The rate constant for a first order reaction | vedclass

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Question

(B) The Arrhenius equation is given by: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.Given: $k_1 = 0.026 \ s^{-1}$ at $T_1 = 290

Vedclass · Accepted Answer

$224.55 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The Arrhenius equation is given by: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.Given: $k_1 = 0.026 \ s^{-1}$ at $T_1 = 290 \ K$,$k_2 = 0.58 \ s^{-1}$ at $T_2 = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $\ln(\frac{0.58}{0.026}) = \frac{E_a}{8.314} [\frac{300 - 290}{300 	imes 290}]$.$\ln(22.307) = \frac{E_a}{8.314} [\frac{10}{87000}]$.$3.1048 = \frac{E_a}{8.314} 	imes 1.1494 	imes 10^{-4}$.$E_a = \frac{3.1048 	imes 8.314}{1.1494 	imes 10^{-4}} \approx 224600 \ J \ mol^{-1} = 224.6 \ kJ \ mol^{-1}$.The closest option is $224.55 \ kJ \ mol^{-1}$.

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

The rate constant for a first order reaction is $0.58 \ s^{-1}$ at $300 \ K$ and $0.026 \ s^{-1}$ at $290 \ K$. What is the energy of activation? $(R=8.314 \ J \ K^{-1} \ mol^{-1})$

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For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

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

Activation energy of any reaction depends on

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

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