Answer to Question #93774 in General Chemistry for dennis blaul

Question

N2O5(g) LaTeX: \longrightarrow ⟶ 2 NO2(g) + ½ O2(g)



N2O5 decomposes by a first-order reaction. Suppose the initial amount of N2O5 in an experiment is 64 g. What mass (g) remains after four half-lives?

Answer 1

The amount of the starting reactant vs. time in each first-order reaction can be expressed using the following equation:

$A=A_{0}e^{-kt}$

In the case, when t is equal to four times t_1/2, then:

$t_{1/2}=\frac{ln2}{k}$

$A=A_{0}e^{-k\frac{4ln2}{k}}=A_{0}e^{-4ln2}=A_{0}2^{-4}=(\frac{1}{2})^4A_{0}$

Thus,

$A(t_{\frac{1}{2}})=64g*\frac{1}{16}=4g$

The amount of the starting reactant (N₂O₅) remained after four half-time periods would be 4g.