The combustion of 3.9 g of liquid benzene ( C ₆ H ₆ ) in an open vessel at 298 K liberates 163.5 kJ of heat. If the standard enthalpies of formation of CO ₂ (g) and H ₂ O(l) are -394 kJ mol ⁻¹ and -286 kJ mol ⁻¹ respectively, the standard enthalpy of formation of liquid benzene is _____ kJ mol ⁻¹ . (Given: Molar mass of C = 12 g mol ⁻¹ , H = 1 g mol ⁻¹ )

Step-by-step solution

First, calculate the number of moles of benzene combusted: Molar mass of C ₆ H ₆ = 6(12) + 6(1) = 78 g mol ⁻¹ Moles of benzene = 3.9 78 = 0.05 mol The heat liberated per mole of benzene (standard enthalpy of combustion, _c H^ ) is: _c H^ = -163.5 kJ 0.05 mol = -3270 kJ mol ⁻¹ The balanced chemical equation for the combustion of benzene is: C ₆ H ₆ (l) + 15 2 O ₂ (g) 6 CO ₂ (g) + 3 H ₂ O(l) Using Hess's Law: _c H^ = [6 _f H^ ( CO ₂) + 3 _f H^ ( H ₂ O )] - _f H^ ( C ₆ H ₆) Substitute the known values: -3270 = [6(-394