The first order rate constant for a certain r | vedclass

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(C) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given $T_1 = 727 + 273 = 1000

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$3.318 	imes 10^{-5} \ s^{-1}$

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(C) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given $T_1 = 727 + 273 = 1000 \ K$,$T_2 = 1571 + 273 = 1844 \ K$,$T_3 = 1150 + 273 = 1423 \ K$.For the first interval: $\log \left( \frac{1.667 	imes 10^{-4}}{1.667 	imes 10^{-6}} ight) = \frac{E_a}{2.303 \ R} \left( \frac{1844 - 1000}{1844 	imes 1000} ight)$$2 = \frac{E_a}{2.303 \ R} \left( \frac{844}{1844000} ight) \dots (1)$For the second interval: $\log \left( \frac{k_3}{1.667 	imes 10^{-6}} ight) = \frac{E_a}{2.303 \ R} \left( \frac{1423 - 1000}{1423 	imes 1000} ight) \dots (2)$Dividing $(2)$ by $(1)$:$\frac{\log \left( \frac{k_3}{1.667 	imes 10^{-6}} ight)}{2} = \frac{423}{1423 	imes 1000} 	imes \frac{1844000}{844} \approx 0.6495$$\log \left( \frac{k_3}{1.667 	imes 10^{-6}} ight) = 1.299$$\frac{k_3}{1.667 	imes 10^{-6}} = 19.9$$k_3 = 19.9 	imes 1.667 	imes 10^{-6} = 3.318 	imes 10^{-5} \ s^{-1}$

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