3 4 1 5 6 7 8 At a certain temperature the rate of this reaction is first order in HI with a rate constant of 0.0797 s 2HI (g) → H₂(g) + 1₂ (8) Suppose a vessel contains HI at a concentration of 0.870 M. Calculate the concentration of HI in the vessel 10.0 seconds later. You may assume no other. reaction is important. Round your answer to 2 significant digits. M ? x10 X S
3 4 1 5 6 7 8 At a certain temperature the rate of this reaction is first order in HI with a rate constant of 0.0797 s 2HI (g) → H₂(g) + 1₂ (8) Suppose a vessel contains HI at a concentration of 0.870 M. Calculate the concentration of HI in the vessel 10.0 seconds later. You may assume no other. reaction is important. Round your answer to 2 significant digits. M ? x10 X S
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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![**Chemical Kinetics Problem**
---
**Problem Statement:**
At a certain temperature, the rate of this reaction is first order in HI with a rate constant of \( 0.0797 \, \text{s}^{-1} \):
\[ 2HI(g) \rightarrow H_2(g) + I_2(g) \]
Suppose a vessel contains HI at a concentration of \( 0.870 \, M \). Calculate the concentration of HI in the vessel 10.0 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Answer Input:**
- Box labeled " \( M \)" for answer input
- Options for exponents (\( \times 10^{x} \)), reset (\( \return \)), and help (\( ? \))
*Interactive Elements and Design:*
At the top of the screen, there are sequential numbered tabs (1, 3, 4, 5, 6, 7, 8) indicating the progression through a set of problems or sections. The design suggests that the user is currently on problem 1, as it is highlighted.
At the bottom, there is a "Continue" button to proceed after entering the answer.
**Steps to Solve:**
1. **Identify the Rate Law:**
Since the reaction is first order in HI, the rate law can be written as:
\[
\text{Rate} = k [\text{HI}]
\]
where \( k = 0.0797 \, \text{s}^{-1} \).
2. **Use the First-Order Integrated Rate Law:**
The integrated rate law for a first-order reaction is:
\[
\ln \left( \frac{[\text{HI}]_t}{[\text{HI}]_0} \right) = -kt
\]
Given:
\[
[\text{HI}]_0 = 0.870 \, M, \quad t = 10.0 \, \text{s}
\]
3. **Plug in Values to Solve for \([\text{HI}]_t\):**
\[
\ln \left( \frac{[\text{HI}]_t}{0.870} \right) = - (0.0797) (10.0)
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Transcribed Image Text:**Chemical Kinetics Problem**
---
**Problem Statement:**
At a certain temperature, the rate of this reaction is first order in HI with a rate constant of \( 0.0797 \, \text{s}^{-1} \):
\[ 2HI(g) \rightarrow H_2(g) + I_2(g) \]
Suppose a vessel contains HI at a concentration of \( 0.870 \, M \). Calculate the concentration of HI in the vessel 10.0 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
**Answer Input:**
- Box labeled " \( M \)" for answer input
- Options for exponents (\( \times 10^{x} \)), reset (\( \return \)), and help (\( ? \))
*Interactive Elements and Design:*
At the top of the screen, there are sequential numbered tabs (1, 3, 4, 5, 6, 7, 8) indicating the progression through a set of problems or sections. The design suggests that the user is currently on problem 1, as it is highlighted.
At the bottom, there is a "Continue" button to proceed after entering the answer.
**Steps to Solve:**
1. **Identify the Rate Law:**
Since the reaction is first order in HI, the rate law can be written as:
\[
\text{Rate} = k [\text{HI}]
\]
where \( k = 0.0797 \, \text{s}^{-1} \).
2. **Use the First-Order Integrated Rate Law:**
The integrated rate law for a first-order reaction is:
\[
\ln \left( \frac{[\text{HI}]_t}{[\text{HI}]_0} \right) = -kt
\]
Given:
\[
[\text{HI}]_0 = 0.870 \, M, \quad t = 10.0 \, \text{s}
\]
3. **Plug in Values to Solve for \([\text{HI}]_t\):**
\[
\ln \left( \frac{[\text{HI}]_t}{0.870} \right) = - (0.0797) (10.0)
\]
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