Question

The entropies of H2(g) and H(g) are 130.6 and 114.6 J/mol K respectively at 298 K. Using the data given below calculate the bond energy of H2 (in kJ/mol):

H2(g)→ 2H(g) ;∆G = 406.62 kJ/mol

Answer 1

Given-
Entropies of H₂(g) = 130.6 J/mol K
Entropies of H(g) = 114.6 J/mol K
∆G = 406.62 kJ/mol
T = 298 K.


Reaction involved-
H₂(g)→ 2H(g)

∆S = ( 2 mol x 114.6 J/mol K) - (1 mol x 130.6 J/mol K)
∆S = 98.6 J/K = 98.6 x 10^-3 kJ/K
∆G = 406.62 kJ/mol
T = 298 K



Equation used-
∆G = ∆H - T∆S
∆H = ∆G + T∆S
∆H = 406.62 + 298 x 98.6 x 10^-3
∆H = 436.00 kJ/mol

Bond energy of H₂ (in kJ/mol) = 436.00 kJ /mol