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

💡 Solution & Explanation

The combustion reaction for 1 mole of solid boron is $\text{B(s)} + \frac{3}{4}\text{O}_2\text{(g)} \to \frac{1}{2}\text{B}_2\text{O}_3\text{(s)}$. The enthalpy change of this process is exactly equal to $\frac{1}{2}$ times the standard molar enthalpy of formation of $\text{B}_2\text{O}_3\text{(s)}$.