The rate coefficient (k) for a particular rea | vedclass

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(B) According to 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: $k_1 = 1.3 	ime

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) According to 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: $k_1 = 1.3 	imes 10^{-4} \ M^{-1} \ s^{-1}$,$T_1 = 100 + 273 = 373 \ K$$k_2 = 1.3 	imes 10^{-3} \ M^{-1} \ s^{-1}$,$T_2 = 150 + 273 = 423 \ K$$\log \left( \frac{1.3 	imes 10^{-3}}{1.3 	imes 10^{-4}} ight) = \frac{E_a}{2.303 	imes 8.314} \left( \frac{423 - 373}{373 	imes 423} ight)$$\log(10) = \frac{E_a}{19.147} \left( \frac{50}{157779} ight)$$1 = \frac{E_a 	imes 50}{3020755.7}$$E_a = \frac{3020755.7}{50} \approx 60415 \ J \ mol^{-1} \approx 60.4 \ kJ \ mol^{-1}$Thus,the closest value is $60 \ kJ \ mol^{-1}$.

The rate coefficient $(k)$ for a particular reaction is $1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$ at $100 \ ^oC$ and $1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$ at $150 \ ^oC$. What is the energy of activation $(E_a)$ (in $kJ \ mol^{-1}$) for this reaction? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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