Question

a chemical engineer is studying the rate of this reaction. 2so₃(g)→2so₂(g)+o₂(g) he fills a reaction vessel with so₃ and measures its concentration as the reaction proceeds. heres a graph of his 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 so₃. 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 so₃ in the engineers reaction vessel after 1.20 seconds have passed. assume no other reaction is important, and continue to assume the rate is first order in so₃. round your answer to 2 significant digits.

Explanation:

Step1: Determine half - life from graph

The half - life ($t_{1/2}$) is the time it takes for the concentration of the reactant ($\text{SO}_3$) to decrease to half of its initial value. From the graph, we find the time when the concentration of $\text{SO}_3$ is half of its starting value.

Step2: Calculate rate constant for first - order reaction

For a first - order reaction, the relationship between the half - life ($t_{1/2}$) and the rate constant ($k$) is given by the formula $k=\frac{\ln2}{t_{1/2}}$. Substitute the value of $t_{1/2}$ into the formula.

Step3: Calculate concentration at a given time for first - order reaction

The integrated rate law for a first - order reaction is $\ln\frac{[A]_t}{[A]_0}=-kt$, where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration, $k$ is the rate constant, and $t$ is the time. Rearrange the formula to solve for $[A]_t$: $[A]_t = [A]_0e^{-kt}$.

Answer:

1. Half - life:

• From the graph, if the initial concentration of $\text{SO}_3$ is around $0.10\ M$, the time when it reaches $0.05\ M$ is approximately $t_{1/2}=0.69\ s$ (rounded to 2 significant digits).

1. Rate constant:

• Using the formula $k = \frac{\ln2}{t_{1/2}}$, with $\ln2\approx0.693$ and $t_{1/2}=0.69\ s$, we have $k=\frac{0.693}{0.69\ s}\approx1.0\ s^{-1}$ (rounded to 2 significant digits).

1. Concentration at $t = 1.20\ s$:

• Let the initial concentration $[A]_0 = 0.10\ M$, $k = 1.0\ s^{-1}$, and $t = 1.20\ s$.
• Using $[A]_t=[A]_0e^{-kt}$, we get $[A]_t=0.10\ M\times e^{-(1.0\ s^{-1}\times1.20\ s)}$.
• $[A]_t=0.10\ M\times e^{- 1.20}$.
• Since $e^{-1.20}\approx0.301$, $[A]_t=0.10\ M\times0.301 = 0.030\ M$ (rounded to 2 significant digits).