For the thermal decomposition of a gaseous compound, the plot of _...

According to the Arrhenius equation:

$k = A e^{-E_a / RT}$

Taking logarithm to the base 10 on both sides:

$\log_{10} k = \log_{10} A - \dfrac{E_a}{2.303 RT}$

Comparing this with the equation of a straight line $y = mx + c$, the slope $m$ of the plot of $\log_{10} k$ versus $\dfrac{1}{T}$ is given by:

$m = -\dfrac{E_a}{2.303 R}$

Given that the slope is $-6.0 \times 10^3 \text{ K}$:

$-\dfrac{E_a}{2.303 R} = -6.0 \times 10^3$

$E_a = 6.0 \times 10^3 \times 2.303 \times R$

Substituting the given value of $2.303 \times R = 19.15 \text{ J K}^{-1} \text{ mol}^{-1}$:

$E_a = 6.0 \times 10^3 \times 19.15 \text{ J mol}^{-1}$

$E_a = 114900 \text{ J mol}^{-1} = 114.9 \text{ kJ mol}^{-1}$

Answer: $114.9 \text{ kJ mol}^{-1}$