Red dye and Hypochlorite lab: I am asked to use the plots that give linear relat

Question

Red dye and Hypochlorite lab: I am asked to use the plots that give linear relationships to determine the rate constant for each trial. This is one trial where there is a linear relationship with the Ln[red dye]. How do I determine the rate constant? The rate law would be -> rate=k{A][B] where it is second order overall. How do I calculate K for rate constant?

Data given with three graphs but the graph of Ln[red dye] is the most linear.

Please help. Thank you!

Here is more information, I'm having trouble answering #2, #4 and #7. Mostly because I don't know what values to use.

Here is the calibration curve:

SAMPLE 6 Time Absorbance In[red dye red dye] 20 40 60 80 100 120 140 160 180 200 220 240 0.266 0.228 0.191 0.161 0.137 0.116 0.097 0.083 0.07 0.059 0.051 0.044 1.324258973.759398496 1.47840965 4.385964912 1.6554818515.235602094 1.8263509146.211180124 1.987774353.299270073 2.1541650888.620689655 2.3330443 10.30927835 2.488914671 2.04819277 2.659260037 14.28571429 2.830217835 16.94915254 2.975929646 19.60784314 3.123565645 22.72727273

Explanation / Answer

ANS 7:

First we look at the graphs. The most linear concentration vs time graph will give the order of the reaction. If the reaaction is of second order than 1/[dye] vs time will be linear but if reaction is of first order than ln{dye} vs time graph will be linear.

From all the graphs here it is evident that ln[dye] vs time graph is the most linear here.So reaction is first order with respect to [dye].

So we will use 1st order rate of reaction formula here.

First we calculate rate of constant here.

We know that-

ln[red dye] = ln[red dye]o - K. t

We will take two times here one t1=20 second t2= 40

Since initial concentration will be same, so our two equations will be

ln[red dye]1 = ln[red dye]o- 20K .....(1)

ln[red dye]2 = ln[red dye]o- 40K .......(2)

Substracting (1) from (2) and putting values of ln[red dye] for respective times

-1.4784-(-1.3242) = -40K+20K

-1.4784+1.3242 = -20K

-0.1542 = -20K

K = 0.1542/20 = 0.00771 sec-1

ANS 2: Similarly applying the first order equation we can calculate initial concentrations by using rate constant for calculating the initial concentration.

For instance, at t=20 ,

ln[dye] = ln[dye]o - kt

at t=20 ln[dye] = -1.3242

-1.3242 = ln[dye]o - 20*0.00771

-1.3242+0.1542 = ln[dye]o

ln[dye]o = -1.17

= exp(-1.17)

= 0.310 mol/litre

Using this approach you can calculate for all trials.

  

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