$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H$ at $298 K = - 285.8 k J$
The molar enthalpy of vaporization of water at 1 atm and $25^{o} C$ is $44 k J$ . The standard enthalpy of formation of 1 mole of water vapour at is ?

- 241.8 k J

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241.8 k J

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329.8 k J

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- 329.8 k J

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Solution

The correct option is C $- 241.8 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H$ at $298 K = - 285.8 k J$ ......(1)
The molar enthalpy of vaporization of water at 1 atm and $25^{o} C$ is $44 k J$ .
$H_{2} O (l) \to H_{2} O (g); Δ H = 44 k J$ ......(2)
Add equations (1) and (2)
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H$ at $298 K = - 285.8 + 44 = - 241.8 k J$
The standard enthalpy of formation of 1 mole of water vapour at is $- 241.8 k J$

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Similar questions

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); {Δ H}_{295 K} = - 285.8 k J .

25^{o} C

25^{o} C

25^{o}

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O (l)

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.77 k J / m o l e

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 241.84 k J / m o l e

H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02 k J

O H^{-}

25^{o} C

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285.77 K J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 241.84 K J m o l^{- 1}

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