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  •  for a chemicalfeaction, the activation Energy (sa) is <img alt="50 \mathrm{~kJ} / \mathrm{mat}." src="https://entrancecorner.oncodecogs.com/gif.latex?50%20%5Cmathrm%7B%7EkJ%7D%20/%20%5Cm

 for a chemicalfeaction, the activation Energy (sa) is 50 \mathrm{~kJ} / \mathrm{mat}. If the temperature is increased form. \mathrm{25^{\circ} \mathrm{C}} to \mathrm{50^{\circ} \mathrm{C}}, what will be the approximate change in the rate constant (k) bused on Arrhenius Equation \mathrm{\left(k=A * e^{-\varepsilon a / k T)}\right.}
 

Option: 1

The rate constant will increased by approx 2 times
 


Option: 2

 The rate constant will increased by area 4 times.
 


Option: 3

The rate constant will mereased by


Option: 4

 The rate constant will increased by corer  16 times. 


Answers (1)

best_answer

\mathrm{\begin{aligned} \log \left(\frac{k_2}{k_1}\right) & =\frac{E_a}{2303 \times R} \times\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ \log \left(\frac{k_2}{k_1}\right) & =\frac{50 \times 10^3}{2.303 \times 19.147} \times\left[\frac{1}{298}-\frac{1}{323}\right] \\ \log \left(\frac{x_2}{k_1}\right) & =\frac{50000}{19.147} \times\left(\frac{25}{298 \times 373}\right) \\ \log \left(\frac{x_2}{k_1}\right) & =0.670 \\ \left(\frac{k_2}{k_1}\right) & =10^{60.670)}=4.6 \\ k_2 & =4.6 \times k_1 \end{aligned} }

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