**Problem 7**[Show your work] The gas phase reaction below obeys second-order kinetics. At a temperature of 20 °C and an initial concentration of NO₃ equal to 0.0500 mol/L, the concentration after 60 minutes is equal to 0.0358 mol/L. Assuming the reverse reaction to be negligible, find the value of the forward rate constant in L mol⁻¹ s⁻¹.\[ \text{NO}_3 \rightleftharpoons \text{NO}_2 + \frac{1}{2} \text{O}_2 \]**Explanation:**This problem requires calculating the rate constant of a reaction that follows second-order kinetics. You are provided with the initial concentration ([NO₃]₀ = 0.0500 mol/L) and the concentration after a specified time period ([NO₃]_t = 0.0358 mol/L after 60 minutes).Key steps involve using the second-order rate equation:\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]Where:- \([A]_t\) is the concentration of reactant A at time t,- \([A]_0\) is the initial concentration,- \(k\) is the rate constant,- \(t\) is the time (in seconds).You need to convert the given time from minutes to seconds and rearrange to solve for \(k\).

Answered: Image /qna-images/answer/bd60427e-9d0e-49f7-90e0-ea1b1be75471.jpg

(Show your work) The gas phase reaction below, obeys second-order kinetics. At a temperature of 20 °C and an initial concentration of NO 3 equal to 0.0500 mol/L, the concentration after 60 minutes is equal to 0.0358 mol/L. Assuming the reverse reaction to be negligible, find the value of the forward rate constant in L mol-¹-¹. NO3 → NO2+½/2 02

(Show your work) The gas phase reaction below, obeys second-order kinetics. At a temperature of 20 °C and an initial concentration of NO 3 equal to 0.0500 mol/L, the concentration after 60 minutes is equal to 0.0358 mol/L. Assuming the reverse reaction to be negligible, find the value of the forward rate constant in L mol-¹-¹. NO3 → NO2+½/2 02

Chemistry

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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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Question

**Problem 7**

[Show your work] The gas phase reaction below obeys second-order kinetics. At a temperature of 20 °C and an initial concentration of NO₃ equal to 0.0500 mol/L, the concentration after 60 minutes is equal to 0.0358 mol/L. Assuming the reverse reaction to be negligible, find the value of the forward rate constant in L mol⁻¹ s⁻¹.

\[ \text{NO}_3 \rightleftharpoons \text{NO}_2 + \frac{1}{2} \text{O}_2 \]

**Explanation:**

This problem requires calculating the rate constant of a reaction that follows second-order kinetics. You are provided with the initial concentration ([NO₃]₀ = 0.0500 mol/L) and the concentration after a specified time period ([NO₃]_t = 0.0358 mol/L after 60 minutes).

Key steps involve using the second-order rate equation:

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

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

You need to convert the given time from minutes to seconds and rearrange to solve for \(k\).

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