The Delta H^o_f for CO_{2(g)},CO_{(g)},and H_ | vedclass

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(B) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(	ext{products}) - \sum \Delta H_f^o(\

Vedclass · Accepted Answer

$41.2$

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(B) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(	ext{products}) - \sum \Delta H_f^o(	ext{reactants})$.For the reaction $CO_{2(g)} + H_{2(g)} 	o CO_{(g)} + H_2O_{(g)}$,the expression is:$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.Given that $\Delta H_f^o(H_{2(g)}) = 0 \ kJ/mol$ (standard state of an element):$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$.$\Delta H^o = -352.3 - (-393.5) = 41.2 \ kJ$.

The $\Delta H^o_f$ for $CO_{2(g)}$,$CO_{(g)}$,and $H_2O_{(g)}$ are $-393.5$,$-110.5$,and $-241.8 \ kJ/mol$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is:

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