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# The rate constant for the decomposition of N2O5 at various temperatures is given below: T/°C → 0 - 20 - 40 - 60 - 80

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The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C → 0 - 20 - 40 - 60 - 80

105 x 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 30 º and 50ºC.

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From the given data, we obtain

Slope of the line
$$\frac{y_2 - y_1}{x_2 - x_1}$$ = -12.301 K
According to Arrhenius equation,
Slope = -$$\frac{E_a}{R}$$
⇒ Ea = -Slope x R
= -(-12.301K) x (8.314 JK-1mol-1)
= 102.27 kJ mol-1
Again,
In k = In A - $$\frac{E_a}{RT}$$
In A = In k - $$\frac{E_a}{RT}$$
When T = 273 K,
Then, In A = -7.147 + $$\frac{102.27 \times 10^3}{8.314 \times 273}$$
= 37.911
Therefore, A = 2.91 x 106
When T = 30 + 273K = 303K
$$\frac{1}{T}$$ = 0.0033K = 3.3 x 10-3 K
Then, at $$\frac{1}{T}$$ = 3.3 x 10-3 K,
ln k = -2.8
Therefore, k = 6.08 x 10-2 s-1
Again, when T = 50 + 273K = 323K
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