For the reaction,2 H_{2(g)} + O_{2(g)} longri | vedclass

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(C) The given reaction is: $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2 O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. This represents the enthalpy of for

Vedclass · Accepted Answer

$286.6$

Vedclass · Answer

(C) The given reaction is: $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2 O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. This represents the enthalpy of formation for $2 \ mol$ of water vapor. To find the heat of decomposition for $1 \ mol$ of water,we reverse the reaction and divide by $2$: $2 H_2 O_{(g)} \longrightarrow 2 H_{2(g)} + O_{2(g)}$,$\Delta H^{\circ} = +573.2 \ kJ$. For $1 \ mol$ of water: $H_2 O_{(g)} \longrightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$,$\Delta H^{\circ} = \frac{573.2}{2} \ kJ = 286.6 \ kJ$. Thus,the heat of decomposition is $286.6 \ kJ \ mol^{-1}$.

For the reaction,$2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2 O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. What is the heat of decomposition of water per mol (in $kJ$)?

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