Question 15.8: Determining Activation Energy The reaction 2N2O5(g) → 4NO2(g...
Determining Activation Energy
The reaction
was studied at several temperatures and the following values of k were obtained:
k ( s ^{- 1 } ) | T (° C ) |
2.0 × 10 ^{-5} | 20 |
7.3 × 10 ^{-5} | 30 |
2.7 × 10 ^{-4} | 40 |
9.1 × 10 ^{-4} | 50 |
2.9 × 10 ^{-3} | 60 |
Calculate the value of E_{a} for this reaction.
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To obtain the value of E_{a} , we need to construct a plot of \ln (k) versus 1/T. First, we must calculate values of \ln (k) and 1/T:
T ( ° C) | T ( K ) | 1 / T ( K ) | k (s ^{-1}) | \ln (k) |
20 | 293 | 3.41 × 10 ^{-3} | 2.0 × 10 ^{-5} | – 10.82 |
30 | 303 | 3.30 × 10 ^{-3} | 7.3 × 10 ^{-5} | – 9.53 |
40 | 313 | 3.19 × 10 ^{-3} | 2.7 × 10 ^{-4} | – 8.22 |
50 | 323 | 3.10 × 10 ^{-3} | 9.1 × 10 ^{-4} | – 7.00 |
60 | 333 | 3.00 × 10 ^{-3} | 2.9 × 10 ^{-3} | – 5.84 |
The plot of \ln (k) versus 1/T is shown in Fig. 15.14 ▼. The slope is found to be – 1.2 × 10^{4} K. Since
Slope = – \frac{E_{a}}{R}
then
E_{a} = – R (slope ) = – (8.3145 J K ^{-1} mol^{-1}) (- 1.2 × 10^{4} K)
= 1.0 × 10^{5} J / mol
Thus the value of the activation energy for this reaction is 1.0 × 10^{5} J / mol

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