Explanation of the Balanced Equation

Oct 13, 2025 by sorusgentrification.com 122 views

Introduction

Chemical reactions involve the transformation of one or more substances into new substances. These reactions can be either exothermic or endothermic, releasing or absorbing energy, respectively. In this article, we will focus on the gaseous reaction between nitrogen pentoxide (N2O5) and fluorine (F2) to form nitrogen trifluoride (NF3) and oxygen (O2). We will derive a balanced equation for this reaction and determine the standard Gibbs free energy change (ΔG°).

Balanced Equation

The reaction between N2O5 and F2 to form NF3 and O2 can be represented by the following unbalanced equation:

N2O5 + F2 → NF3 + O2

To balance this equation, we need to ensure that the number of atoms of each element is the same on both the reactant and product sides. The balanced equation is:

N2O5 + 3F2 → 2NF3 + 3O2

Explanation of the Balanced Equation

In the balanced equation, 1 mole of N2O5 reacts with 3 moles of F2 to produce 2 moles of NF3 and 3 moles of O2. The coefficients in the balanced equation indicate the stoichiometry of the reaction, which is the ratio of the amounts of reactants and products.

Determination of ΔG°

The standard Gibbs free energy change (ΔG°) is a measure of the spontaneity of a reaction at standard conditions. It can be calculated using the following equation:

ΔG° = ΔH° - TΔS°

where ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change.

To determine ΔG°, we need to calculate ΔH° and ΔS° for the reaction. The standard enthalpy change (ΔH°) can be calculated using the following equation:

ΔH° = Σ(ΔHf°,products) - Σ(ΔHf°,reactants)

where ΔHf° is the standard enthalpy of formation of a substance.

The standard enthalpy of formation of NF3 is -111.3 kJ/mol, and the standard enthalpy of formation of O2 is 0 kJ/mol. The standard enthalpy of formation of N2O5 is -102.5 kJ/mol, and the standard enthalpy of formation of F2 is 0 kJ/mol.

Using these values, we can calculate ΔH° as follows:

ΔH° = 2(ΔHf°,NF3) + 3(ΔHf°,O2) - (ΔHf°,N2O5) - 3(ΔHf°,F2) = 2(-111.3 kJ/mol) + 3(0 kJ/mol) - (-102.5 kJ/mol) - 3(0 kJ/mol) = -221.6 kJ/mol

The standard entropy change (ΔS°) can be calculated using the following equation:

ΔS° = Σ(ΔSf°,products) - Σ(ΔSf°,reactants)

where ΔSf° is the standard entropy of formation of a substance.

The standard entropy of formation of NF3 is 202.8 J/mol·K, and the standard entropy of formation of O2 is 205.14 J/mol·K. The standard entropy of formation of N2O5 is 209.2 J/mol·K, and the standard entropy of formation of F2 is 202.8 J/mol·K.

Using these values, we can calculate ΔS° as follows:

ΔS° = 2(ΔSf°,NF3) + 3(ΔSf°,O2) - (ΔSf°,N2O5) - 3(ΔSf°,F2) = 2(202.8 J/mol·K) + 3(205.14 J/mol·K) - (209.2 J/mol·K) - 3(202.8 J/mol·K) = 405.6 J/mol·K

Now that we have calculated ΔH° and ΔS°, we can calculate ΔG° using the following equation:

ΔG° = ΔH° - TΔS°

Assuming a temperature of 298 K, we can calculate ΔG° as follows:

ΔG° = -221.6 kJ/mol - (298 K)(405.6 J/mol·K) = -221.6 kJ/mol - 121.37 kJ/mol = -343 kJ/mol

Conclusion

In this article, we have derived a balanced equation for the gaseous reaction between N2O5 and F2 to form NF3 and O2. We have also calculated the standard Gibbs free energy change (ΔG°) for this reaction using the standard enthalpy change (ΔH°) and standard entropy change (ΔS°). The calculated value of ΔG° indicates that the reaction is spontaneous at standard conditions.

