At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.0968 s 2N,O; (g) → 2N,O, (g) + O, (g) Suppose a vessel contains N,0, at a concentration of 1.07M. Calculate the concentration of N,O, in the vessel 6.40 seconds late You may assume no other reaction is important. Round your answer to 2 significant digits.

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### Reaction Kinetics Problem

**Problem Statement:**

At a certain temperature, this reaction follows first-order kinetics with a rate constant of \(0.0968 \, \text{s}^{-1}\):

\[ 2 \text{N}_2\text{O}_5 (g) \rightarrow 2 \text{N}_2\text{O}_4 (g) + \text{O}_2 (g) \]

Suppose a vessel contains \(\text{N}_2\text{O}_5\) at a concentration of \(1.07 \, M\). Calculate the concentration of \(\text{N}_2\text{O}_5\) in the vessel 6.40 seconds later. You may assume no other reaction is important.

Round your answer to 2 significant digits.

**Calculation:**

Use the formula for first-order reactions:

\[ [\text{A}]_t = [\text{A}]_0 \cdot e^{-kt} \]

Where:
- \([\text{A}]_t\) is the concentration of the reactant at time \(t\).
- \([\text{A}]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
- \(t\) is the time.

Given:
- \( [\text{N}_2\text{O}_5]_0 = 1.07 \, M \)
- \( k = 0.0968 \, \text{s}^{-1} \)
- \( t = 6.40 \, \text{s} \)

Calculate \([\text{N}_2\text{O}_5]_t\):

\[ 
[\text{N}_2\text{O}_5]_{6.40} = 1.07 e^{-0.0968 \times 6.40} 
\]

Solve for the exponent first:

\[ 
0.0968 \times 6.40 = 0.61952 
\]

Now, find \( e^{-0.61952} \):

\[ 
e^{-0.61952} \approx 0.538 
\]

Finally, calculate the concentration:

\[ 
[\text{N}_2\text{O}_5]_{6.40} = 1.07 \times 0.538 = 0.575
Transcribed Image Text:### Reaction Kinetics Problem **Problem Statement:** At a certain temperature, this reaction follows first-order kinetics with a rate constant of \(0.0968 \, \text{s}^{-1}\): \[ 2 \text{N}_2\text{O}_5 (g) \rightarrow 2 \text{N}_2\text{O}_4 (g) + \text{O}_2 (g) \] Suppose a vessel contains \(\text{N}_2\text{O}_5\) at a concentration of \(1.07 \, M\). Calculate the concentration of \(\text{N}_2\text{O}_5\) in the vessel 6.40 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. **Calculation:** Use the formula for first-order reactions: \[ [\text{A}]_t = [\text{A}]_0 \cdot e^{-kt} \] Where: - \([\text{A}]_t\) is the concentration of the reactant at time \(t\). - \([\text{A}]_0\) is the initial concentration of the reactant. - \(k\) is the rate constant. - \(t\) is the time. Given: - \( [\text{N}_2\text{O}_5]_0 = 1.07 \, M \) - \( k = 0.0968 \, \text{s}^{-1} \) - \( t = 6.40 \, \text{s} \) Calculate \([\text{N}_2\text{O}_5]_t\): \[ [\text{N}_2\text{O}_5]_{6.40} = 1.07 e^{-0.0968 \times 6.40} \] Solve for the exponent first: \[ 0.0968 \times 6.40 = 0.61952 \] Now, find \( e^{-0.61952} \): \[ e^{-0.61952} \approx 0.538 \] Finally, calculate the concentration: \[ [\text{N}_2\text{O}_5]_{6.40} = 1.07 \times 0.538 = 0.575
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