Interpretation:
The standard free energy change ΔGorxn has to be calculated for given reaction at 298K.
4NH3(g) + 5O2(g) → 2NO(g) + 6H2O(g)
Concept introduction:
- Enthalpy (H): it is the total amount of heat in a particular system.
- Entropy (S) : it is used to describe the disorder. It is the amount of arrangements possible in a system at a particular state. ΔSuniv = ΔSsys + ΔSsurr
- Free energy change (ΔGo): change in the free energy takes place while reactant converts to product where both are in standard state.
Enthalpy is the amount energy absorbed or released in a process.
The enthalpy change in a system (ΔΗsys) can be calculated by the following equation.
ΔHorxn = ΔH°produdcts- ΔH°reactants
Where,
ΔH°reactants is the standard entropy of the reactants
ΔH°produdcts is the standard entropy of the products
Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants (ΔS°rxn) can be calculated by the following equation.
ΔS°rxn = S°Products- S°reactants
Where,
S°reactants is the standard entropy of the reactants
S°Products is the standard entropy of the products
Standard free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter Go. All spontaneous process is associated with the decrease of free energy in the system. The equation given below helps us to calculate the change in free energy in a system.
ΔGorxn = ΔΗorxn- TΔSorxn
Where,
ΔGorxn is the change in standard free energy of the system
ΔΗorxn is the change in standard enthalpy of the system
T is the absolute value of the temperature
ΔSorxn is the change in entropy in the system
Answer to Problem 20.4BFP
The standard free-energy change of given reaction is −959 kJ_.
Explanation of Solution
Given,
Standared free energy equation is,
4NH3(g) + 5O2(g) → 2NO(g) + 6H2O(g)
The balanced equation rightfully, multiplying the reactant NH3, O2 and the products NO and H2O with the coefficents 4,5,2 and 6 respectively.
Calculation of enthalpy ΔH°rxn value at 298 K
Standard enthalpy change equation is,
ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)
ΔH°rxn = [(4 mol NO) ( ΔH°fof NO)+(6 mol H2O)(ΔH°fof H2O)] −[(4 mol NH3)(ΔH°f of NH3) + (5 mol O2)(ΔH°f of O2)] ΔH°rxn = [(4 mol NO)(90.29 kJ/mol ) +(6 mol H2O)(−241.826 kJ/mol)]− [(4 mol NH3)(−45.9 kJ/mol ) +(5 mol O2)(0 kJ/mol )] ΔH°rxn = −906.196 kJ
The enthalpy change is expected to be negative because in gas mole is negative.
Therefore, the enthalpy (ΔH°rxn) value is −906.196 kJ_
Calculation of entropy ΔS°rxn value at 298 K
Standard entropy change equation is,
ΔS°rxn = ∑m S°Products- ∑n S°reactants
ΔS°rxn =[(4 mol NO) ( Soof NO)+(6 mol H2O)(Soof H2O)] −[(4 mol NH3)(So of NH3) + (5 mol O2)(So of O2)]ΔS°rxn = [(4 mol NO)(210.65 J/mol×K)+(6 mol H2O)(188.72 J/mol×K) ]− [(4 mol NH3)(193.0 J/mol×K) +(5 mol O2)(205.0 J/mol×K)] ΔS°rxn = 177.92 J/K
Hence, the entropy (ΔS°rxn) of the reaction is 177.92 J/K_
Calculation of standard entropy value ΔGorxn
The standard entropy equation is,
ΔGorxn = ΔΗorxn- TΔSorxn
Enthalpy and entropy values are substituted in above equation.
ΔGorxn= −906.196 kJ−[(298 K)(177.92 J/K)(1 kJ/103J)]ΔGorxn= −959.21616 kJΔGorxn= −959 kJ (Rounded value)
Hence, the standard free-energy change of the reaction is ΔGorxn= −959 kJ_
The standard free energy of the reaction is negative. So, the reaction is spontaneous.
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Chapter 20 Solutions
Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics
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