The half-life for the first-order reaction N_ | vedclass

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(C) Initial moles of $N_2O_5 = \frac{10.8 	ext{ g}}{108 	ext{ g/mol}} = 0.1 	ext{ mol}$.Number of half-lives $n = \frac{t}{t_{1/2}} = \frac{9.6 	e

Vedclass · Accepted Answer

$1.05$

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(C) Initial moles of $N_2O_5 = \frac{10.8 	ext{ g}}{108 	ext{ g/mol}} = 0.1 	ext{ mol}$.Number of half-lives $n = \frac{t}{t_{1/2}} = \frac{9.6 	ext{ h}}{2.4 	ext{ h}} = 4$.Moles of $N_2O_5$ remaining after $4$ half-lives $= 0.1 	imes (\frac{1}{2})^4 = \frac{0.1}{16} = 0.00625 	ext{ mol}$.Moles of $N_2O_5$ reacted $= 0.1 - 0.00625 = 0.09375 	ext{ mol}$.From the reaction $N_2O_5 	o 2NO_2 + \frac{1}{2} O_2$,$1 	ext{ mole}$ of $N_2O_5$ produces $0.5 	ext{ mole}$ of $O_2$.Moles of $O_2$ produced $= 0.5 	imes 0.09375 = 0.046875 	ext{ mol}$.Volume of $O_2$ at $STP = 0.046875 	ext{ mol} 	imes 22.4 	ext{ L/mol} = 1.05 	ext{ L}$.

The half-life for the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2} O_2$ is $2.4 \text{ hours}$. Starting with $10.8 \text{ g}$ of $N_2O_5$ at $STP$,how many liters of oxygen will be obtained after a period of $9.6 \text{ hours}$? (Given: $10^{0.2} = 1.584$)

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