Question

The heat of combustion of methane at 298° K is expressed
by
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂Oand ∆H = 890.2 kJ.
Magnitude of ∆E of reaction at this temperature is
(a) infinity (b) equal to ∆H
(c) less than ∆H (d) greater than ∆H

Answer 1

answer : option (d) greater than ∆H

The heat of combustion of methane at 298 K is expressed by CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O ; ∆H = 890.2 kJ

magnitude of w = ∆n_gRT

here ∆n_g = number of moles of gaseous products - number of moles of gaseous reactants

= 1 - (2 + 1) = -2

so, w = -2RT

= -2mol × 25/3 J/mol.K × 298K

= -7450 J

= -7.45 kJ

from first law of thermodynamics,

∆H = ∆E + w = ∆E + ∆n_gRT

⇒890.2 kJ = ∆E - 7.45 kJ

⇒∆E = 897.65 kJ > ∆H

hence, option (d) greater than ∆H