When the curve starts to look “flat,” stop collecting. Save the data as described in the SpectroVis Plus Instructions.
  2. Follow the User Guide, which will be available in the laboratory, to plot ln A vs. time and 1/A vs. time. Determine the reaction order and effective rate constant, keff. Determine the standard deviation of the rate constant. The RMSE listed in the curve fit window is the “root- mean squared error” of the y-values. The y-values correspond to either ln A or 1/A depending on the curve fit. To get a rough estimate of the standard deviation of the slope, divide the RMSE by the range of the x values. In this experiment the x-values are the time data points. This rate constant is the effective rate constant, see the Time-Course Data Analysis section below. Print copies of the two plots that you create for your lab notebook and your lab report. Record the temperature.
  3. You only need to complete one run successfully. If your curve fitting didn’t work out well, because the reaction ran either too fast or too slow, repeat the experiment with a different amount

(d). Give the average and standard deviation of the trials for the rate constant from the five initial rate studies and the rate constant from the time-course study with the standard deviation. (e). Consider the effect of random and systematic errors on the difference in Part d. (i). Using the observed standard deviations, determine if the difference between the average rate constant from the initial rate studies and the time course measurement are explained by random error (do the standard deviation ranges overlap?). Discuss if the difference in temperature played a role in the difference between the average from the initial rate study and the time-course based measurement (assume the reaction rate doubles for a 10°C temperature change). (ii). Suggest a significant source of systematic error in the time-course study. Remember that student mistakes are neither random nor systematic errors; student mistakes are just student mistakes. Base your choice on the measurements: the calibration of the micro-pipettors, the concentration of the stock solutions, and the absorbance accuracy of the spectrophotometer. What effect does the source of systematic error have on the rate constant determined from the time-course studies? (For example, does the systematic error cause curvature in the plots or increase or decrease the value determined for the time-course based rate constant or have no effect?) (f). To summarize the experiment answer the following questions: (i). Give an advantage or disadvantage that initial rate studies have in comparison to time course studies in determining the order of the reaction with respect to a given reactant. (ii). Eqs. 5 and 7 apply if only one reactant undergoes significant changes in concentration. What conditions were used in this experiment to allow Eqs. 5 and 7 to be used in determining the order of the reaction? (iii). Is the order of the reaction with respect to CV the same as the stoichiometry? Is the order of the reaction necessarily the same as the stoichiometry? (This question is repeated from last week’s lab, but with respect to CV.)

Literature Cited : Give all literature cited, numbered according to the references in the body of your report. See the previous lab reports for the format of the reference to the on-line General Chemistry lab manual. Include last accessed dates for on-line resources.

Attach the two plots used to determine the order with respect to crystal violet.