Please show work the student conducts two different runs. in RUN 1, 10.0 mL of 5
ID: 712384 • Letter: P
Question
Please show work the student conducts two different runs. in RUN 1, 10.0 mL of 5.00 M NaOH solution and 80.0 mL of water. The absorbance i Beer's law is used to convert this is combined with 10.0 ml of 0.010 M Crystal Orange monitored as a function of time, and absorbance to a Crystal Orange concentration. Graphs of (Crystal Orangel vs time, In/Crystal Orange] vs time, and 1/Crystal are made. Orangel vs time In RUN 2, 20.0 mL of 5.00 M NaOH is combined with 10.0 ml of 0.010 M Crystal Orange solution and 70.0 mL of water. The absorbance is monitored as a function of time, and Beer's law is used to convert this absorbance to a Crystal Orange concentration. Only one graph of this data for RUN 2 is made: a Graph 1/Crystal Orange] vs time. All three graphs for RUN 1, and 1/Crystal Orange] vs time for RUN 2 are shown below [Crystal Orange] vs time (RUN in(Crystal Orangej vs time (RUN 1) 0.0007 S00012 00002 1/TCrystal Orange] vs time (RUN 1) /TCrystal Orange) vs (RUN 2) y2000+1000 r' 4000 12000 10000 5000 4000 2000 Time (hour Time Chour Applying the methods used in the Crystal Violet-NaOH experiment, use this data to determine the rate law for the reaction of Crystal Orange with NaOH. k, with units Also include a value for Rate- REKFICB Edits 17 15
Explanation / Answer
From the plots shown above
The rate of reaction was doubled when the concentration of NaOH was doubled
order for NaOH = 1
order for crystal violet = 1
So,
rate = k[CV][OH-]
CV = crysta violet
rate constant (k)
k = 2000 M-1.h-1