Pls solve no 2 sum.
Reaction of combustion of benzene is-
C6H6 (l) + 15/2O2 (g) → 6CO2 (g) + 3H2O (l)
∆n = 6 – 15/2
= -3/2 mol
Heat produced by combusting 1.0 g benzene is 41.80 kJ
Heat produced by combusting 1 mole benzene = 78 x 41.80 kJ (1 mol of benzene = 78 g/mol)
= 3260.4 kJ/mol
Therefore, ∆Ucombustion = - 3260.4 kJ/mol
Enthalpy of combustion can be calculated by applying following relation
∆Hcombustion = ∆U + ∆nRT
= - 3260.4 kJ/mol + (-3/2 mol)(8.314 x 10-3 kJK-1mol-1)(298 K)
= - 3260.4 – 3.716
= - 3264.116 kJ/mol
C6H6 (l) + 15/2O2 (g) → 6CO2 (g) + 3H2O (l)
∆n = 6 – 15/2
= -3/2 mol
Heat produced by combusting 1.0 g benzene is 41.80 kJ
Heat produced by combusting 1 mole benzene = 78 x 41.80 kJ (1 mol of benzene = 78 g/mol)
= 3260.4 kJ/mol
Therefore, ∆Ucombustion = - 3260.4 kJ/mol
Enthalpy of combustion can be calculated by applying following relation
∆Hcombustion = ∆U + ∆nRT
= - 3260.4 kJ/mol + (-3/2 mol)(8.314 x 10-3 kJK-1mol-1)(298 K)
= - 3260.4 – 3.716
= - 3264.116 kJ/mol