PROBLEM SOLVING NUMBER 2 only Therefore, at 25°C (298.15 K),-AHASsurrrxn-(-1648400 D5528.8 J/K%3D%3DT298.15 KTo get the total entropy of the process (ASuniv),[ASsys +ASurr 1=-549.5 J/K + 5528.8 J/K= 4979.3 J/K- Notice that the change in entropy of the reaction is negative, but the entropy change in the surroundingsis large enough such that the total entropy of the process is positive. This indicates that the oxidation ofiron is a spontaneous process at standard conditions.Activity 2 Problem SolvingDirections: Calculate the total entropy change ASuniv and determine if it is spontaneous or not.1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H,O2). Is the reactionspontaneous or not? Hint: 1kj = 1000JAUm(1)Page | 4 H,O20109.6-187.8Substance205.069.9(kJ/mol)S (J/mol-K)AH°,-285.8Solutions:Lesson 4 Gibbs Free Energy and SpontaneityV Gibbs free energy also determines the spontaneity of a reactionThe spontaneity of a process depends on two thermodynamic quantities: enthalpy and entropy. Combining thesetwo into a single equation introduces a third thermodynamic quantity that ultimately determines spontaneity. This quantityis referred to as the Gibbs free energy (G), named after Josiah Willard Gibbs. Free energy is a portion of the total energyof a system that is available to do useful work. The three thermodynamic quantities are related by the equationa.G=H-TSLike enthalpy and entropy, Gibbs free energy is also a state function. As such, considering the final and initialstates invol in a process, the equation above can be modified to give the Gibbs-Helmholtz equation expressed asAG= AH- TASreaction is spontaneous if AG is negative; it is non-spontaneous if AG is positive. This follows that for a reaction to bespontaneous, either of these two conditions must be met:AH must be negative, while AS is positive.If AH is positive, AS must be large or T must be high enough so that AG will be negative.Under standard conditions of 1 atm and 25°C, the standard free energy change (AG") of a chemical reaction canbe obtaine from the standard enthalpy change (AH) and standard entropy change (AS). The Gibbs-Helmholtz equationthen becomesAG =AH°-TAShe standard free energy change of a reaction can also be calculated from the standard free energy of formation(AG°) of the reactants and products.EAG, (products) - EAG"; (reactants)%3DSimilar to enthalpy, the AG of elements is zero.Sample Probl 1. Consider the dissolution of silver chloride (AgCl)AgCl ()→(be)(be) D +The AG of the reaction can be calculated from the standard molar free energies of formation of the substances.CI (ag)SubstanceAG (kJ/mol)-131.3-109. 8AG = [AG'; (Ag') + AGr(CI)] –[AG, (AgCl) ]=[(1 mol)(77.1 kJ/mol) + (1 mol)(- 131.3 kJ/mol)]-[(1 mol)(- 109.8 kJ/mol) ]The positive value of AG° indicates that the reaction is non-spontaneous at 25°C. This means that the reversereaction is more spontaneous.= 55.7kJ 55700 JActivity 3 Problem SolvingDirections: Calculate the AH, AS", AG°. Show your complete solution.1. Determine the spontaneity of the dissociation of ammonium nitrate (NH,NO3), the compound used in cold packs.Interpret the values obtained from the calculations. The temperature is 298.15 K.4.+ NO3 (aq)NH,NO3 ()+.(be)NO3 (ag)SubstanceAH (kJ/mol)S (J/mol-K)NH,NO3 (9)-365.6151.1-207.4-132.5113.0146.4a.Calculate the AH°,

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PROBLEM SOLVING NUMBER 2 only Therefore, at 25°C (298.15 K),-AHASsurrrxn-(-1648400 D5528.8 J/K%3D%3DT298.15 KTo get the total entropy of the process (ASuniv),[ASsys +ASurr 1=-549.5 J/K + 5528.8 J/K= 4979.3 J/K- Notice that the change in entropy of the reaction is negative, but the entropy change in the surroundingsis large enough such that the total entropy of the process is positive. This indicates that the oxidation ofiron is a spontaneous process at standard conditions.Activity 2 Problem SolvingDirections: Calculate the total entropy change ASuniv and determine if it is spontaneous or not.1. Calculate the total entropy change for the decomposition of hydrogen peroxide (H,O2). Is the reactionspontaneous or not? Hint: 1kj = 1000JAUm(1)Page | 4 H,O20109.6-187.8Substance205.069.9(kJ/mol)S (J/mol-K)AH°,-285.8Solutions:Lesson 4 Gibbs Free Energy and SpontaneityV Gibbs free energy also determines the spontaneity of a reactionThe spontaneity of a process depends on two thermodynamic quantities: enthalpy and entropy. Combining thesetwo into a single equation introduces a third thermodynamic quantity that ultimately determines spontaneity. This quantityis referred to as the Gibbs free energy (G), named after Josiah Willard Gibbs. Free energy is a portion of the total energyof a system that is available to do useful work. The three thermodynamic quantities are related by the equationa.G=H-TSLike enthalpy and entropy, Gibbs free energy is also a state function. As such, considering the final and initialstates invol in a process, the equation above can be modified to give the Gibbs-Helmholtz equation expressed asAG= AH- TASreaction is spontaneous if AG is negative; it is non-spontaneous if AG is positive. This follows that for a reaction to bespontaneous, either of these two conditions must be met:AH must be negative, while AS is positive.If AH is positive, AS must be large or T must be high enough so that AG will be negative.Under standard conditions of 1 atm and 25°C, the standard free energy change (AG&#34;) of a chemical reaction canbe obtaine from the standard enthalpy change (AH) and standard entropy change (AS). The Gibbs-Helmholtz equationthen becomesAG =AH°-TAShe standard free energy change of a reaction can also be calculated from the standard free energy of formation(AG°) of the reactants and products.EAG, (products) - EAG&#34;; (reactants)%3DSimilar to enthalpy, the AG of elements is zero.Sample Probl 1. Consider the dissolution of silver chloride (AgCl)AgCl ()→(be)(be) D +The AG of the reaction can be calculated from the standard molar free energies of formation of the substances.CI (ag)SubstanceAG (kJ/mol)-131.3-109. 8AG = [AG'; (Ag') + AGr(CI)] –[AG, (AgCl) ]=[(1 mol)(77.1 kJ/mol) + (1 mol)(- 131.3 kJ/mol)]-[(1 mol)(- 109.8 kJ/mol) ]The positive value of AG° indicates that the reaction is non-spontaneous at 25°C. This means that the reversereaction is more spontaneous.= 55.7kJ 55700 JActivity 3 Problem SolvingDirections: Calculate the AH, AS&#34;, AG°. Show your complete solution.1. Determine the spontaneity of the dissociation of ammonium nitrate (NH,NO3), the compound used in cold packs.Interpret the values obtained from the calculations. The temperature is 298.15 K.4.+ NO3 (aq)NH,NO3 ()+.(be)NO3 (ag)SubstanceAH (kJ/mol)S (J/mol-K)NH,NO3 (9)-365.6151.1-207.4-132.5113.0146.4a.Calculate the AH°,

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