Calculate the activation energy of a reaction | vedclass

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Question

(A) The relation between temperature and activation energy is given by the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left

Vedclass · Accepted Answer

$3.45 \ kJ / mol$

Vedclass · Answer

(A) The relation between temperature and activation energy is given by the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}ight)$.Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 600 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $\log(2) = \frac{E_a}{2.303 	imes 8.314} \left(\frac{1}{300} - \frac{1}{600}ight)$.$0.3010 = \frac{E_a}{19.147} \left(\frac{2-1}{600}ight)$.$0.3010 = \frac{E_a}{19.147 	imes 600}$.$E_a = 0.3010 	imes 19.147 	imes 600 \approx 3458 \ J / mol = 3.46 \ kJ / mol$.

$Case$	$E_a \ (kcal/mol)$	$Temp. \ Change \ (K)$
$I$	$40$	$200 - 210$
$II$	$80$	$200 - 210$
$III$	$40$	$300 - 310$
$IV$	$80$	$300 - 310$

Calculate the activation energy of a reaction,whose rate constant doubles on raising the temperature from $300 \ K$ to $600 \ K$.

Similar Questions

In which of the following cases is the percentage increase in rate constant maximum?
$Case$ $E_a \ (kcal/mol)$ $Temp. \ Change \ (K)$
$I$ $40$ $200 - 210$
$II$ $80$ $200 - 210$
$III$ $40$ $300 - 310$
$IV$ $80$ $300 - 310$

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In the presence of a catalyst,the heat evolved or absorbed during the reaction . . . . . . .

Activation energy of any chemical reaction can be calculated if one knows the value of

An increase in temperature by $10\,^{\circ}C$,generally increases the rate of a reaction by .......... times.

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