The time taken for half of the initial amount | vedclass

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(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.At $T_1 = 310 \ K$,$t_{1/2} = 12 \ min$,so $k_1 = \frac{0.693}{12

Vedclass · Accepted Answer

$177.76$

Vedclass · Answer

(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.At $T_1 = 310 \ K$,$t_{1/2} = 12 \ min$,so $k_1 = \frac{0.693}{12} \ min^{-1}$.At $T_2 = 300 \ K$,$t_{1/2} = 2 \ hrs = 120 \ min$,so $k_2 = \frac{0.693}{120} \ min^{-1}$.Using the Arrhenius equation: $\ln\left(\frac{k_1}{k_2}ight) = \frac{E_a}{R} \left(\frac{T_1 - T_2}{T_1 T_2}ight)$.$\ln\left(\frac{0.693/12}{0.693/120}ight) = \ln(10) = 2.303$.$2.303 = \frac{E_a}{8.3} \left(\frac{310 - 300}{310 	imes 300}ight)$.$2.303 = \frac{E_a}{8.3} \left(\frac{10}{93000}ight) = \frac{E_a}{8.3 	imes 9300}$.$E_a = 2.303 	imes 8.3 	imes 9300 \approx 177760 \ J \ mol^{-1} = 177.76 \ kJ \ mol^{-1}$.

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

The time taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

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