H 2 O l - H 2 g -1-2 O 2 g - H -890-36 KJ - moleWhat is - H for formation of H 2 O 1 from its constituent elements-A- 890-36 KJ - moleB- 980-36 KJ - moleC- 908-36 KJ - moleD- -890-36 KJ - mole

The correct option is D 890.36 KJ/moleH_2O_(l)→ H_2(g)+1/2(g) Δ H=+890.36KJ/mole H_2(s)+1/2O_2(g)→ H_2O_(l) Δ H= 890.36KJ/mole ∴ (Δ H_f)_H_2O= 890.36KJ/mole ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

H 2 O l → H 2...

Question

$H_{2} O_{(l)} \to H_{2 (g)} + \frac{1}{2} O_{2 (g)} Δ H = + 890.36 K J / m o l e$
What is $Δ H$ for formation of $H_{2} O_{(l)}$ from its constituent elements?

890.36 KJ/mole

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980.36 KJ/mole

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908.36 KJ/mole

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- 890.36 KJ/mole

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Solution

The correct option is D - 890.36 KJ/mole
$H_{2} O_{(l)} \to H_{2} (g) + \frac{1}{2} (g)$
$Δ H = + 890.36 K J / m o l e$
$H_{2 (s)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}$
$Δ H = - 890.36 K J / m o l e$
$∴ (Δ H_{f})_{H_{2} O} = - 890.36 K J / m o l e$

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Similar questions

Q. The enthalpy of vaporisation of liquid water using data:

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

Q. Consider the following reactions :

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

II)

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

III)

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

IV)

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

Enthalpy of formation of

H_{2} O (l)

is:

2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g), Δ H = - 787 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H = - 286 k J

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l), Δ H = - 1310 k J

From the above data, heat of formation of acetylene is:

Q. When 23g Na reacts with 1 mole of propyne then how many mole of

H_{2}

gas will be released:-

Q. Calculate enthalpy for formation of ethylene from the following data:

(I)

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J

(II)

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.2 k J

(III)

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l); Δ H = - 1410.8 k J

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H2O(l)→H2(g)+12O2(g) ΔH=+890.36KJ/mole What is ΔH for formation of H2O(l) from its constituent elements?

The correct option is D - 890.36 KJ/moleH2O(l)→H2(g)+12(g) ΔH=+890.36KJ/mole H2(s)+12O2(g)→H2O(l) ΔH=−890.36KJ/mole ∴(ΔHf)H2O=−890.36KJ/mole

$H_{2} O_{(l)} \to H_{2 (g)} + \frac{1}{2} O_{2 (g)} Δ H = + 890.36 K J / m o l e$
What is $Δ H$ for formation of $H_{2} O_{(l)}$ from its constituent elements?

The correct option is D - 890.36 KJ/mole
$H_{2} O_{(l)} \to H_{2} (g) + \frac{1}{2} (g)$
$Δ H = + 890.36 K J / m o l e$
$H_{2 (s)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}$
$Δ H = - 890.36 K J / m o l e$
$∴ (Δ H_{f})_{H_{2} O} = - 890.36 K J / m o l e$