   Chapter 10, Problem 89CP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# Calculate for the reactionmg src=Images/HTML_99425-10-89CP_image002.jpg alt="" align="top"/> Given the following data:l> Equation  −1010 −317 −143 −286

Interpretation Introduction

Interpretation:

The enthalpy change for the following reaction should be determined.

N2H4(l)+O2(g)N2(g)+2H2O(l)

Concept Introduction:

Enthalpy of a reaction is a state function thus; change in enthalpy of a reaction does not depend on the path of the reaction.

The change in enthalpy is same if the same reaction takes place in one step or series of steps. This is known as Hess’s law. There are following characteristics of change in enthalpy which are important to calculate the change in enthalpy of a reaction using Hess’s law.

1. For a reversed reaction, the sign of enthalpy change of the reaction also reversed.
2. If any reactant or product is multiplied by any integer in the chemical reaction, enthalpy change of that reactant or product is also multiplied by the same integer.
Explanation

Given Information:

The following reactions are given:

2NH3(g)+3N2O(g)4N2(g)+3H2O(l)     ΔH=1010 kJ N2O(g)+3H2(g)N2H4(l)+H2O(l)            ΔH=317 kJ 2NH3(g)+12O2(g)N2H4(l)+H2O(l)        ΔH=143 kJH2(g)+12O2(g)H2O(l)                               ΔH=286 kJ

Calculation:

The net reaction is as follows:

N2H4(l)+O2(g)N2(g)+2H2O(l)

To obtain the above reaction,

Reverse the 2nd reaction as follows:

N2H4(l)+H2O(l)N2O(g)+3H2(g)    ΔH=+317 kJ

Multiply the above reaction by 2

2N2H4(l)+2H2O(l)2N2O(g)+6H2(g)    ΔH=+634 kJ

Adding the above reaction to 1st reaction thus,

3N2H4(l)+3H2O(l)3N2O(g)+9H2(g)    ΔH=+951 kJ2NH3(g)+3N2O(g)4N2(g)+3H2O(l)     ΔH=1010 kJ_3N2H4(l)+2NH3(g)4N2(g)+9H2(g)      ΔH=59 kJ

Now, reverse the 3r

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