(2) Given:compound DH°, (KJ/mol)H,O,(1) - 187.7H,O(1)- 285.5(a) Calculate the standard heat of reaction, (H°rexn.for2 H,O,(1) ----> 2 H,O(1) + 0,(g),(b) Use IG° = DH° -TIIS° to predict whether theabove reaction will be spontaneous ornonspontaneous at standard state.

Given:The given balanced chemical equation is represented as follows: 2H2O2l→2H2Ol+O2ga. The enthalpy of reaction can be expressed as the difference of sum of enthalpy of products and sum of enthalpy of the reactants. The expression for enthalpy of the reaction is represented as follows: ∆H∘rxn=∑∆HProducts-∑∆HReactantsThus, the standard enthalpy of reaction will be calculated as follows: ∆H∘rxn=2∆HH2O l+1∆HO2g-2∆HH2O2l=2×-285.5+1×0-2×-187.7 kJ/mol=-571+375.4 kJ/mol∆H∘rxn=-195.6 kJ/mol Therefore, the standard enthalpy of the reaction is -195.6 kJ/mol. b. First, the entropy change for the given reaction is calculated as follows: ∆S∘=2∆S∘H2O l+1∆S∘O2g-∆S∘H2O2l=2×70+1×205.2-2×109.6 J/mol.K=345.2-219.2 J/mol.K∆S∘=126 J/mol.K Now, the value of Gibb's free energy at 298.15 K is calculated as follows: ∆G=∆H-T∆S=-195.6 kJ/mol-298.15 K×126 J/mol.K=-195.6 kJ/mol-37566.9 J/mol.K×1 kJ1000 J=-195.6 kJ/mol-37.5669 kJ/mol∆G=-233.1669 kJ/mol As the value of Gibb's free energy is negative thus, the reaction is spontaneous.Answer: a. The standard enthalpy of the reaction is -195.6 kJ/mol. b. The reaction will be spontaneous.