Combustion of Hydrogen in a fuel cell at 300 K is represented as 2 H2 (g) + O2 (g) arrow 2 H2 O (g). If Δ H and Δ G are -241.60 kJ mol- 1 and -228.40 kJ mol- 1 respectively for H2O then the value of Δ S for the above process is:

Question

Combustion of Hydrogen in a fuel cell at 300 K is represented as 2 H2 (g) + O2 (g) arrow 2 H2 O (g). If Δ H and Δ G are -241.60 kJ mol- 1 and -228.40 kJ mol- 1 respectively for H2O then the value of Δ S for the above process is: (A) 44 J K- 1 (B) -88 J K- 1 (C) +88 J K- 1 (D) -44 J K- 1

Tardigrade Admin · Accepted Answer

Δ S=(Δ H - Δ G/T)=(- 241.6 - (- 228.4)/300) For 1 mole of H2O=-44J K- 1 Hence, for 2 moles =-88J K- 1

Q. Combustion of Hydrogen in a fuel cell at $300 K$ is represented as $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g) .$ If $Δ H$ and $Δ G$ are $- 241.60 k J m o l^{- 1}$ and $- 228.40 k J m o l^{- 1}$ respectively for $H_{2} O$ then the value of $Δ S$ for the above process is:

$44 J K^{- 1}$

$- 88 J K^{- 1}$

$+ 88 J K^{- 1}$

$- 44 J K^{- 1}$

$Δ S = \frac{Δ H - Δ G}{T} = \frac{- 241.6 - ( - 228.4 )}{300}$
For $1$ mole of $H_{2} O = - 44 J K^{- 1}$
Hence, for $2$ moles $= - 88 J K^{- 1}$

Q. Combustion of Hydrogen in a fuel cell at 300K is represented as 2H2​(g)+O2​(g)→2H2​O(g). If ΔH and ΔG are −241.60kJmol−1 and −228.40kJmol−1 respectively for H2​O then the value of ΔS for the above process is:

44JK−1

−88JK−1

+88JK−1

−44JK−1