At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0618 M 2NH3 (g) → N₂ (g) + 3H₂(g) Suppose a vessel contains NH3 at a concentration of 1.25 M. Calculate the concentration of NH3 in the vessel 68.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM S ? 0.9 X

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### Second-Order Reaction Kinetics

At a certain temperature, the reaction described follows second-order kinetics with a rate constant of \(0.0618 \, M^{-1} \, s^{-1}\):

\[2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\]

Suppose a vessel contains \(NH_3\) at a concentration of \(1.25 \, M\). Calculate the concentration of \(NH_3\) in the vessel 68.0 seconds later. You may assume no other reaction is important.

Round your answer to 2 significant digits.

**Input:**  
\[ \boxed{\,\,\,\,\,\,} \, M \]

For more assistance:

\[ \square \times 10 \, \boxed{x} \, \boxed{\Rightarrow} \, \boxed{?} \]

**Explanation:**

This problem presents a second-order reaction where the rate law is given by:

\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]

- \([A]\) is the concentration of \(NH_3\) at time \(t\).
- \([A_0]\) is the initial concentration of \(NH_3\) (1.25 M).
- \(k\) is the rate constant (0.0618 \(M^{-1} s^{-1}\)).
- \(t\) is the time (68.0 seconds).

### Calculation Steps

1. Plug in the values into the second-order rate equation:
   \[
   \frac{1}{[A]} - \frac{1}{1.25} = (0.0618)(68.0)
   \]

2. Solve for \( [A] \):
   \[
   \frac{1}{[A]} = \frac{1}{1.25} + (0.0618 \times 68.0)
   \]

3. Simplify:
   \[
   \frac{1}{[A]} = 0.8 + 4.2024 = 5.0024
   \]

4. Take the reciprocal to find the concentration at \( t = 68.0 \) seconds:
   \[
   [A] = \frac{1}{5.0024} \approx 0.2 \, M
Transcribed Image Text:### Second-Order Reaction Kinetics At a certain temperature, the reaction described follows second-order kinetics with a rate constant of \(0.0618 \, M^{-1} \, s^{-1}\): \[2NH_3(g) \rightarrow N_2(g) + 3H_2(g)\] Suppose a vessel contains \(NH_3\) at a concentration of \(1.25 \, M\). Calculate the concentration of \(NH_3\) in the vessel 68.0 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. **Input:** \[ \boxed{\,\,\,\,\,\,} \, M \] For more assistance: \[ \square \times 10 \, \boxed{x} \, \boxed{\Rightarrow} \, \boxed{?} \] **Explanation:** This problem presents a second-order reaction where the rate law is given by: \[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \] - \([A]\) is the concentration of \(NH_3\) at time \(t\). - \([A_0]\) is the initial concentration of \(NH_3\) (1.25 M). - \(k\) is the rate constant (0.0618 \(M^{-1} s^{-1}\)). - \(t\) is the time (68.0 seconds). ### Calculation Steps 1. Plug in the values into the second-order rate equation: \[ \frac{1}{[A]} - \frac{1}{1.25} = (0.0618)(68.0) \] 2. Solve for \( [A] \): \[ \frac{1}{[A]} = \frac{1}{1.25} + (0.0618 \times 68.0) \] 3. Simplify: \[ \frac{1}{[A]} = 0.8 + 4.2024 = 5.0024 \] 4. Take the reciprocal to find the concentration at \( t = 68.0 \) seconds: \[ [A] = \frac{1}{5.0024} \approx 0.2 \, M
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