I am not sure how to finish this problem? Determine the activation energy for th

Question

I am not sure how to finish this problem?

Determine the activation energy for the reaction investigated in this experiment. Use a graphical analysis of all the necessary data using the Arrhenius equation. Make a plot of 1/T v.s lnK. Calculate the slope of the line (=Ea/R), then determine Ea. use this equation Ln(k1/k2) = (Ea/R) (1/T1-1/T2)

I have allready found this out.
lnk1 =-10.48867
lnk2 = -11.20397
lnk3 = -10.22507
lnk4 = -9.0675

and
1/T =.00336304
2/T =.0034710
3/T =.00325574
4/T =.003148615

I am not sure how to find the activation energy, the slope, and Ea.

Explanation / Answer

When you plot the values of lnK for various values of 1/T,then you will get a straight line..Select any two points on the line, say(x1,y1) and (x2,y2)..Then The slope is (y2-y1)/(x2-x1)

The Arrhenius Equation is : k= Ae-E/RT

or lnK = -Ea/RT + lnA

Comparing this with y=mx+c(where y=lnK,m= -Ea/R,x=1/T,c=ln A)

Slope = m = -Ea/R = Slope calculated above)

Ea = -Slope * R(R = 8.314)

Ea is activation Energy

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