Chapter 20, Problem 44GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# The formation of NO from N2 and O2 is unfavorable at 298 K, but it becomes increasingly favored at high temperatures such as those in an automobile cylinder. Using data (ΔfH°, S° and ΔfG°) in Appendix L, calculate the equilibrium constant for the reaction ½ N2(g) + ½ O2(g) → NO(g) at 298 K and at 1000 K.

Interpretation Introduction

Interpretation: The equilibrium constant for the reaction 12NO2(g)+12O2(g)NO(g) at 298Kand1000K should be determined.

Concept introduction:

Equilibrium constant in terms of concentration[KC]: Equilibrium constant can be expressed in terms of concentration.

aA(g)+bB(g)cC(g)+dD(g)Kc=[C]c×[D]d[A]a×[B]b

Equilibrium constant and Gibbs free energy:

ΔG0=RTlnK

Relation between Gibbs free energy , Enthalpy and Entropy:

ΔG=ΔH-TΔSΔG-GibbsfreeenergyΔH-EnthalpyΔS-Entropy

Explanation

Given:

Î”fH0â€‰â€‰=â€‰90.29â€‰kJ/molÎ”fG0â€‰â€‰=â€‰86.58â€‰kJ/molÎ”Sâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰210.76â€‰J/K.mol

We know the relation:Î”G0â€‰â€‰â€‰=â€‰âˆ’â€‰RTlnâ€‰K. Using this we can calculate the equilibrium constant at 298â€‰K

Î”G0â€‰â€‰â€‰=â€‰âˆ’RTlnâ€‰K86.58Ã—103â€‰=â€‰âˆ’8

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