The rate constant of a first order reaction i | vedclass

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(B) Given: $K_1 = 3.46 	imes 10^{-2} \ s^{-1}$,$T_1 = 298 \ K$,$T_2 = 350 \ K$,$E_a = 50.1 \ kJ \ mol^{-1} = 50100 \ J \ mol^{-1}$,$R = 8.314 \ J \ K

Vedclass · Accepted Answer

$0.692$

Vedclass · Answer

(B) Given: $K_1 = 3.46 	imes 10^{-2} \ s^{-1}$,$T_1 = 298 \ K$,$T_2 = 350 \ K$,$E_a = 50.1 \ kJ \ mol^{-1} = 50100 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$.Substituting the values:$\log \frac{K_2}{3.46 	imes 10^{-2}} = \frac{50100}{2.303 	imes 8.314} \left[ \frac{350 - 298}{298 	imes 350} ight]$.$\log \frac{K_2}{3.46 	imes 10^{-2}} = \frac{50100}{19.147} 	imes \left[ \frac{52}{104300} ight]$.$\log \frac{K_2}{3.46 	imes 10^{-2}} = 2616.59 	imes 0.0004985 \approx 1.304$.$\frac{K_2}{3.46 	imes 10^{-2}} = 10^{1.304} \approx 20.14$.$K_2 = 20.14 	imes 3.46 	imes 10^{-2} \approx 0.696 \ s^{-1}$.The closest option is $0.692 \ s^{-1}$.

$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$

The rate constant of a first order reaction is $3.46 \times 10^{-2} \ s^{-1}$ at $298 \ K$. What is the rate constant of the reaction at $350 \ K$ if its activation energy is $50.1 \ kJ \ mol^{-1}$ (in $s^{-1}$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$(\log 2 = 0.3010)$

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The rate constant of a first order reaction is $3.46 \times 10^{-2} \ s^{-1}$ at $298 \ K$. What is the rate constant of the reaction at $350 \ K$ if its activation energy is $50.1 \ kJ \ mol^{-1}$ (in $s^{-1}$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$$(\log 2 = 0.3010)$