The standard enthalpy of combustion of solid boron is numerically equal to — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The combustion reaction for solid boron is: B(s) + 3/4 $O_2$(g) -> 1/2 B2$O_3$(s). The standard enthalpy of formation of B2$O_3$(s) refers to: 2B(s) + 3/2 $O_2$(g) -> B2$O_3$(s). Thus, the enthalpy of combustion of B(s) is exactly half of the enthalpy of formation of B2$O_3$(s), i.e., 1/2 delta_f H°_B2$O_3$(s).