Enthalpy of formation of CO_{(g)},CO_{2(g)},N | vedclass

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

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$-778 \ kJ \ mol^{-1}$

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(D) The enthalpy of reaction $\Delta_r H$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(	ext{products}) - \sum \Delta_f H^{\circ}(	ext{reactants})$.Given values are: $\Delta_f H(CO) = -110 \ kJ \ mol^{-1}$,$\Delta_f H(CO_2) = -393 \ kJ \ mol^{-1}$,$\Delta_f H(N_2O) = 81 \ kJ \ mol^{-1}$,and $\Delta_f H(N_2O_4) = 9.7 \ kJ \ mol^{-1}$.For the reaction $N_2O_{4(g)} + 3 CO_{(g)} \longrightarrow N_2O_{(g)} + 3 CO_{2(g)}$:$\Delta_r H = [\Delta_f H(N_2O) + 3 	imes \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 	imes \Delta_f H(CO)]$.Substituting the values:$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)]$.$\Delta_r H = [81 - 1179] - [9.7 - 330]$.$\Delta_r H = (-1098) - (-320.3)$.$\Delta_r H = -1098 + 320.3 = -777.7 \ kJ \ mol^{-1}$.Rounding off,we get $\Delta_r H \approx -778 \ kJ \ mol^{-1}$.

Enthalpy of formation of $CO_{(g)}$,$CO_{2(g)}$,$N_2O_{(g)}$ and $N_2O_{4(g)}$ are $-110$,$-393$,$81$,and $9.7 \ kJ \ mol^{-1}$ respectively. Calculate $\Delta_r H$ for the following reaction: $N_2O_{4(g)} + 3 CO_{(g)} \longrightarrow N_2O_{(g)} + 3 CO_{2(g)}$

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