# The rate constant for the decomposition of

Question:

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

Draw a graph between ln and 1/and calculate the values of and Ea.

Predict the rate constant at 30º and 50ºC.

Solution:

From the given data, we obtain

Slope of the line,

According to Arrhenius equation,

Slope $=-\frac{E_{a}}{R}$

$\Rightarrow E_{a}=-S$ lope $\times \mathrm{R}$

$=-(-12.301 \mathrm{~K}) \times\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

$=102.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Again,

$\ln k=\ln A-\frac{E_{a}}{\mathrm{R} T}$

$\ln A=\ln k+\frac{E_{a}}{\mathrm{R} T}$

When $T=273 \mathrm{~K}$,

$\ln k=-7.147$

Then, $\ln A=-7.147+\frac{102.27 \times 10^{3}}{8.314 \times 273}$

$=37.911$

Therefore, $A=2.91 \times 10^{6}$

When $T=30+273 \mathrm{~K}=303 \mathrm{~K}$,

$\frac{1}{T}=0.0033 \mathrm{~K}=3.3 \times 10^{-3} \mathrm{~K}$

Then, at $\frac{1}{T}=3.3 \times 10^{-3} \mathrm{~K}$,

$\ln k=-2.8$

Therefore, $k=6.08 \times 10^{-2} \mathrm{~s}^{-1}$

Again, when $T=50+273 \mathrm{~K}=323 \mathrm{~K}$,

$\frac{1}{T}=0.0031 \mathrm{~K}=3.1 \times 10^{-3} \mathrm{~K}$

Then, at $\frac{1}{T}=3.1 \times 10^{-3} \mathrm{~K}$,

In $k=-0.5$

Therefore, $k=0.607 \mathrm{~s}^{-1}$