Question

The half life for the first order reaction, $N_2{O}_5\to 2N{O}_2+1/2O_2$ is $24$ hours at ${30}^o$C. Starting with $10$g of $N_2{O}_5$, how many grams of $N_2{O}_5$ will remain after a period of $96$ hours?

• A. $1.25$ g
• B. $0.63$ g
• C. $1.77$ g
• D. $0.5$ g

Answer 1

The correct option is B $0.63$ g

Given $t_{1/2}=24hrs$ thus, we can re write

$k=\frac{0.693}{t_{1/2}}=0.02875h{r}^{-1}$

so using $ln\left(\frac{A_o}{A}\right)=kt$ where $A=$ Unknown

we get

$A=A_o{e}^{-kt}$

$\Rightarrow A=10.e^{-0.02875\times 96}$

$=0.6329g$