The standard enthalpies of formation (Delta H | vedclass

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(B) The reaction is: $CO_{2(g)} + H_{2(g)} \rightarrow CO_{(g)} + H_2O_{(g)}$The standard enthalpy change of the reaction is calculated using the form

Vedclass · Accepted Answer

$41.2$

Vedclass · Answer

(B) The reaction is: $CO_{2(g)} + H_{2(g)} ightarrow CO_{(g)} + H_2O_{(g)}$The standard enthalpy change of the reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(	ext{products}) - \sum \Delta H_f^o(	ext{reactants})$Substituting the given values:$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$Since the standard enthalpy of formation of an element in its standard state,such as $H_{2(g)}$,is $0 \ kJ \ mol^{-1}$,we have:$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$$\Delta H^o = -352.3 + 393.5 = +41.2 \ kJ \ mol^{-1}$

The standard enthalpies of formation $(\Delta H_f^o)$ for $CO_{2(g)}$,$CO_{(g)}$,and $H_2O_{(g)}$ are $-393.5$,$-110.5$,and $-241.8 \ kJ \ mol^{-1}$ respectively. What is the standard enthalpy change (in $kJ \ mol^{-1}$) for the reaction: $CO_{2(g)} + H_{2(g)} \rightarrow CO_{(g)} + H_2O_{(g)}$?

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