The decomposition of hydrocarbon follows the equation
k = (4.5 x 10¹¹ s^–1) e^–28000K/T
Calculate E_a. 

K₁ = 4.5 x 10³ S^–1 ,  T₁ = 273 + 10 = 283 K
Ea = 60 kJ mol^–1 T₂ = ?
K₂ = 1.5 x 10⁴S^–1
We know that

$$\begin{gathered} \log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right) = \frac{{\mathrm{E}}_{\mathrm{a}}({\mathrm{T}}_2-{\mathrm{T}}_1)}{2.303\mathrm{R} {\mathrm{T}}_1{\mathrm{T}}_2} \\ \log \left(\frac{1.5 \times {10}^4}{4.5\times {10}^3}\right) = \frac{60({\mathrm{T}}_2-283)}{2.303 \times 8.314 \times 283 \times {\mathrm{T}}_2} \\ \log \left(\frac{150}{45}\right) = \frac{60({\mathrm{T}}_2\times 283)}{5418.64 {\mathrm{T}}_2} \\ \log 150 - \log 45 = \frac{60{\mathrm{T}}_2-16980}{5418.64 {\mathrm{T}}_2} \\ 2.1761 - 1.6532 = \frac{60{\mathrm{T}}_2-16980}{5418.64 {\mathrm{T}}_2} \\ 0.5229 = \frac{60{\mathrm{T}}_2-16980}{5418.64 {\mathrm{T}}_2} \\ 2833.4 {\mathrm{T}}_2 = 60{\mathrm{T}}_2-16980 \\ 2833.4 {\mathrm{T}}_2 = 60{\mathrm{T}}_2 = -16980 \\ 2773.4 {\mathrm{T}}_2 = 16980 \\ {\mathrm{T}}_2 = \frac{16980}{2733.4} = 6.1 \mathrm{K} \end{gathered}$$