The standard heat of combustion for naphthalene, C₁₀H₈(s), is -5156.8 kJ mol^-1. Using this information and the standard enthalpies of formation, $\Delta$H°_f: H₂O(l)= -285.9 kJ mol^-1; CO₂(g)= ?393.5 kJ mol^-1, calculate the standard enthalpy of formation of C₁₀H₈(s), in kJ mol^-1.Hint: Write out an equation and remember products minus reactants for your calculations.

Question

Quizplus · Accepted Answer

The standard enthalpy of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its constituent elements in their standard states. The standard enthalpy of combustion is the enthalpy change when 1 mole of a compound is burned completely in oxygen to form carbon dioxide and water.The standard enthalpy of formation of C10H8(s) can be calculated using the following equation:$$\Delta H°_f(C_{10}H_8(s)) = \Delta H°_c(C_{10}H_8(s)) - 10\Delta H°_f(CO_2(g)) - 4\Delta H°_f(H_2O(l))$$Substituting the given values into this equation, we get:$$\Delta H°_f(C_{10}H_8(s)) = -5156.8 kJ mol^{-1} - 10(-393.5 kJ mol^{-1}) - 4(-285.9 kJ mol^{-1})$$$$\Delta H°_f(C_{10}H_8(s)) = -5156.8 kJ mol^{-1} + 3935 kJ mol^{-1} + 1143.6 kJ mol^{-1}$$$$\Delta H°_f(C_{10}H_8(s)) = -78.2 kJ mol^{-1}$$Therefore, the standard enthalpy of formation of C10H8(s) is -78.2 kJ mol-1.

The Standard Heat of Combustion for Naphthalene, C10H8(s), Is -5156 Δ\DeltaΔ

The Standard Heat of Combustion for Naphthalene, C₁₀H₈(s), Is -5156 $\Delta$