Calculate ∆ H for the reaction N 2 H 4 ( l ) + O 2 ( g ) → N 2 ( g ) + 2 H 2 O ( l ) given the following data: Equation ∆ H (KJ) 2NH 3 ( g ) + 3 N 2 O ( g ) → 4 N 2 ( g ) + 3 H 2 O ( l ) −1010 N 2 O ( g ) + 3 H 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) −317 2 NH 3 ( g ) + 1 2 O 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l ) −143 H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) −286

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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 an numeral, the ΔH value is multiplied by the similar numeral. Explanation Explanation 2NH 3(g) +3N 2 O (g) → 4N 2(g) +3H 2 O (l) ΔΗ = -1010 KJ ...... (1) N 2 O (g) +3H 2(g) → N 2 H 4(l) +H 2(l) ΔΗ = -317 KJ ...... (2) 2NH 3(g) + 1 2 O 2(g) → N 2 H 4(l) +H 2 O (l) ΔH = -143KJ ...... (3) H 2(g) + 1 2 O 2(g) → H 2 O (l) ΔΗ= -286KJ ...... ( 4 ) ΔH= ΔH 1 + ΔH 2 +ΔH 3 +ΔH 4 ..... (2) ΔH= -1010+9(-286)+-3(-317)+-(-143) The given reaction enthalpy change is: 4N 2 H (l) +4O 2(g) → 4N 2(g) +8H 2 O (l) ΔH = -2490 KJ N2H (l) +O 2(g) ®N 2(g) </

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Chemistry: An Atoms First Approach

2nd Edition

Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Chemistry: An Atoms First Approach

2nd Edition

Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Answer 6

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 an numeral, the ΔH value is multiplied by the similar numeral.

Explanation of Solution

Explanation

2NH3(g)+3N2O(g)→4N2(g)+3H2O(l) ΔΗ = -1010 KJ......(1)N2O(g)+3H2(g)→N2H4(l)+H2(l) ΔΗ = -317 KJ......(2)2NH3(g)+12O2(g)→N2H4(l)+H2O(l) ΔH = -143KJ......(3)H2(g)+12O2(g)→H2O(l) ΔΗ= -286KJ......(4)

ΔH= ΔH1+ ΔH2+ΔH3+ΔH4.....(2)

ΔH= -1010+9(-286)+-3(-317)+-(-143)

The given reaction enthalpy change is:

4N2H(l)+4O2(g)→4N2(g)+8H2O(l) ΔH = -2490 KJ N2H(l)+O2(g)®N2(g)</

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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Equation	∆H(KJ)
2NH 3 ( g ) + 3 N 2 O ( g ) → 4 N 2 ( g ) + 3 H 2 O ( l )	−1010
N 2 O ( g ) + 3 H 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l )	−317
2 NH 3 ( g ) + 1 2 O 2 ( g ) → N 2 H 4 ( l ) + H 2 O ( l )	−143
H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l )	−286