Answer to Question #134503 in General Chemistry for Errol

Question

The combustion of 1.000g of benzene, C6H6 (l), in O2 (g) liberates 41.84kJ of heat and yields CO2(g) and H2O (l) . Write the thermochemical equation for the combustion of one mole of C6H6 (l) .

Answer 1

At first the balanced chemical equation should be written:

2C6H6+15O2=12CO2+6H2O.

It is seen that heat here should be written for 2 moles of benzene. So the given number of moles should be found:

n=m/M

M=6×12.011+6×1.008=72.066+6.048=78.114(g/mol)

n=1.000/78.114=0.013(mol).

The proportion for 2 moles should be used:

∆H=2×41.84/0.013=6436.92(kJ).

The heat is released, so the reaction is exothermic, and ∆H is negative.

2C6H6 + 15O2 = 12CO2 + 6H2O, ∆H= -6436.92kJ