The calorific value of hydrogen gas is -143 | vedclass

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(C) The combustion reaction of hydrogen is: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$.Since the enthalpy of formation of elements in their stan

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$-286 \ kJ \ mol^{-1}$

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(C) The combustion reaction of hydrogen is: $H_2(g) + \frac{1}{2}O_2(g) ightarrow H_2O(l)$.Since the enthalpy of formation of elements in their standard state is zero,the enthalpy of combustion of $H_2$ is equal to the enthalpy of formation of $H_2O$.$\Delta H_{comb}^{o}(H_2) = \Delta H_{f}^{o}(H_2O)$.The calorific value is given as $-143 \ kJ \ g^{-1}$.Since the molar mass of $H_2$ is $2 \ g \ mol^{-1}$,the enthalpy of formation per mole is:$\Delta H_{f}^{o}(H_2O) = 	ext{Calorific value} 	imes 	ext{Molar mass of } H_2$.$\Delta H_{f}^{o}(H_2O) = -143 \ kJ \ g^{-1} 	imes 2 \ g \ mol^{-1} = -286 \ kJ \ mol^{-1}$.

Reaction	Energy Change (in $kJ$)
$Li_{(s)} \to Li_{(g)}$	$161$
$Li_{(g)} \to Li^{+}_{(g)}$	$520$
$\frac{1}{2} F_{2(g)} \to F_{(g)}$	$77$
$F_{(g)} + e^- \to F^{-}_{(g)}$	(Electron gain enthalpy)
$Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$	$-1047$
$Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$	$-617$

The calorific value of hydrogen gas is $-143 \ kJ \ g^{-1}$. The standard enthalpy of formation of $H_2O$ will be $—$

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