Question

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


[tex]\maltese \: H_2
(g) +
½O_2
(g) \longrightarrow H_2O(g)[∆H = 241.60KJ, ∆G= 288.40KJ][/tex]

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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:

$\bold{Combustion of hydrogen in}$

$\bold{a fuel cell at 298}$

$\bold{K. H2 (g) + 12 1202 (g) → H2O (g)}$

$\bold{AH = -> 241.60 kJ AG = 288.40}$

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Explanation:

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