Tungsten carbide is very hard and is used to make cutting tools and rock drills. What is the enthalp — Thermodynamics and Thermochemistry Chemistry Question
Question
Tungsten carbide is very hard and is used to make cutting tools and rock drills. What is the enthalpy of formation (in kJ/mol) of tungsten carbide? The enthalpy change for this reaction is difficult to measure directly, because the reaction occurs at $1400^\circ\text{C}$. However, the enthalpies of combustion of the elements and of tungsten carbide can be measured easily.<br>$2\text{W(s)} + 3\text{O}_2\text{(g)} \to 2\text{WO}_3\text{(s)}; \Delta H = -1680.6\text{ kJ}$<br>$\text{C(graphite)} + \text{O}_2\text{(g)} \to \text{CO}_2\text{(g)}; \Delta H = -393.5\text{ kJ}$<br>$2\text{WC(s)} + 5\text{O}_2\text{(g)} \to 2\text{WO}_3\text{(s)} + 2\text{CO}_2\text{(g)}; \Delta H = -2391.6\text{ kJ}$
💡 Solution & Explanation
The goal is $\text{W} + \text{C} \to \text{WC}$. From the given data: $\Delta H_c(\text{W}) = -1680.6/2 = -840.3$. $\Delta H_c(\text{C}) = -393.5$. $\Delta H_c(\text{WC}) = -2391.6/2 = -1195.8$. $\Delta H_f(\text{WC}) = \Delta H_c(\text{W}) + \Delta H_c(\text{C}) - \Delta H_c(\text{WC}) = -840.3 - 393.5 - (-1195.8) = -1233.8 + 1195.8 = -38.0\text{ kJ/mol}$.