Question

Calculate the entropy change ∆S per mole for following reaction. Cumbustion of Hydrogen in a
fuel cell at 298K.



● H_2
(g) +
½O2
(g) − − − −−→ H2O(g)[∆H = 241.60KJ, ∆G= 288.40KJ]​

Answer 1

Answer:

Combustion of hydrogen in a fuel cell at 298

K. H2 (g) + 1 2 12O2 (g) → H2O (g) ΔH = 241.60 kJ ΔG = 288.40

Answer 2

Answer:

ΔG = ΔH - TΔS

Putting in the values,

−288.40 = −241.6 − 298 × ΔS

$ΔS = \frac{ - 288.4 + 241.6}{ - 298}$

= 0.157 kJ K−¹ mol−¹