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Heat of Reaction

Standard Enth...

Question

Standard Enthalpy (Heat) of formation of liquid water at $25^{o}$ C is around :
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O (l)$

A

- 237

kJ/mol

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B

237

kJ/mol

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C

- 286

kJ/mol

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D

286

kJ/mol

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Solution

The correct option is B $- 286$ kJ/mol
Standard Enthalpy(Heat) of formation of liquid water at $25^{o}$ C is around $- 286$ kJ/mol.

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O (l)$ .
$(Δ H_{f}^{0})$ water $= Δ H_{H_{2} O (l)}^{0} - [Δ H_{H_{2} (g)}^{0} + 0.5 Δ H_{O_{2} (g)}^{0}]$
$(Δ H_{f}^{0})$ water $= - 286 kJ/mol - [0 kJ/mol + 0.5 \times 0 kJ/mol]$
$(Δ H_{f}^{0})$ water $= - 286 kJ/mol$ .

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