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Ymx Part 2 Calculations lnk=-Ea R .) + constant 1. Using the slope of the line of the Arrhenius plot, calculate the activatio
Temperature, K 205.15 296.15 293.15 296.15 301.15 305.15 IT,K) * 35|410 3.xio” 3.46310 23.38x10$ 3.82x16° 3.28x101 Imela LM
R. M S 179x10 1.656x16 2.19810 3.6000 16.7 No 1.30x10. b) Y 114,84 -6.41 15.84 E500 -5.00 -4.34
Using the appropriate data from Table 4, construct an Arrhenius plot, In k vs. 1/T(K), using Excel where the plot type is Sca

first create a graph to solve the arrhenius equation for activation energy

also need the graph please
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Answer #1

As per the given data, a lnK vs 1/T plot is created using excel and is given in the image below:

Since slope of lnK vs 1/T plot = - Ea / R, the slope is also given in the said graph.

From the graph, Slope = -4094.9 = - Ea / R

So, Ea = 4094.9 K-1 X 8.313 J K-1mol-1 = 34040.90 J mol-1 or 34.04 kJ mol-1

In K vs 1/T 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035 0.00355 In K y = -4094.9x + 8.5485 R = 0.21

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