   Chapter 21, Problem 105GQ

Chapter
Section
Textbook Problem

Use ΔfH° data in Appendix L to calculate the enthalpy change of the reaction2 N2(g) + 5 O2(g) + 2 H2O(ℓ) → 4 HNO3(aq)Speculate on whether such a reaction could be used to “fix” nitrogen. Would research to find ways to accomplish this reaction be a useful endeavor?

Interpretation Introduction

Interpretation:

To calculate enthalpy change (ΔrH) of the given reaction.

Concept introduction:

The change of enthalpy to form one mole of a substance from its constituent elements when all the substances in the standard form is known as standard enthalpy of formation.

The standard enthalpy of reaction (ΔrH) in terms of standard enthalpy of formation (ΔfH) is written as,

ΔrH=ΔfHproductsΔfHreactants (1)

The change of Gibbs free energy to form one mole of a substance from its constituent elements when all the substances in the standard form is known as standard Gibbs free energy of formation (ΔfG°).

If the value of standard Gibbs free energy of reaction (ΔrG) is positive then the reaction is reactant favored. If the value of standard Gibbs free energy of reaction (ΔrG) is negative then the reaction is product favoured and spontaneous.

Explanation

The enthalpy change (ΔrH) of the given reaction is calculated below.

Given:

Refer to the appendix L for the value of standard enthalpy of formation (ΔfH).

ΔfH(H2O)=285.83kJmol1ΔfH(HNO3)=207.36kJmol1

The entropy (So) of following molecules are written as,

So(H2O)=69.95JK1So(N2)=191.56JK1So(O2)=205.07JK1So(HNO3)=146.4JK1

The balanced chemical equation of nitrogen fixation is written as,

2N2(g)+5O2(g)+2H2O(l)4HNO3(aq)

The standard enthalpy of formation (ΔrH) is written as,

ΔrH=4(ΔfH(HNO3))[2(ΔfH(N2))+5(ΔfHo(O2))+2(ΔfHo(H2O))] (2)

Substitute the value of standard enthalpy of formation (ΔfH) in equation (2) to calculate standard enthalpy of reaction (ΔrH)

ΔrH=4(207.36kJmol1)2(285.83kJmol1)=257.78kJmol1

Therefore, the standard enthalpy of reaction (ΔrH) is 257

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