For the decomposition of ammonia on a platinum surface at 856 °C 2 NH3N2 +3 H2 the average rate of disappearance of NH3 over the time period from t = 0 s to t= 4102 s is found to be 1.50×10-6 M s'. The average rate of formation of H2 over the same time period is |Ms!.
For the decomposition of ammonia on a platinum surface at 856 °C 2 NH3N2 +3 H2 the average rate of disappearance of NH3 over the time period from t = 0 s to t= 4102 s is found to be 1.50×10-6 M s'. The average rate of formation of H2 over the same time period is |Ms!.
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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![### Decomposition of Ammonia on a Platinum Surface
#### Reaction Details:
- **Temperature:** 856 °C
- **Reaction Equation:**
\[
2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2
\]
#### Rate of Reaction:
- The average rate of disappearance of \(\text{NH}_3\) over the time period from \( t = 0 \, \text{s} \) to \( t = 4102 \, \text{s} \) is found to be \( 1.50 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1} \).
#### Problem:
- Calculate the average rate of formation of \(\text{H}_2\) over the same time period.
\[
\text{The average rate of formation of } \text{H}_2 \text{ over the same time period is } \boxed{\phantom{1}} \, \text{M} \cdot \text{s}^{-1}.
\]
To find the average rate of formation of \(\text{H}_2\), use the stoichiometry of the reaction. For every 2 moles of \(\text{NH}_3\) that decompose, 3 moles of \(\text{H}_2\) are formed. Thus, the rate of formation of \(\text{H}_2\) is:
\[
\text{Rate of formation of } \text{H}_2 = \left( \frac{3}{2} \right) \times \text{Rate of disappearance of } \text{NH}_3
\]
\[
= \left( \frac{3}{2} \right) \times 1.50 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1}
\]
\[
= 2.25 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1}
\]
So,
\[
\text{The average rate of formation of } \text{H}_2 \text{ over the same time period is } 2.25 \times 10^{-6} \, \text{M} \cdot \text{s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9cce6311-c94c-4650-90f7-705d2febbb40%2F545ee2c6-58fa-4482-87b4-c402685cb0e6%2Flcww7xh_processed.png&w=3840&q=75)
Transcribed Image Text:### Decomposition of Ammonia on a Platinum Surface
#### Reaction Details:
- **Temperature:** 856 °C
- **Reaction Equation:**
\[
2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2
\]
#### Rate of Reaction:
- The average rate of disappearance of \(\text{NH}_3\) over the time period from \( t = 0 \, \text{s} \) to \( t = 4102 \, \text{s} \) is found to be \( 1.50 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1} \).
#### Problem:
- Calculate the average rate of formation of \(\text{H}_2\) over the same time period.
\[
\text{The average rate of formation of } \text{H}_2 \text{ over the same time period is } \boxed{\phantom{1}} \, \text{M} \cdot \text{s}^{-1}.
\]
To find the average rate of formation of \(\text{H}_2\), use the stoichiometry of the reaction. For every 2 moles of \(\text{NH}_3\) that decompose, 3 moles of \(\text{H}_2\) are formed. Thus, the rate of formation of \(\text{H}_2\) is:
\[
\text{Rate of formation of } \text{H}_2 = \left( \frac{3}{2} \right) \times \text{Rate of disappearance of } \text{NH}_3
\]
\[
= \left( \frac{3}{2} \right) \times 1.50 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1}
\]
\[
= 2.25 \times 10^{-6} \, \text{M} \cdot \text{s}^{-1}
\]
So,
\[
\text{The average rate of formation of } \text{H}_2 \text{ over the same time period is } 2.25 \times 10^{-6} \, \text{M} \cdot \text{s
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