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Reward This is What Is Added To X 102 Manual.pdf X W Persulfate Wikipedia X C O

ID: 525597 • Letter: R

Question

Reward This is What Is Added To X 102 Manual.pdf X W Persulfate Wikipedia X C O ars-chemia net Manual.pdf 102 Manual.pdf 90 131 90 4. The following data was obtained for a reaction in which a chemical, X, decomposed. Concentration of X (in Molarity) Time (in seconds) 5.00 5.00 x 102 3.52 2.48 0.00 x 102 15.00 x 102 1.75 20.00 x 102 1.23 Chem 102 students are expected to prepare proper graphs (or lose points!) Appendix B of this document is a reprint of the Chem 101 lab manual's graphing exercise which includes instructions for proper graph constructionby hand and using Excel a. Prepare plots of concentration versus time using the provided data in a manner appropriate for zero, first and second order processes. You must include all three graphs with your report. b. Based on your graphs, is this reaction zero, first, or second order forX? c. Determine the slope for the straight line graph. Show how you arrived at the value for the slope of the line. Calculate the rate constant for this reaction from the slope. d. Write the complete rate law for the reaction including the value k (with units) XE P 5 3:46 PM 5/5/2017

Explanation / Answer

for any order reaction, -dCA/dt= KCAn, n is ther order of reaction.

for 1st order, -dCA/dt= KCA, when integrated lnCA= lnCAO- Kt where CAO= initial concentration of A and CA= concentration of A at any time and K= rate constant

So a plot of ln CA vs t gives straight line whose slope is -K, id this is the case, then the reaction is 1st order.

The plot is shown below along with data points.

The reaction is 1st order since the plot is straight line.

define t'= t/100

hence lnCA= lnCAO- K*t'*100

the equation of best fit is lnCA= lnCAO- Kt' = lnCAO- K*t*100

hence slope -K*100 =-0.070, K= 0.070/100 sec-1=0.0007/sec

-dCA/dt= 0.007*CA

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