Combustion of 1 mole of benzene is expressed as C6H6(l) + 152O2(g) 6CO2(g) + 3H2O(l). The standard enthalpy of combustion of 2 mol of benzene is −'x' kJ. x = . Given : 1. Standard Enthalpy of formation of 1 mol of C6H6(l), for the reaction 6C(graphite) + 3H2(g) C6H6(l) is 48.5 kJ mol-1. 2. Standard Enthalpy of formation of 1 mol of CO2(g), for the reaction C(graphite) + O2(g) CO2(g) is -393.5 kJ mol-1. 3. Standard Enthalpy of formation of 1 mol of H2O(l), for the reaction H2(g) + 12O2(g) H2O(l) is -286 kJ mol-1.
Answer
6535
Solution
cH = fH (products) - fH (reactants)
= 6(-393.5) + 3(-286) - 48.5 ∵ fH (O2) = 0
= -2361 - 858 - 48.5 = -3267.5 kJ mol-1
For 2 mol: 2 (-3267.5) = -6535 kJ
∴ x = 6535
Answer: 6535