Rate Law Determination of the Crystal Violet Reaction Objectives
RateLaw Determination of the Crystal Violet Reaction
Objectives
Theaim of the experiment was to study the relationship between, reactionof crystal violet and sodium hydroxide.
BackgroundInformation
Thereaction between crystal violet and sodium hydroxide is defined usingthe equation for rate law for a reaction given by
Rate=k [CV^{–}]^{m­­­}[OH^{–}]^{n},
Wherek is the rate constant for the reaction, m is the order with respectto crystal violet (CV^{+}),and n is the order with respect to the hydroxide ion. The reactionbetween sodium hydrox­ide and crystal violet is expressed by
Whichcan be solved into an ionic equation as
WhereCV is crystal violet ion and OH is hydroxide ion.
Duringthe reaction of crystal violet and sodium hydroxide, the latter isover a thousand times higher in concentration as compared to theformer. The hydroxide ion used during the reaction is thusnegligible, and it`s concentration hardly changes. The equation ofrate solves for the order with respect to crystal violet (m) and thepseudo rate constant K^{’}is given by K[OH^{–}]^{n}.
Rate=k^{’}[CV^{–}]^{m}
Duringthe experiment, the Willchange from violet to colorless. Green light source forcomputerinterfaced calorimeter is the used to measure the absorbanceof the crystal against time. According to Beer’s Law, theobservance of crystal violet changes with time. The concentration ofthe crystal violet is then calculated from absorbance. From plottingthe figures the derived from the experiment the order of reaction isthen determined.
Theexperiment is then repeated with different concentration of sodiumhydroxide. Using the new data derived from the new concentration ofhydroxide ions, the value of K_{2}^{’}.
Thetwo rates K_{2}^{’}andK^{’}_{1}are compared using the formula
Apparatus

Vernier Go! Link senor Interface

Vernier colorimeter and plastic cuvette

25ml Volumetric

25ml Beaker

Two10mL graduated cylinder

Stirring rod
Procedure
Part1
10mLof 0.02M sodium hydroxide solution was taken. The computer waspowered and the power strip was then turned out. The calorimeter wasalso turned on and set to 565nm and allowed to warm up. The computerwas then prepared for data collection by opening 30 rate crystalviolet.cmbl. The calorimeter was then calibrated. The reaction wasthen initiated simultaneously by adding 10mL of crystal violet andsodium hydroxide into a beaker. The reaction was then stirred using astirring rod. The cuvette was then rinsed with 1ml of the reactionmixture and the filled to three quarters full. After 1.5min thecuvette was placed in the slot of the calorimeter. The keep buttonwas then clicked after closing the lid of the cuvette. The absorbanceand time data were then saved. The cuvette was removed after fortyfive seconds and repeat the procedure to get second value and more.After 20 minutes and when the recorded values of absorbance are below0.10 stop the experiment and discard the mixture. The same procedurewas repeated for part 2 except for the fact that10mL of 0.01M NaOHand 10mL of crystal violet. 0.01M NaOH was prepared by mixing 5.0mLof 0.020M NaOH solution and 5.0mL of deionized water. The liquidswere pipetted directly into a 250mL beaker. The computer was theprepared for data collection by opening 30 rate crystal violet.cmblfrom the computer. The other steps were followed as part one.
Discussion
Question11
Part1
Accordingto the three graphs plotted from the data derived from the experimentfor,and against time, it is seen that the graph for lnagainst time has the highest correlation at 0.994. The graph’slineoffit is the closest to the point. The reaction for crystalviolet vs sodium hydroxide is that of first order.
Part2
Accordingto the three graphs plotted from the data derived from the experimentfor,and against time, it is seen that the graph for lnagainst time has the highest correlation at 0.9958. The graph’slineoffit is the closest to the point. The reaction for crystalviolet vs sodium hydroxide is that of first order.
Question12
Part1
PseudoRate constant K^{’}is given by the gradient of the chosen line. The chosen line is ln[CV^{+}]vs time. The equation of the best line of fit is given as y =0.0762x – 0.4709.
Thegradient = 0.0762
Butfor first order, the K^{’}= Slope
K^{’}= – (0.0762)
K^{’}= 0.0762 min^{1}
Part2
PseudoRate constant K^{’}_{2}is given by the gradient of the chosen line. The chosen line is ln[CV^{+}]vs time. The equation of the best line of fit is given as y =0.0385x – 0.5307.
Thegradient = 0.0385
Butfor first order, the K^{’}= Slope
K^{’}_{2}= – (0.0385)
K^{’}_{2}= 0.0385 min^{1}
Question13
ForT=5min
Thelinear equation y =mx+c is represented by
ln[CV+] = kt + ln[CV+]_{0}
y= 0.0762x – 0.4709
ln[CV+]= (0.0762x 5)+ 0.4709
ln[CV+]=0.8519
[CV+]=0.42660
Rate=k^{’}[CV+]
Rate= 0.42660x 0.0762
Rate= 0.0325
ForT=10min
Thelinear equation y =mx+c is represented by
ln[CV+]= kt + ln[CV+]_{0}
y= 0.0762t – 0.4709
ln[CV+]= (0.0762x 10)+ 0.4709
ln[CV+]=1.233
[CV+]=0.2914
Rate=k^{’}[CV+]
Rate= 0.2914x 0.0762
Rate= 0.0222
ForT=15min
Thelinear equation y =mx+c is represented by
ln[CV+]= kt + ln[CV+]_{0}
y= 0.0762t – 0.4709
ln[CV+]= (0.0762x 15)+ 0.4709
ln[CV+]=1.6139
[CV+]=0.1991
Rate=k^{’}[CV+]
Rate= 0.1991×0.0762
Rate= 0.0152
Question16
Integratedform for Run 1 Rate Law is
ln[CV] = kt + ln [CV_{o}]
Question17
Halflifefor Run one
Butk = 0.0762
Question18
n= ln (k_{2}/k1)/ log ([OH–]1/[OH–]2)
n=ln(0.0762/0.0385)/(ln(0.685/0.0262)
n=0.2094
Conclusion
Inconclusion, the objectives of the report were and from the experimentthe relationship between concentration and time elapsed was deduced.It was found that the amount of hydroxide used during the reaction isnegligible. The reaction was confirm be of the first order and therate constant for the reaction was also determined. The experimentwas thus a success.
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