Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0066 M-s¹: 2NH₂(g) → N₂(g) + 3H₂(g) Suppose a 450. mL flask is charged under these conditions with 150. mmol of ammonia. How much is left 10. s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. 0 olo
Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0066 M-s¹: 2NH₂(g) → N₂(g) + 3H₂(g) Suppose a 450. mL flask is charged under these conditions with 150. mmol of ammonia. How much is left 10. s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. 0 olo
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Educational Content: Understanding Zero Order Reaction Kinetics**
Under specific conditions, the rate of a chemical reaction can be described as zero order in a particular reactant. This means that the rate of reaction is independent of the concentration of that reactant. An example of a zero-order reaction involving ammonia is shown below:
\[ 2 \text{NH}_3 \text{(g)} \rightarrow \text{N}_2 \text{(g)} + 3 \text{H}_2 \text{(g)} \]
In this case, the rate of the reaction is zero order in ammonia with a rate constant \( k \) of 0.0066 \( \text{M} \cdot \text{s}^{-1} \).
Suppose a 450 mL flask is initially charged with 150 mmol of ammonia. To determine how much ammonia remains after 10 seconds, follow these steps:
### Calculation:
1. **Convert the initial amount (mmol) to moles**:
- \( 150 \text{ mmol} = 0.150 \text{ mol} \)
2. **Convert the volume from mL to L**:
- \( 450 \text{ mL} = 0.450 \text{ L} \)
3. **Calculate the initial concentration of ammonia**:
\[
\text{Initial concentration} = \frac{\text{Amount of NH}_3}{\text{Volume}} = \frac{0.150 \text{ mol}}{0.450 \text{ L}} = 0.333 \text{ M}
\]
4. **Use the zero-order kinetics equation**:
\[
[\text{NH}_3] = [\text{NH}_3]_0 - kt
\]
Where \( [\text{NH}_3]_0 \) is the initial concentration, \( k \) is the rate constant, and \( t \) is the time.
5. **Substitute the values into the equation**:
\[
[\text{NH}_3] = 0.333 \text{ M} - (0.0066 \text{ M} \cdot \text{s}^{-1})(10 \text{ s})
\]
6. **Perform the calculation**:
\[
[\text{NH}_3] =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20b515c9-5e3a-448e-90a2-ac3b562e1523%2F6f07797d-b35e-4c48-a781-afdd4930dca0%2F2ajo9ho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Understanding Zero Order Reaction Kinetics**
Under specific conditions, the rate of a chemical reaction can be described as zero order in a particular reactant. This means that the rate of reaction is independent of the concentration of that reactant. An example of a zero-order reaction involving ammonia is shown below:
\[ 2 \text{NH}_3 \text{(g)} \rightarrow \text{N}_2 \text{(g)} + 3 \text{H}_2 \text{(g)} \]
In this case, the rate of the reaction is zero order in ammonia with a rate constant \( k \) of 0.0066 \( \text{M} \cdot \text{s}^{-1} \).
Suppose a 450 mL flask is initially charged with 150 mmol of ammonia. To determine how much ammonia remains after 10 seconds, follow these steps:
### Calculation:
1. **Convert the initial amount (mmol) to moles**:
- \( 150 \text{ mmol} = 0.150 \text{ mol} \)
2. **Convert the volume from mL to L**:
- \( 450 \text{ mL} = 0.450 \text{ L} \)
3. **Calculate the initial concentration of ammonia**:
\[
\text{Initial concentration} = \frac{\text{Amount of NH}_3}{\text{Volume}} = \frac{0.150 \text{ mol}}{0.450 \text{ L}} = 0.333 \text{ M}
\]
4. **Use the zero-order kinetics equation**:
\[
[\text{NH}_3] = [\text{NH}_3]_0 - kt
\]
Where \( [\text{NH}_3]_0 \) is the initial concentration, \( k \) is the rate constant, and \( t \) is the time.
5. **Substitute the values into the equation**:
\[
[\text{NH}_3] = 0.333 \text{ M} - (0.0066 \text{ M} \cdot \text{s}^{-1})(10 \text{ s})
\]
6. **Perform the calculation**:
\[
[\text{NH}_3] =
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