   Chapter 12, Problem 108CP

Chapter
Section
Textbook Problem

# The decomposition of NO2(g) occurs by the following bimolecular elementary reaction: 2 NO 2 ( g ) → 2 NO ( g ) + O 2 ( g ) The rate constant at 273 K is 2.3 × 10−12 L/mol · s, and the activation energy is 111 kJ/mol. How long will it take for the concentration of NO2(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500. K? Assume ideal gas behavior.

Interpretation Introduction

Interpretation: The time taken for the given concentration of NO2(g) to decrease from an initial pressure to final pressure is to be calculated. Ideal gas behavior is to be assumed.

Concept introduction: Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant.

Arrhenius equation is the relation which describes the temperature dependence of rate constant.

Activation energy is the minimum energy required to start a chemical reaction.

Elementary reaction is the reaction in which the reactant directly converts to the product only with a single step.

In elementary reaction the order of reaction is equal to the number of reactants involve in the reaction.

A gas that follows ideal gas laws is known as an ideal gas.

To determine: The time taken by the NO2 molecule to decrease the concentration of NO2 from an initial pressure of 2.5atm to 1.5atm at 500K by assuming ideal gas behavior.

Explanation

Explanation

Given

The given reaction is stated as,

2NO2(g)2NO(g)+O2(g)

The rate constant at 273K for the reaction is 2.3×1012L/mols.

The activation energy is 111kJ/mol.

As the rate constant is dependent on temperature therefore, the new rate constant at the temperature 500K is determined by the Arrhenius equation,

lnk2k1=EaR(1T11T2)

Where,

• The constants k1,k2 are the rate constants and T1,T2 are the temperatures.
• The term Ea is known as activation energy.
• The term R is known as gas constant with the universal value of 8.314J/mol
• The temperature 273K is considered as T1 so, its rate constant is termed as k1 and is equal to 2.3×1012L/mol.s.
• Similarly, the temperature 500K is considered as T2 so, its rate constant is termed as k2.

Substitute the value of k2,Ea,T1,T2 and R to calculate the rate constant k2.

lnk2k1=EaR(1T11T2)lnk22.3×1012L/mols=111kJ/mol8.314J/molK(1273K1500K)

Simplify the above expression.

lnk22.3×1012L/mols=0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 