References

CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
NIST Chemistry WebBook, National Institute of Standards and Technology, 2020.
Thermodynamic Properties of Chemical Substances, I. Barin, VCH Publishers, 1989.
Q&A: Balancing Equations and Thermodynamic Properties of Chemical Reactions ====================================================================

Q: What is the balanced equation for the reaction between N2O5 and F2 to form NF3 and O2?

A: The balanced equation for the reaction between N2O5 and F2 to form NF3 and O2 is:

N2O5 + 3F2 → 2NF3 + 3O2

Q: How do I determine the standard Gibbs free energy change (ΔG°) for a reaction?

A: To determine the standard Gibbs free energy change (ΔG°) for a reaction, you need to calculate the standard enthalpy change (ΔH°) and standard entropy change (ΔS°) for the reaction. The standard Gibbs free energy change (ΔG°) can be calculated using the following equation:

ΔG° = ΔH° - TΔS°

Q: What is the standard enthalpy change (ΔH°) for the reaction between N2O5 and F2 to form NF3 and O2?

A: The standard enthalpy change (ΔH°) for the reaction between N2O5 and F2 to form NF3 and O2 can be calculated using the following equation:

ΔH° = Σ(ΔHf°,products) - Σ(ΔHf°,reactants)

where ΔHf° is the standard enthalpy of formation of a substance.

Using the standard enthalpies of formation of NF3, O2, N2O5, and F2, we can calculate ΔH° as follows:

ΔH° = 2(ΔHf°,NF3) + 3(ΔHf°,O2) - (ΔHf°,N2O5) - 3(ΔHf°,F2) = 2(-111.3 kJ/mol) + 3(0 kJ/mol) - (-102.5 kJ/mol) - 3(0 kJ/mol) = -221.6 kJ/mol

Q: What is the standard entropy change (ΔS°) for the reaction between N2O5 and F2 to form NF3 and O2?

A: The standard entropy change (ΔS°) for the reaction between N2O5 and F2 to form NF3 and O2 can be calculated using the following equation:

ΔS° = Σ(ΔSf°,products) - Σ(ΔSf°,reactants)

where ΔSf° is the standard entropy of formation of a substance.

Using the standard entropies of formation of NF3, O2, N2O5, and F2, we can calculate ΔS° as follows:

ΔS° = 2(ΔSf°,NF3) + 3(ΔSf°,O2) - (ΔSf°,N2O5) - 3(ΔSf°,F2) = 2(202.8 J/mol·K) + 3(205.14 J/mol·K) - (209.2 J/mol·K) - 3(202.8 J/mol·K) = 405.6 J/mol·K

Q: What is the standard Gibbs free energy change (ΔG°) for the reaction between N2O5 and F2 to form NF3 and O2?

A: The standard Gibbs free energy change (ΔG°) for the reaction between N2O5 and F2 to form NF3 and O2 can be calculated using the following equation:

ΔG° = ΔH° - TΔS°

Using the calculated values of ΔH° and ΔS°, we can calculate ΔG° as follows:

ΔG° = -221.6 kJ/mol - (298 K)(405.6 J/mol·K) = -221.6 kJ/mol - 121.37 kJ/mol = -343 kJ/mol

Q: Is the reaction between N2O5 and F2 to form NF3 and O2 spontaneous at standard conditions?

A: Yes, the reaction between N2O5 and F2 to form NF3 and O2 is spontaneous at standard conditions, as indicated by the negative value of ΔG°.

Q: What are some common mistakes to avoid when balancing equations and calculating thermodynamic properties?

A: Some common mistakes to avoid when balancing equations and calculating thermodynamic properties include:

Not balancing the equation correctly
Not using the correct values for the standard enthalpies and entropies of formation
Not calculating the standard Gibbs free energy change correctly
Not considering the units and signs of the values used in the calculations

Q: How can I apply the concepts of balancing equations and thermodynamic properties to real-world problems?

A: The concepts of balancing equations and thermodynamic properties can be applied to real-world problems in a variety of ways, such as:

Designing and optimizing chemical processes
Predicting the behavior of complex systems
Developing new materials and technologies
Understanding the environmental impact of chemical reactions

By mastering the concepts of balancing equations and thermodynamic properties, you can gain a deeper understanding of the underlying principles of chemistry and apply them to a wide range of real-world problems.