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? 2W(s) + 3(g) - 2WO3(s); δ H = -1680.6 kJ; C(graphit

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? 2W(s) + 3$O_2$(g) -> 2WO3(s); δ H = -1680.6 kJ; C(graphite) + $O_2$(g) -> $CO_2$(g); δ H = -393.5 kJ; 2WC(s) + 5$O_2$(g) -> 2WO3(s) + 2$CO_2$(g); δ H = -2391.6 kJ

Prashant Kotian · Accepted Answer

Correct answer: A. We want W(s) + C(graphite) -> WC(s). Divide all three equations: (1) W(s) + 1.5(g) -> WO3(s); dH = -840.3 kJ. (2) C(graphite) + (g) -> (g); dH = -393.5 kJ. (3) WC(s) + 2.5(g) -> WO3(s) + (g) reversed: WO3(s) + (g) -> WC(s) + 2.5(g); dH = +1195.8 kJ. Sum: W(s) + C(graphite) -> WC(s); delta_f H = -840.3 + (-393.5) + 1195.8 = -38.0 kJ/mol.

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