Please help! Thank you! Table: Thermodynamic DataS°, J/molAH:° KJ/mol| -201.2AG, KJ/molCH3OH237.6-161.9H2130.58C2H6-84.68229.5-32.89H2O-241.82188.83-228.572CH3OH(g) + H2(8)C2H6(8) + 2H,O(g)1. Use Table 3 to calculate AH°, AS°, and AG° for the reaction above2. Use AH° and AS° calculated in questions 1 and 2 to calculate AG° using Eq. 2. Assumethe temperature is 298K.w wDraw Lewis structures for the system under study in this experiment

For a general reaction aA + bB → cc + dD Entropyb change ∆S°rxn =( Σv S°f )product - ( Σ vS°f )reactant ∆S°rxn = [c (S°f )c + d (S°f )d] - [a(S°f )a + b(S°f )b] enthalpy change ∆H°rxn =( Σv ∆H°f )product - ( Σ v∆H°f )reactant ∆H°rxn = [c (∆H°f )c + d (∆H°f )d] - [a(∆H°f )a + b(∆H°f )b] and gibbs free enegy change ∆G°rxn =( Σv G°f )product - ( Σ vG°f )reactant ∆G°rxn = [c (∆G°f )c + d (∆G °f )d] - [a(∆G °f )a + b(∆G °f )b]For the reaction , 2CH3OH (g) + H2(g) ⇌ C2H6 (g) + 2H2O(g) enthalpy change ∆H°rxn =( Σv ∆H°f )product - ( Σ v∆H°f )reactant ∆H°rxn = [2 (∆H°f ) H2O + (∆H°f ) C2H6] - [2(∆H°f ) CH3OH + (∆H°f ) H2] ∆H°rxn = [2 x (-241.82 ) + (-84.68) ] - [2 x (-201.2) + 0] ∆H°rxn =-165.92 kJ/mol Entropy change ∆S°rxn =( Σv S°)product - ( Σ vS°)reactant ∆ S°rxn = [2 (S° ) H2O + (S° ) C2H6] - [2(S° ) CH3OH + (S° ) H2] ∆ S°rxn = [2 x (188.83) + (229.5) ] - [2 x (237.6) + 130.58] ∆ S°rxn = 1.18 J/mol gibbs free enegy change ∆G°rxn =( Σv G°f )product - ( Σ vG°f )reactant ∆ G°rxn = [2 (∆G°f ) H2O + (∆G°f ) C2H6] - [2(∆G°f ) CH3OH + (∆G°f ) H2] ∆ G°rxn = [2 x (-228.57 ) + (-32.89 ) ] - [2 x (-161.9 ) + 0] ∆ G°rxn = -166.23 kJ/mol At T = 298 K ∆G°rxn = ∆H°rxn -T ∆S°rxn ∆G°rxn = -165.92 kJ/mol -298 K x1.18 J/Kmol ∆G°rxn = -165920 J/mol -351.64 J/mol ∆G°rxn = -166271.64 J/mol ∆G°rxn = -166.272 kJ/mol