The values of rate constants for the decomposition of N2O5 at varioustemperatures are given below :Slope of the line = tan θ $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$= \frac{-10.98 - (-14.08)}{3.4 - 3.6} \times 10^{3}$$= -15.5 \times 10^{3}$$E_{a} = -$ slope $\times R$$= - ( -15.5 \times 10^{3} \times 8.314 )$$= 128.86\ \text{kJ}\ \text{K}^{-1}\ \text{mol}^{-1}$$\text{Again}\ \ln A = \ln k + \frac{E_{a}}{R T}$$= -14.06 + \frac{128.86 \times 10^{3}\ \mathrm{J\,K}^{-1}\,\mathrm{mol}^{-1}}{8.314 \times 273}$$= -14.06 + 56.77 = 42.71$ or, $\log A = 18.53$or, $A =$ antilog $18.53$$= 0.3388 \times 10^{19}$ or,$A = 3.388 \times 10^{18}$Value of rate constant k at 303 K and 323 K can be obtained from graph.First of all ln k is obtained corresponding to$\frac{1}{303}\ \mathrm{K}$and $\frac{1}{323}\ \mathrm{K}$and thenk is calculated.