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Question
The rate constant for the decomposition of N2O5 at various temperatures is given below:
| T/°C | 0 | 20 | 40 | 60 | 80 |
| 105 × k/s−1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and `1/T` and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.
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Solution
The rate constants for the decomposition of N2O5 at different temperatures are shown below.
| T(°C) | T(K) | 1/T | k(s−1) | In k (= 2.303 log k) |
| 0 | 273 | 3.6 × 10−3 | 7.87 × 10−7 | −14.06 |
| 20 | 293 | 3.4 × 10−3 | 1.70 × 10−5 | −10.98 |
| 40 | 313 | 3.19 × 10−3 | 25.7 × 10−5 | −8.266 |
| 60 | 333 | 3.00 × 10−3 | 178 × 10−5 | −6.332 |
| 80 | 353 | 2.8 × 10−3 | 2140 × 10−5 | −3.844 |
Slope of the line = tan θ
= `(y_2 - y_1)/(x_2 - x_1)`
= `(-10.98 - (-14.06))/(3.4 - 3.6) xx 10^3`
= `3.08/-0.2 xx 10^3`
= −15.4 × 103
Ea = −slope × R
= −(−15.4 × 103 × 8.314)
= 128.035 kJ K−1 mol−1
Again,
In A = In k + `E_a/(RT)`
= `-14.06 + (128.035 xx 10^3)/(8.314 xx 273)`
= `-14.06 + 128035/2269.722`
= −14.06 + 56.40
ln A = 42.34
⇒ A = antilog 42.34
= 0.3388 × 1019
Values of the rate constant k at 303 K and 323 K can be obtained from the graph.
First, k is obtained corresponding to `1/(303 K) and 1/(323 K)`, and then k is calculated.
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