If you were going to graphically determine the activation energy of this reactio

Question

If you were going to graphically determine the activation energy of this reaction, what points would you plott? Consider this reaction data: Al products Number Number T (K) K(S 275 0 399 825 0.815 point 1 275 0.399 To avoid rounding l errors, use at least n three significant Number Number figures in all values point 2: 825 Determine the rise, run, and slope of the line formed by these points. rise run slope Number Number Number 416 550 7.56 x 10-4 What is the activation energy of this reaction? Number 25%2ol J/ mol Explanation Give Up & View Solution # Try Again O Next tl En- Previous

Explanation / Answer

First Question: Points to graph in Arrhenius graph (lnK vs 1 / T):

Point 1:

x1 = 1 / T1 = 1/275 = 0.00364

y1 = lnK1 = ln (0.399) = -0.919

Point 2:

x2 = 1 / T2 = 1/875 = 0.00114

y2 = lnK2 = ln (0.815) = -0.205

Slope:

(y2-y1) / (x2-x1) = - (0.205-0.919) /0.00114-0.00364 = -285.6

Run:

x2-x1 = 0.00114 - 0.00364 = -0.0025

Rise:

y2-y1 = 0.205-0.919 = -0.714

Activation energy:

2450.74 J / mol.

Second Question: D2 + H2 = 2HD

Kp = [HD] ^ 2 / [D2] * [H2]

Clearing [H2]:

[H2] = [HD] ^ 2 / [D2] * Kp = (3.1x10 ^ -3 atm) ^ 2 / (1.7 x10 ^ -3 atm) * 1.8 = 3.14 x10 ^ -3 atm.

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