-1 -1 At a certain temperature this reaction follows second-order kinetics with a rate constant of 4.91 M 'S : 2HI (g) → H, (g) +I, (g) Suppose a vessel contains HI at a concentration of 1.33M. Calculate the concentration of HI in the vessel 2.40 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM ?

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Chapter1: Chemical Foundations
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### Second-Order Reaction Kinetics Educational Module

#### Reaction Overview
At a certain temperature, this reaction follows second-order kinetics with a rate constant of \(4.91 \, M^{-1} \cdot s^{-1}\):

\[ 2\text{HI (g)} \rightarrow \text{H}_2 \text{(g)} + \text{I}_2 \text{(g)} \]

#### Problem Statement
Suppose a vessel contains HI at a concentration of \(1.33 \, M\). Calculate the concentration of HI in the vessel \(2.40\) seconds later. You may assume no other reaction is important.

Round your answer to 2 significant digits.

#### Inputs and Interface
- **Initial Concentration (\( [HI]_0 \)**): \(1.33 \, M\)
- **Time (t): 2.40 \, \text{s}**

- **Rate Constant (k):** \(4.91 \, M^{-1} \cdot s^{-1}\)

- **Required Output:** Concentration of HI \( [HI] \, \text{(M)} \) after \(2.40 \, \text{s}\)

#### Solving the Problem
To calculate the concentration of HI after a given time period for a second-order reaction, the integrated rate law for second-order kinetics is used:

\[
\frac{1}{[HI]_t} = \frac{1}{[HI]_0} + kt
\]

Where:
- \([HI]_t\) is the concentration of HI at time t
- \([HI]_0\) is the initial concentration of HI
- \(k\) is the rate constant
- \(t\) is the elapsed time

Rearranging for \([HI]_t\):

\[
[HI]_t = \frac{1}{ \frac{1}{[HI]_0} + kt }
\]

Substituting the given values:

\[
[HI]_t = \frac{1}{ \frac{1}{1.33} + (4.91 \times 2.40) }
\]

Perform the calculations to find the concentration of HI:

\[
[HI]_t = \frac{1}{ 0.7519 + 11.784 } = \frac{1}{ 12.5359 } \approx 0.079
Transcribed Image Text:### Second-Order Reaction Kinetics Educational Module #### Reaction Overview At a certain temperature, this reaction follows second-order kinetics with a rate constant of \(4.91 \, M^{-1} \cdot s^{-1}\): \[ 2\text{HI (g)} \rightarrow \text{H}_2 \text{(g)} + \text{I}_2 \text{(g)} \] #### Problem Statement Suppose a vessel contains HI at a concentration of \(1.33 \, M\). Calculate the concentration of HI in the vessel \(2.40\) seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. #### Inputs and Interface - **Initial Concentration (\( [HI]_0 \)**): \(1.33 \, M\) - **Time (t): 2.40 \, \text{s}** - **Rate Constant (k):** \(4.91 \, M^{-1} \cdot s^{-1}\) - **Required Output:** Concentration of HI \( [HI] \, \text{(M)} \) after \(2.40 \, \text{s}\) #### Solving the Problem To calculate the concentration of HI after a given time period for a second-order reaction, the integrated rate law for second-order kinetics is used: \[ \frac{1}{[HI]_t} = \frac{1}{[HI]_0} + kt \] Where: - \([HI]_t\) is the concentration of HI at time t - \([HI]_0\) is the initial concentration of HI - \(k\) is the rate constant - \(t\) is the elapsed time Rearranging for \([HI]_t\): \[ [HI]_t = \frac{1}{ \frac{1}{[HI]_0} + kt } \] Substituting the given values: \[ [HI]_t = \frac{1}{ \frac{1}{1.33} + (4.91 \times 2.40) } \] Perform the calculations to find the concentration of HI: \[ [HI]_t = \frac{1}{ 0.7519 + 11.784 } = \frac{1}{ 12.5359 } \approx 0.079
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