The enthalpy of formation of reaction -2H2g-O2g-2H2Ol is -r Ho-572 kJmol-1What will be the enthalpy of formation of H2Ol-

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Heat of Formation

The enthalpy ...

Question

The enthalpy of formation of reaction :
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}$
What will be the enthalpy of formation of $H_{2} O (l)$ ?

- 286 k J m o l^{- 1}

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286 k J m o l^{- 1}

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- 1144 k J m o l^{- 1}

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1144 k J m o l^{- 1}

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2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}

H_{2} O (l)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}

H_{2} O (l)

S n O_{2} (s) + 2 H_{2} (g) ⟶ S n (s) + 2 H_{2} O (l)

S n O_{2} (s)

H_{2} O (l)

- 580.7 k J

- 285.8 k J

C (s) + O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{\circ} = - 393.5 k J

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

Δ_{r} H^{\circ} = - 571.8 k J

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

Δ_{r} H^{\circ} = - 890.3 k J

H_{2} O_{2} (l),

N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) \to N_{2} (g) + 4 H_{2} O (l);

△_{r} H_{1}^{o} = - 818 k J / m o l

N_{2} H_{4} (l) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (l);

△_{r} H_{2}^{o} = - 622 k J / m o l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{f} H_{3}^{o} = - 285 k J / m o l

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