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).

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 A $436.0$
Reaction $H_{2} (g) \to 2 H (g)$
entropy of $H_{2} \to 130.6 J / m o l K$
entropy of $H \to 114.6 J / m o l K$
$H_{2} (g) \to 2 H (g); △ G = 406.62 k J / m o l △ G = 406.62 k J / m o l T = 298 K$
We know, $△ G = △ H - T △ S; △ H = ?$
For the reaction,
$△ S =$ 2(entropy of $H$ ) - 1(entropy of $H_{2}$ )
$= 2 (114.6) - 1 (130.6) = 98.6 J / K = 98.6 \times 10^{- 3} k J / K △ G = △ H - T △ S 406.62 = △ H - (298) (98.6 \times 10^{- 3}) △ H = 436$
Hence bond energy $= △ H = 436 k J / m o l$

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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

k J / m o l

H_{2} (g) \to 2 H (g); Δ G^{0} = 406.6 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.

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

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.