Question

The heat of combustion of naphthalene (C10H8(s)) at constant volume was measured to be −5133kJ/mol at 298K. Calculate the value of enthalpy change (R=8.314J/Kmol)

(a)5128.04kJ

(b)−5128.04kJ

(c)513.7955kJ

(d)−5137955.14J

Answer 1

The combustion equation of naphthalene is

C10H8(s)+12O2(g)→10CO2(g)+4H2O(l);ΔE=−5133kJ

Δn=10−12=−2mol

Now, applying the relation, ΔH=ΔE+(Δn)RT

=−51335128.04kJ103J+(−2mol)(8.314J/molK)(298K)

=−5133000J–4955.14J=−5137955.14J

Ans : (d)