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

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

Vedclass · Accepted Answer

$41.2$

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(B) The standard enthalpy of reaction $\Delta_r H^o$ is calculated using the formula: $\Delta_r H^o = \sum \Delta H_f^o(	ext{products}) - \sum \Delta H_f^o(	ext{reactants})$.For the reaction $CO_{2(g)} + H_{2(g)} 	o CO_{(g)} + H_2O_{(g)}$:$\Delta_r H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.Given $\Delta H_f^o(H_{2(g)}) = 0 \ kJ \ mol^{-1}$ (standard state of an element).$\Delta_r H^o = [(-110.5) + (-241.8)] - [(-393.5) + 0]$.$\Delta_r H^o = [-352.3] - [-393.5]$.$\Delta_r H^o = -352.3 + 393.5 = +41.2 \ kJ$.

The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. 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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