The entropies of $H_{2} (g)$ and $H (g)$ are $130.6$ and $114.6$ $J m o l^{- 1} K^{- 1}$ respectively at $298 K$ . Using the data given below calculate the bond energy of $H_{2}$ (in $k J / m o l$ ).
$H_{2} (g) \to 2 H (g); Δ G^{0} = 406.6 k J$

377.2

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436.0

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425.5

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430.5

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Solution

The correct option is B $436.0$
$Δ G = Δ H - T Δ S$

$Δ H = Δ G + T Δ S$

$Δ H = 406600 + 298 \times 98.6$ $= 435.98 k J = 436 k J$

[Note: $Δ S = 2 \times 114.6 - 130.6 = 98.6 J K^{- 1} m o l^{- 1}$

As the bond energy is endothermic in nature so the answer will be $436$ .

Option B is correct.

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Similar questions

Q. The entropies of

H_{2} (g)

and

H (g)

are

130.6

and

114.6 J m o l^{- 1} K^{- 1}

respectively at

298 K

. Using the data given below calculate the bond energy of

H_{2}

(in kJ/mol).

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.

$H (g) \to H (g) + I (g)$ ; $Δ G^{0} = 271.8 k J$

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.