At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.219 s¹: 2NH, (g) N₂ (g) + 3H₂(g) Suppose a vessel contains NH, at a concentration of 0.140 M. Calculate how long it takes for the concentration of NH, to decrease by 88.0%. You may assume no other reaction is important. Round your answer to 2 significant digits. 0.2
At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.219 s¹: 2NH, (g) N₂ (g) + 3H₂(g) Suppose a vessel contains NH, at a concentration of 0.140 M. Calculate how long it takes for the concentration of NH, to decrease by 88.0%. You may assume no other reaction is important. Round your answer to 2 significant digits. 0.2
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![### Chemical Kinetics Problem
At a certain temperature, this reaction follows first-order kinetics with a rate constant of 0.219 s<sup>-1</sup>:
\[2\text{NH}_3(\text{g}) \rightarrow \text{N}_2(\text{g}) + 3\text{H}_2(\text{g})\]
Suppose a vessel contains NH<sub>3</sub> at a concentration of 0.140 M. Calculate how long it takes for the concentration of NH<sub>3</sub> to decrease by 88.0%. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Input Box:**
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**Unit Selection Dropdown:**
[s]
[s<sup>-1</sup>]
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**Explanation:**
First-order reactions can be described by the equation:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \]
Where:
- \([A]_0\) is the initial concentration (0.140 M in this case).
- \([A]\) is the final concentration.
- \(k\) is the rate constant (0.219 s<sup>-1</sup> here).
- \(t\) is the time.
Given that the concentration decreases by 88.0%, the final concentration \([A]\) is:
\[ [A] = [A]_0 \times (1 - 0.88) = 0.140 \times 0.12 = 0.0168 \text{ M} \]
Plugging the values into the first-order kinetics equation:
\[ \ln \left( \frac{0.140}{0.0168} \right) = 0.219 \times t \]
Solving for \(t\):
\[ t = \frac{\ln \left( 0.140 / 0.0168 \right)}{0.219} \]
Proceed to calculate the exact value to obtain the time \(t\) in seconds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F127f63d9-8f89-4c37-8cc0-122cae65c6f1%2F64d8a075-a3c2-4b9a-af8a-15dcb03b1560%2Fflhvxe4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Kinetics Problem
At a certain temperature, this reaction follows first-order kinetics with a rate constant of 0.219 s<sup>-1</sup>:
\[2\text{NH}_3(\text{g}) \rightarrow \text{N}_2(\text{g}) + 3\text{H}_2(\text{g})\]
Suppose a vessel contains NH<sub>3</sub> at a concentration of 0.140 M. Calculate how long it takes for the concentration of NH<sub>3</sub> to decrease by 88.0%. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Input Box:**
[ ]
**Unit Selection Dropdown:**
[s]
[s<sup>-1</sup>]
---
**Explanation:**
First-order reactions can be described by the equation:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \]
Where:
- \([A]_0\) is the initial concentration (0.140 M in this case).
- \([A]\) is the final concentration.
- \(k\) is the rate constant (0.219 s<sup>-1</sup> here).
- \(t\) is the time.
Given that the concentration decreases by 88.0%, the final concentration \([A]\) is:
\[ [A] = [A]_0 \times (1 - 0.88) = 0.140 \times 0.12 = 0.0168 \text{ M} \]
Plugging the values into the first-order kinetics equation:
\[ \ln \left( \frac{0.140}{0.0168} \right) = 0.219 \times t \]
Solving for \(t\):
\[ t = \frac{\ln \left( 0.140 / 0.0168 \right)}{0.219} \]
Proceed to calculate the exact value to obtain the time \(t\) in seconds.
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