The Delta H_f for CO_{2(g)},CO_{(g)} and H_2O | vedclass

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(B) The standard enthalpy change of reaction $\Delta_r H^{\circ}$ is calculated using the formula:$\Delta_r H^{\circ} = \sum \Delta_f H^{\circ} (	ext

Vedclass · Accepted Answer

$41.2 \ kJ \ mol^{-1}$

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

The $\Delta H_f$ 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 for the reaction is:
$CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$

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The $\Delta H_f$ 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 for the reaction is:$CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$