Question

Part A Calculate AH (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(1) + 1502 (g)+12C02(g) + 6H2O(1) AH° = -6534kJ AH (CO2) = -393.5kJ/mol AH (H2O) = - 285.8kJ/mol O ALQ 6 A 0 2 ? AH = kJ/mol Submit Request Answer

Answer 1

Heat of reaction,

ΔΗ° = Ση,ΔΗ?(products) - ΣΩ,ΔΗ (reactants)

where, n_p and n_r stands for the moles of products and moles of reactants respectively.

Thus,

OHV
(moles of CO₂) x $\Delta H_{f}^{0}$ (CO₂) + (moles of H₂O) x $\Delta H_{f}^{0}$ (H₂O) - (moles of C₆H₆) x $\Delta H_{f}^{0}$ (C₆H₆) - (moles of O₂) x $\Delta H_{f}^{0}$ (O₂)

or, - 6534 kJ = 12 mol x (- 393.5 kJ/mol) + 6 mol x (- 285.8 kJ/mol) - (moles of C₆H₆) x $\Delta H_{f}^{0}$ (C₆H₆) - 0          [$\Delta H_{f}^{0}$(O₂) = 0 as oxygen is in its standard elemental state)

or, - 6534 kJ = - 4722 kJ - 1714.8 kJ - 2 mol x $\Delta H_{f}^{0}$ (C₆H₆)

or, $\Delta H_{f}^{0}$ (C₆H₆) = 48.6 kJ/mol

Hence, $\Delta H_{f}^{0}$ of benzene = 48.6 kJ/mol