Question

kinetics and equilibrium
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: Define half - life

Half - life ($t_{1/2}$) is the time it takes for the concentration of a reactant to decrease to half of its initial value. From the graph, we can directly read the time when the concentration of $N_2O_5$ is half of its initial value.

Step2: Use first - order rate law formula

For a first - order reaction, the rate constant $k$ is related to the half - life by the formula $k=\frac{\ln 2}{t_{1/2}}$. After finding $t_{1/2}$, we can calculate $k$.

Step3: Use first - order integrated rate law

The integrated rate law for a first - order reaction is $\ln\frac{[A]_0}{[A]_t}=kt$. We know the initial concentration $[A]_0$, the rate constant $k$, and the time $t = 240\ s$. We can solve for $[A]_t$ (the concentration of $N_2O_5$ at time $t$).

Answer:

(1) To find $t_{1/2}$, assume the initial concentration of $N_2O_5$ is $[N_2O_5]_0$. From the graph, when the concentration is $\frac{[N_2O_5]_0}{2}$, the time $t_{1/2}$ is read off the $x$ - axis. Let's assume from the graph $t_{1/2}=150\ s$ (this value is assumed as we don't have the actual graph values precisely).
(2) Using the formula $k = \frac{\ln 2}{t_{1/2}}$, substituting $t_{1/2}=150\ s$, we have $k=\frac{\ln 2}{150\ s}\approx4.6\times10^{-3}\ s^{-1}$.
(3) Let the initial concentration of $N_2O_5$ be $[N_2O_5]_0 = 1.0\ M$ (assumed). Using the integrated rate law $\ln\frac{[N_2O_5]_0}{[N_2O_5]_t}=kt$, with $k = 4.6\times10^{-3}\ s^{-1}$ and $t = 240\ s$.
$\ln\frac{1.0\ M}{[N_2O_5]_t}=(4.6\times10^{-3}\ s^{-1})\times240\ s$
$\ln\frac{1.0\ M}{[N_2O_5]_t}=1.104$
$\frac{1.0\ M}{[N_2O_5]_t}=e^{1.104}\approx3.01$
$[N_2O_5]_t=\frac{1.0\ M}{3.01}\approx0.33\ M$

So, $t_{1/2}=150\ s$, $k = 4.6\times10^{-3}\ s^{-1}$, $[N_2O_5]=0.33\ M$ (values are approximate based on assumed graph - reading and initial conditions). You need to adjust according to the actual graph data.