CH3OH = H2 + H2CO

methanol ⟶ hydrogen + formaldehyde

Balance the chemical equation algebraically: ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 ⟶ c_2 + c_3 Set the number of atoms in the reactants equal to the number of atoms in the products for C, H and O: C: | c_1 = c_3 H: | 4 c_1 = 2 c_2 + 2 c_3 O: | c_1 = c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | ⟶ +

⟶ +

methanol ⟶ hydrogen + formaldehyde

| methanol | hydrogen | formaldehyde molecular enthalpy | -238.7 kJ/mol | 0 kJ/mol | -108.6 kJ/mol total enthalpy | -238.7 kJ/mol | 0 kJ/mol | -108.6 kJ/mol | H_initial = -238.7 kJ/mol | H_final = -108.6 kJ/mol | ΔH_rxn^0 | -108.6 kJ/mol - -238.7 kJ/mol = 130.1 kJ/mol (endothermic) | |

| methanol | hydrogen | formaldehyde molecular free energy | -166.3 kJ/mol | 0 kJ/mol | -102.5 kJ/mol total free energy | -166.3 kJ/mol | 0 kJ/mol | -102.5 kJ/mol | G_initial = -166.3 kJ/mol | G_final = -102.5 kJ/mol | ΔG_rxn^0 | -102.5 kJ/mol - -166.3 kJ/mol = 63.77 kJ/mol (endergonic) | |

| methanol | hydrogen | formaldehyde Hill formula | CH_4O | H_2 | CH_2O name | methanol | hydrogen | formaldehyde IUPAC name | methanol | molecular hydrogen | formaldehyde

| methanol | hydrogen | formaldehyde molar mass | 32.042 g/mol | 2.016 g/mol | 30.026 g/mol phase | liquid (at STP) | gas (at STP) | gas (at STP) melting point | -98 °C | -259.2 °C | -92 °C boiling point | 64.7 °C | -252.8 °C | -19.1 °C density | 0.791 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 1.09 g/cm^3 (at 25 °C) solubility in water | miscible | | miscible dynamic viscosity | 5.44×10^-4 Pa s (at 25 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | pungent | odorless | none-slight