   # Calculate ∆ H for the reaction 2 NH 3 ( g ) + 1 2 O 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) given the following data: 2 NH 3 ( g ) + 3 N 2 O ( g ) → 4 N 2 ( g ) + 3 H 2 O ( l ) Δ H = − 1010. KJ N 2 O ( g ) + 3 H 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) Δ H = − 317 KJ N 2 H 4 ( l ) + O 2 ( g ) → N 2 ( g ) + 2 H 2 O ( l ) Δ H = − 623 KJ H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) Δ H = − 286 K J ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 72E
Textbook Problem
5 views

## Calculate ∆H for the reaction 2 NH 3 ( g ) + 1 2 O 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) given the following data: 2 NH 3 ( g ) + 3 N 2 O ( g ) → 4 N 2 ( g ) + 3 H 2 O ( l ) Δ H = − 1010.   KJ N 2 O ( g ) + 3 H 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) Δ H = − 317   KJ N 2 H 4 ( l ) + O 2 ( g ) → N 2 ( g ) + 2 H 2 O ( l ) Δ H = − 623   KJ H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) Δ H = − 286   K J

Interpretation Introduction

Interpretation: The enthalpy change ΔH of given reaction has to be calculated and the feasibility of ammonia synthesis should be discussed.

Concept Introduction:

Hess's Law:

• The enthalpy change along with a chemical reaction is self-determining of the route by which the chemical reaction take place.
• The enthalpy change of a reaction is equal to were the reaction take place in a single step or multiple steps.

Formula:

ΔH(total)ΔH(number of steps)......(1)

Rules:

• The sign of enthalpy is inverted when the reaction is inverted.
• The amounts of reactants and products is straight comparative to the level of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by numeral, the ΔH value is multiplied by the similar numeral.

### Explanation of Solution

Explanation

To calculate: The enthalpy change ΔH

2NH3+3N2O4N2+H2OΔH1=-1010KJ......(1)3N2H4+3H2O3N2O+9H2ΔH2=-3(-317KJ)......(2)9H2+92O29H2OΔH3=9(-286KJ)......(3)4N2+8H2O4N2H4+4O2ΔH4=-4(-623KJ)......(4)

Given:

2NH3(g)+3N2O(g)4N2(g)+3H2O(l)ΔH1=-1010KJ......(1)N2O(g)+3H2(g)N2H4(l)+H2O(l)ΔH2=-317KJ

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