Question

finding half - life and rate constant from a graph of concentration versus...
use this graph to answer the following questions:
what is the half - life of the reaction?
round your answer to 2 significant digits.
suppose the rate of the reaction is known to be first order in n₂o₅. calculate the value of the rate constant k.
round your answer to 2 significant digits. also be sure you include the correct unit symbol.
predict the concentration of n₂o₅ in the engineers reaction vessel after 240. seconds have passed.
assume no other reaction is important and continue to

Explanation:

Step1: Determine half - life from graph

The half - life $t_{1/2}$ is the time when the concentration of $N_2O_5$ is half of its initial value. From the graph, if the initial concentration is around $1.0\times10^{-5}\ M$ and it reaches half of that value at $t = 60\ s$. So $t_{1/2}=60\ s$.

Step2: Calculate rate constant for first - order reaction

For a first - order reaction, the relationship between the half - life and the rate constant $k$ is $t_{1/2}=\frac{\ln2}{k}$. Rearranging for $k$, we get $k=\frac{\ln2}{t_{1/2}}$. Substituting $t_{1/2}=60\ s$, $k = \frac{\ln2}{60\ s}\approx0.012\ s^{-1}$.

Step3: Predict concentration after a time

The integrated rate law for a first - order reaction is $\ln\frac{[A]_t}{[A]_0}=-kt$. Let $[A]_0$ be the initial concentration (assume $[A]_0 = 1.0\times10^{-5}\ M$), $k = 0.012\ s^{-1}$ and $t = 240\ s$. Then $\ln\frac{[A]_t}{[A]_0}=-0.012\ s^{-1}\times240\ s=- 2.88$. So $\frac{[A]_t}{[A]_0}=e^{-2.88}\approx0.056$. Then $[A]_t=0.056\times1.0\times10^{-5}\ M = 5.6\times10^{-7}\ M$.

Answer:

$t_{1/2}=60\ s$
$k = 0.012\ s^{-1}$
$[N_2O_5]=5.6\times10^{-7}\ M$