Input interpretation
aluminum + barium oxide ⟶ aluminum oxide + barium
Balanced equation
Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, Ba and O: Al: | c_1 = 2 c_3 Ba: | c_2 = c_4 O: | c_2 = 3 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 3 ⟶ + 3
Structures
+ ⟶ +
Names
aluminum + barium oxide ⟶ aluminum oxide + barium
Reaction thermodynamics
Enthalpy
| aluminum | barium oxide | aluminum oxide | barium molecular enthalpy | 0 kJ/mol | -548 kJ/mol | -1676 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -1644 kJ/mol | -1676 kJ/mol | 0 kJ/mol | H_initial = -1644 kJ/mol | | H_final = -1676 kJ/mol | ΔH_rxn^0 | -1676 kJ/mol - -1644 kJ/mol = -32 kJ/mol (exothermic) | | |
Entropy
| aluminum | barium oxide | aluminum oxide | barium molecular entropy | 28.3 J/(mol K) | 70 J/(mol K) | 51 J/(mol K) | 67 J/(mol K) total entropy | 56.6 J/(mol K) | 210 J/(mol K) | 51 J/(mol K) | 201 J/(mol K) | S_initial = 266.6 J/(mol K) | | S_final = 252 J/(mol K) | ΔS_rxn^0 | 252 J/(mol K) - 266.6 J/(mol K) = -14.6 J/(mol K) (exoentropic) | | |
Chemical names and formulas
| aluminum | barium oxide | aluminum oxide | barium Hill formula | Al | BaO | Al_2O_3 | Ba name | aluminum | barium oxide | aluminum oxide | barium IUPAC name | aluminum | oxobarium | dialuminum;oxygen(2-) | barium