### Decomposition of Nitrous Oxide at 565 °C**Reaction:**\[ \text{N}_2\text{O(g)} \rightarrow \text{N}_2\text{(g)} + \frac{1}{2} \text{O}_2\text{(g)} \]The reaction is **second order** in N₂O with a rate constant of \( 2.20 \times 10^{-2} \, \text{M}^{-1} \text{s}^{-1} \).**Problem Statement:**If the initial concentration of N₂O is 0.400 M, what will the concentration of N₂O (in M, to three significant figures) be after 25.0 seconds have passed?---**Solution Approach:**To find the concentration of N₂O after 25.0 seconds for a second-order reaction, the formula used is:\[\frac{1}{[\text{N}_2\text{O}]_t} = \frac{1}{[\text{N}_2\text{O}]_0} + kt\]where:- \([\text{N}_2\text{O}]_t\) is the concentration at time \( t \).- \([\text{N}_2\text{O}]_0\) is the initial concentration (0.400 M).- \( k \) is the rate constant \( (2.20 \times 10^{-2} \, \text{M}^{-1} \text{s}^{-1}) \).- \( t \) is the time (25.0 s).Please proceed by substituting the values and solving for \([\text{N}_2\text{O}]_t\).

The mathematical expression for the rate constant of a second-order reaction is: 1[A]=kt+1[A]0

The decomposition of nitrous oxide at 565 °C N20(g) → N2(g) + ½ O2(g) is second order in N20 with a rate constant of 2.20x10-2 M-1 s-1. If the initial concentration of N20 is 0.400 M, what will the concentration of N20 (in M, to three significant figures) remain after 25.0 seconds have passed.

The decomposition of nitrous oxide at 565 °C N20(g) → N2(g) + ½ O2(g) is second order in N20 with a rate constant of 2.20x10-2 M-1 s-1. If the initial concentration of N20 is 0.400 M, what will the concentration of N20 (in M, to three significant figures) remain after 25.0 seconds have passed.

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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