C4H802 + Naoh -> ( Y ) + Ch3OH. Este ban đầu là?

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj

The answer is “groups” or families. periods go horizontal. : )