Delta H and Delta E for the reaction,Fe_{2}O_ | vedclass

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(D) For any chemical reaction,the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by:$\Delta H = \D

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$\Delta H = \Delta E - 3RT$

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(D) For any chemical reaction,the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by:$\Delta H = \Delta E + \Delta n_{g}RT$Where $\Delta n_{g}$ is the change in the number of moles of gaseous species.$\Delta n_{g} = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants})$For the reaction: $Fe_{2}O_{3(s)} + 3H_{2(g)} ightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$Note that $Fe_{2}O_{3}$ and $Fe$ are solids,and $H_{2}O$ is a liquid. Only $H_{2}$ is in the gaseous state.$\Delta n_{g} = 0 - 3 = -3$Substituting this value into the equation:$\Delta H = \Delta E + (-3)RT$$\Delta H = \Delta E - 3RT$

$\Delta H$ and $\Delta E$ for the reaction,$Fe_{2}O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$ at constant temperature are related as

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