   Chapter 12, Problem 34E

Chapter
Section
Textbook Problem

# The reaction 2NO ( g )   +   O 2 ( g ) → 2 NO 2 ( g ) was studied. and the following data were obtained where Rate   =   − Δ [ O 2 ] Δ t [NO]0(molecules/cm3) [O2]0(molecules/cm3) Initial Rate (molecules/cm3 · s) 1.00 × 1018 1.00 × 1018 2.00 × 1016 3.00 × 1018 1.00 × 1018 1.80 × 1017 2.50 × 1018 2.50 × 1018 3.13 × 1017 What would be the initial rate for an experiment where [NO]0 = 6.21 × 1018 molecules/cm3 and [O2]0 = 7.36 × 1018 molecules/cm3?

Interpretation Introduction

Interpretation: A reaction between NO and O2 certain data related to this reaction is given. The initial rate for an experiment with the given initial concentration value of NO and O2 is to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

To determine: The rate law for the given reaction.

Explanation

Given

The stated reaction is,

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

The rate law for the given reaction is calculated by the expression,

Rate=k[NO]2[O2]1

Where,

• [NO] and [O2] are the concentrations of NO and O2 .
• k is the rate constant.

To determine: The value of the rate constant, k , for the given reaction.

Given

The stated reaction is,

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

The rate law for the given reaction is calculated by the expression,

Rate=k[NO]2[O2]1

The given values for experiment 1 ,

[NO]ο=1.00×1018molecules/cm3[O2]ο=1.00×1018molecules/cm3

Substitute the value of rate and the concentration of NO and O2 in the rate expression.

Rate=k[NO]2[O2]12.00×1016molecules/cm3s=k[(1.00×1018)2(1.00×1018)(molecules/cm3)3]2.00×1016molecules/cm3s=k[1.00×1054molecules3/cm9]

Simplify the above expression.

k=2×1016molecules/cm3s[1

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