Question

a chemical engineer is studying the rate of this reaction. 2h₃po₄(aq)→p₂o₅(aq)+3h₂o(aq) he fills a reaction vessel with h₃po₄ 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 h₃po₄. 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 h₃po₄ in the engineers reaction vessel after 3.00 seconds have passed. assume no other reaction is important, and continue to assume the rate is first order in h₃po₄. round your answer to 2 significant digits.

Explanation:

Step1: Determine half - life from the graph

The initial concentration of $H_3PO_4$, $[H_3PO_4]_0 = 0.8M$. The half - life $t_{1/2}$ is the time when the concentration of $H_3PO_4$ is half of its initial value, i.e., $[H_3PO_4]=\frac{[H_3PO_4]_0}{2}=0.4M$. From the graph, when $[H_3PO_4] = 0.4M$, $t_{1/2}=0.40s$.

Step2: Calculate the rate constant for a first - order reaction

For a first - order reaction, the half - life formula is $t_{1/2}=\frac{\ln2}{k}$. Rearranging for $k$, we get $k = \frac{\ln2}{t_{1/2}}$. Substituting $t_{1/2}=0.40s$, $k=\frac{\ln2}{0.40s}\approx1.7s^{-1}$.

Step3: Predict the concentration after a certain time for a first - order reaction

The integrated rate law for a first - order reaction is $\ln\frac{[A]_t}{[A]_0}=-kt$. Here, $[A]_0 = 0.8M$, $k = 1.7s^{-1}$, and $t = 3.00s$. First, calculate $-kt=-1.7s^{-1}\times3.00s=- 5.1$. Then, $\ln\frac{[H_3PO_4]_t}{0.8M}=-5.1$. Exponentiating both sides, $\frac{[H_3PO_4]_t}{0.8M}=e^{-5.1}$. So, $[H_3PO_4]_t=0.8M\times e^{-5.1}\approx0.014M$.

Answer:

$t_{1/2}=0.40s$
$k = 1.7s^{-1}$
$[H_3PO_4]=0.014M$