How the shapes of the plots [dye] vs time, ln[dye] vs time, and 1/[dye] vs time are used to determine the order of the reaction with respect to the dye concentration?

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Feb 7, 2018

Answer:

Warning! Long Answer. They come from the integrated rate laws for zero-, first-, and second-order reactions.

Explanation:

Chemists like to plot graphs that are straight lines.

Zero-order reactions

The integrated rate law is

#["A"] = "-"kt + ["A"]_0#

Compare this with the equation for a straight line.

#["A"] = "-"kt + ["A"]_0#
#color(white)(l)ycolor(white)(ll) = mx +color(white)(l) b#

If we let #y = ["A"]# and #x = t#, a plot of # ["A"]# vs #t# is a straight line with slope #m = "-"k# and #x#-intercept #["A"]_0#

Thus, if a plot of # ["A"]# vs #t# is a straight line with a negative slope, the reaction is zero-order.

www.chem.purdue.edu

First-order reactions

The integrated rate law is

#ln["A"] = -kt + ln["A"]_0#

Compare this with the equation for a straight line.

#ln["A"] = "-"kt + ["A"]_0#
#color(white)(m)ycolor(white)(ll) = color(white)(l)mx +b#

If we let #y = ln["A"]# and #x = t#, a plot of #ln["A"]# vs #t# is a straight line with slope #m = "-"k# and #x#-intercept #ln["A"]_0]#

Thus, if a plot of #ln["A"]# vs #t# is a straight line with a negative slope, the reaction is first-order.

www.chem.purdue.edu

Second-order reactions

The integrated rate law is

#1/(["A"]) = kt + 1/(["A"]_0)#

Compare this with the equation for a straight line.

#1/(["A"]) = color(white)(l)kt + 1/(["A"]_0)#
#color(white)(ll)ycolor(white)(ll) = color(white)()mx + color(white)(ll)b#

If we let #y = 1/(["A"])# and #x = t#, a plot of #1/(["A"])# vs #t# is a straight line with slope #m = k# and #x#-intercept #1/(["A"]_0#.

Thus, if a plot of #ln1/(["A"])# vs #t# is a straight line with a positive slope, the reaction is second-order.

www.chem.purdue.edu

Thus, you plot [dye] vs time, ln[dye] vs time, and 1/[dye] vs time.

The graph that is a straight line gives you the order of the reaction.

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