Question

The decomposition of N2O5(g) —> NO2(g) + NO3(g) proceeds as a first order reaction with a half life 30.0s at a certain temperature.
If the initial concentration [N2O5]0 = 0.400 M, what is the concentration after 120 seconds?
A)0.050
B)0.200
C)0.025
D)0.100

Answer 1

Answer:

$\large \boxed{\text{0.025 mol/L}}$

Explanation:

The half-life (30.0 s) is the time it takes for half of the N₂O₅ to react.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

$\begin{array}{crcc}\textbf{No. of} && \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/s} & \textbf{remaining} &\rm \mathbf{{[N_{2}O_{5}] /(mol/L)}}\\0 & 0 & 1 & 0.400\\1 & 30.0 & 1/2 & 0.200\\2 & 60.0 & 1/4 & 0.100\\3 & 90.0 &1/8 & 0.050\\4 & 120.0 & 1/16 & 0.025\\5& 150.0 & 1/32 & 0.012\\\end{array}$

$\text{We see that the concentration has dropped to \large \boxed{\textbf{0.025 mol/L}} after 120 s}$