Combustion of hydrogen in a fuel cell at 300 KJ is represented as 2H2g - O2g - 2H2Og- If - H and - G are -241-60 KJ mol -1 and -228-40 KJ mol -1 of H2O respectively- then the value of - S for the above process is

The correct option is B 88 J Δ H=2× 241.60 Δ G=Δ H TΔ S Δ G=2× ( 228.40) Δ S=Δ H Δ GT =2×( 241.60+228.40)300KJ =2×( 13.2)× 100030 ...

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Standard XII

Chemistry

Sucrose

Combustion of...

Question

Combustion of hydrogen in a fuel cell at $300 K J$ 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 KJ mol $^{- 1}$ and -228.40 KJ mol $^{- 1}$ of $H_{2} O$ respectively, then the value of $Δ S$ for the above process is

4.4 J

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-88 J

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+88 J

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-44 J

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Solution

The correct option is B -88 J
$Δ H = 2 \times 241.60$
$Δ G = Δ H - T Δ S$
$Δ G = 2 \times (- 228.40)$
$Δ S = \frac{Δ H - Δ G}{T}$
$= 2 \times \frac{(- 241.60 + 228.40)}{300} K J$
$= 2 \times \frac{(- 13.2) \times 1000}{300}$
$= - 88 J / K$

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Combustion of hydrogen in a fuel cell at 300 KJ is represented as 2H2(g)+O2(g)→2H2O(g). If ΔH and ΔG are -241.60 KJ mol−1 and -228.40 KJ mol−1 of H2O respectively, then the value of ΔS for the above process is

The correct option is B -88 JΔH=2×241.60ΔG=ΔH−TΔSΔG=2×(−228.40)ΔS=ΔH−ΔGT=2×(−241.60+228.40)300KJ=2×(−13.2)×1000300=−88 J/K

Combustion of hydrogen in a fuel cell at $300 K J$ 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 KJ mol $^{- 1}$ and -228.40 KJ mol $^{- 1}$ of $H_{2} O$ respectively, then the value of $Δ S$ for the above process is

The correct option is B -88 J
$Δ H = 2 \times 241.60$
$Δ G = Δ H - T Δ S$
$Δ G = 2 \times (- 228.40)$
$Δ S = \frac{Δ H - Δ G}{T}$
$= 2 \times \frac{(- 241.60 + 228.40)}{300} K J$
$= 2 \times \frac{(- 13.2) \times 1000}{300}$
$= - 88 J / K$