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!.

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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
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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