Question

a chemical engineer is studying the rate of this reaction. 2cl₂o₅(g)→2cl₂(g)+5o₂(g) she fills a reaction vessel with cl₂o₅ and measures its concentration as the reaction proceeds. heres a graph of her data: 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 cl₂o₅. 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 cl₂o₅ in the engineers reaction vessel after 6.0 seconds have passed. assume no other reaction is important, and continue to assume the rate is first order in cl₂o₅. round your answer to 2 significant digits.

Explanation:

Step1: Determine half - life from graph

The half - life ($t_{1/2}$) of a first - order reaction is the time it takes for the concentration of the reactant to decrease to half of its initial value. From the graph, if the initial concentration of $\text{Cl}_2\text{O}_5$ is approximately $0.14\ M$, half of this value is $0.07\ M$. Reading from the graph, the time at which the concentration of $\text{Cl}_2\text{O}_5$ is $0.07\ M$ is $t_{1/2}=0.80\ s$.

Step2: Calculate the rate constant for first - order reaction

For a first - order reaction, the relationship between the half - life and the rate constant ($k$) is given by the formula $t_{1/2}=\frac{\ln 2}{k}$. Rearranging for $k$, we get $k = \frac{\ln 2}{t_{1/2}}$. Substituting $t_{1/2}=0.80\ s$ into the formula, $k=\frac{\ln 2}{0.80\ s}\approx0.87\ s^{-1}$.

Step3: Predict the concentration after a certain 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 (approx. $0.14\ M$), $k = 0.87\ s^{-1}$, and $t = 6.0\ s$. First, calculate $-kt=- 0.87\ s^{-1}\times6.0\ s=-5.22$. Then, $\ln\frac{[A]_t}{[A]_0}=-5.22$. Exponentiating both sides gives $\frac{[A]_t}{[A]_0}=e^{-5.22}$. So, $[A]_t=[A]_0\times e^{-5.22}=0.14\ M\times e^{-5.22}\approx0.0012\ M$.

Answer:

$t_{1/2}=0.80\ s$
$k = 0.87\ s^{-1}$
$[\text{Cl}_2\text{O}_5]=0.0012\ M$