Diborane is a potential rocket fuel which undergoes combustion according to the reaction: B2H6(g) + — Thermodynamics and Thermochemistry Chemistry Question
Question
Diborane is a potential rocket fuel which undergoes combustion according to the reaction: B2H6(g) + 3$O_2$(g) -> B2$O_3$(s) + 3$H_2O$(g). From the following data, calculate the enthalpy change for the combustion of diborane: 2B(s) + 3/2 $O_2$(g) -> B2$O_3$(s); δ H = -1273 kJ/mol; $H_2$(g) + 1/2 $O_2$(g) -> $H_2O$(l); δ H = -286 kJ/mol; $H_2O$(l) -> $H_2O$(g); δ H = 44 kJ/mol; 2B(s) + 3$H_2$(g) -> B2H6(g); δ H = 36 kJ/mol
💡 Solution & Explanation
Using Hess's Law: (1) 2B(s) + 1.5$O_2$(g) -> B2$O_3$(s); dH = -1273 kJ. (2) 3$H_2$(g) + 1.5$O_2$(g) -> 3$H_2O$(g); dH = 3*(-286+44) = 3*(-242) = -726 kJ. (3) B2H6(g) -> 2B(s) + 3$H_2$(g); dH = -36 kJ. Sum: B2H6(g) + 3$O_2$(g) -> B2$O_3$(s) + 3$H_2O$(g); δ H = -1273 + (-726) + (-36) = -2035 kJ/mol.