The standard heat enthalpy of a compound represents heat of formation of that compound at $25^{\circ} C$ and 1 atm pressure.

True

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False

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Solution

The correct option is A True
The standard heat enthalpy of a compound is equal to its standard heat of formation.
When a compound is formed from its constituent elements in their standard states ( $25^{o} C$ and 1 atm pressure), the heat of the reaction is the standard heat enthalpy or the standard heat of the formation.

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Similar questions

25^{o} C

1

C_{3} H_{8} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g) + C H_{4} (g)

25^{\circ} C

C_{3} H_{8} (g)

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

Compound:	$H_{2} (g)$	$C H_{4} (g)$	$C_{2} H_{6} (g)$	$C (g r a p h i t e)$
$Δ H^{⊖} (k J m o l^{- 1}) :$	$- 285.8$	$- 890.0$	$- 1560.0$	$- 393.5$

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H_{298 K} = - 68.32 K c a l .

^{o}

^{o}

(Δ H_{f})

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