For the first-order reaction N_2O_5 to 2NO_2 | vedclass

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Question

(A) For a first-order reaction,the number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}}$.
Given $t = 96 \ 	ext{hours}$ and $t_{1/2} =

Vedclass · Accepted Answer

$0.625$

Vedclass · Answer

(A) For a first-order reaction,the number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}}$.
Given $t = 96 \ 	ext{hours}$ and $t_{1/2} = 24 \ 	ext{hours}$,we have $n = \frac{96}{24} = 4$.
The amount remaining after $n$ half-lives is given by the formula $N = N_0 	imes (\frac{1}{2})^n$.
Substituting the values,$N = 10 \ g 	imes (\frac{1}{2})^4 = 10 	imes \frac{1}{16} = 0.625 \ g$.
Thus,the amount of $N_2O_5$ remaining after $96 \ 	ext{hours}$ is $0.625 \ g$.

For the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2}O_2$,the half-life at $30 \ ^\circ C$ is $24 \ \text{hours}$. Starting with $10 \ g$ of $N_2O_5$,how many grams of $N_2O_5$ will remain after $96 \ \text{hours}$ (in $g$)?

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