Enthalpies 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. Find the value of $\Delta_r H$ for the reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
(D) $\Delta_r H$ for a reaction is defined as the difference between the sum of $\Delta_f H$ values of products and the sum of $\Delta_f H$ values of reactants.
$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$
For the given reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
$\Delta_r H = [\Delta_f H(N_2O) + 3 \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 \Delta_f H(CO)]$
Substituting the given values:
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)] \, kJ \, mol^{-1}$
$\Delta_r H = [81 - 1179] - [9.7 - 330] \, kJ \, mol^{-1}$
$\Delta_r H = -1098 - (-320.3) \, kJ \, mol^{-1}$
$\Delta_r H = -777.7 \, kJ \, mol^{-1}$
$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$
For the given reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
$\Delta_r H = [\Delta_f H(N_2O) + 3 \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 \Delta_f H(CO)]$
Substituting the given values:
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)] \, kJ \, mol^{-1}$
$\Delta_r H = [81 - 1179] - [9.7 - 330] \, kJ \, mol^{-1}$
$\Delta_r H = -1098 - (-320.3) \, kJ \, mol^{-1}$
$\Delta_r H = -777.7 \, kJ \, mol^{-1}$
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Similar Questions
Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
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