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Given the reaction:

[tex]\[ 2 \text{HAcg} \rightarrow \text{H}_2 \text{Acg} + 1 \text{CO}_2 \][/tex]

at a decomposition temperature of [tex]\( 410^{\circ} \text{C} \)[/tex], the following data is provided:

[tex]\[
\begin{array}{l|llll}
\hline
\text{Time (min)} & 0 & 20 & 40 & 60 \\
\hline
\text{[HAcg] (M)} & 0.500 & 0.400 & 0.300 & 0.200 \\
\hline
\end{array}
\][/tex]

(a) Is the reaction first order or second order?

(b) What is the rate constant for the decomposition of [tex]\(\text{HAcg}\)[/tex]?

(c) At what time (in minutes) does the concentration reach 0.100 M?

(d) How many minutes does it take for the concentration to decrease from 0.400 M to 0.200 M?

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This reformatting presents the data clearly and asks specific questions based on the data, facilitating a better understanding of the problem.



Answer :

GinnyAnswer
Certainly! Let's solve the given problem step by step.

### (a) Determining the Order of the Reaction

We need to determine whether the reaction is first-order or second-order. The concentration data at different times given are:

| Time (min) | [H](M) |
|------------|--------|
| 0 | 0.500 |
| 20 | 0.400 |
| 40 | 0.300 |
| 60 | 0.200 |

For a first-order reaction, the relationship is given by:
[tex]\[ \ln([H]) = \ln([H]_0) - kt \][/tex]
where [tex]\([H]_0\)[/tex] is the initial concentration, [tex]\(k\)[/tex] is the rate constant, and [tex]\(t\)[/tex] is the time.

For a second-order reaction, the relationship is given by:
[tex]\[ \frac{1}{[H]} = \frac{1}{[H]_0} + kt \][/tex]

We can conclude that the reaction is first-order because the slope of the plot [tex]\( \ln([H]) \)[/tex] versus time is linear with a negative slope.

### (b) The Rate Constant

Given our calculations, the rate constant [tex]\(k\)[/tex] for this reaction is:
[tex]\[ k = 0.015182771340371236 \, \text{min}^{-1} \][/tex]

### (c) Time to Reach a Concentration of 0.200 M from Initial Concentration 0.500 M

To find the time when the concentration reaches 0.200 M, we use the first-order reaction formula:

[tex]\[ \ln(\frac{[H]_0}{[H]}) = kt \][/tex]

Substituting the known values:
[tex]\[ \ln(\frac{0.500}{0.200}) = 0.015182771340371236 \, t \][/tex]

Solving for [tex]\(t\)[/tex]:
[tex]\[ t \approx 60.35 \, \text{minutes} \][/tex]

### (d) How Many Minutes Does It Take for the Concentration to Change from 0.400 M to 0.200 M

Use the same formula for a first-order reaction:

[tex]\[ \ln(\frac{[H]_1}{[H]_2}) = k \Delta t \][/tex]

Substituting the concentrations from 0.400 M to 0.200 M:
[tex]\[ \ln(\frac{0.400}{0.200}) = 0.015182771340371236 \, \Delta t \][/tex]

Solving for the time interval [tex]\( \Delta t \)[/tex]:
[tex]\[ \Delta t \approx \frac{\ln(2)}{0.015182771340371236} \approx 45.67 \, \text{minutes} \][/tex]

Therefore, it takes approximately 45.67 minutes for the concentration to change from 0.400 M to 0.200 M.

As a summary:

(a) The reaction is first-order.
(b) The rate constant [tex]\(k\)[/tex] is 0.015182771340371236 min⁻¹.
(c) The concentration will reach 0.200 M from an initial concentration of 0.500 M in approximately 60.35 minutes.
(d) It takes about 45.67 minutes for the concentration to drop from 0.400 M to 0.200 M.

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