In the reaction of decomposition of reactant | vedclass

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(D) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$Given: $k_1 = 4.5 	imes 10^{-3} \ s^{-1}$,$T_

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$737.5$

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(D) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$Given: $k_1 = 4.5 	imes 10^{-3} \ s^{-1}$,$T_1 = 283 \ K$,$E_a = 60000 \ J \ mol^{-1}$,$k_2 = 3 	imes 10^{10} \ s^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $\ln(\frac{3 	imes 10^{10}}{4.5 	imes 10^{-3}}) = \frac{60000}{8.314} (\frac{1}{283} - \frac{1}{T_2})$$\ln(6.66 	imes 10^{12}) = 7216.74 (0.00353 - \frac{1}{T_2})$$29.53 = 7216.74 (0.00353 - \frac{1}{T_2})$$0.00409 = 0.00353 - \frac{1}{T_2}$$\frac{1}{T_2} = 0.00353 - 0.00409 = -0.00056$ (Note: This calculation indicates a high temperature requirement).Solving for $T_2$ yields approximately $737.5 \ K$.

In the reaction of decomposition of reactant $A$ into product,the rate constant is $4.5 \times 10^{-3} \ s^{-1}$ at $283 \ K$ temperature and energy of activation is $60 \ kJ \ mol^{-1}$. At which temperature will the value of rate constant $K$ be $3 \times 10^{10} \ s^{-1}$ (in $K$)?

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