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
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \dots \dots(I) \quad \Delta H = -393 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} \dots \dots(II) \quad \Delta H = -287.3 \, kJ \, mol^{-1}$
$2CO_{2(g)} + 3H_{2}O_{(l)}$ $\rightarrow C_{2}H_{5}OH_{(l)} + 3O_{2(g)} \dots \dots(III) \quad \Delta H = 1366.8 \, kJ \, mol^{-1}$
Find the standard enthalpy of formation of $C_{2}H_{5}OH_{(l)}$.
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} \dots \dots(II) \quad \Delta H = -287.3 \, kJ \, mol^{-1}$
$2CO_{2(g)} + 3H_{2}O_{(l)}$ $\rightarrow C_{2}H_{5}OH_{(l)} + 3O_{2(g)} \dots \dots(III) \quad \Delta H = 1366.8 \, kJ \, mol^{-1}$
Find the standard enthalpy of formation of $C_{2}H_{5}OH_{(l)}$.
MediumAIIMS 2019
View SolutionFind the standard enthalpy of formation of $SO_{3(g)}$ from the following reactions:
$S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$; $\Delta H_1 = -2775 \ kJ/mol$
$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$; $\Delta H_2 = -198 \ kJ/mol$
$S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$; $\Delta H_1 = -2775 \ kJ/mol$
$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$; $\Delta H_2 = -198 \ kJ/mol$
Medium
View SolutionIf the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:
Medium
View SolutionGiven: $2Fe + \frac{3}{2} O_2 \to Fe_2O_3$,$\Delta H = -193.4 \ kJ$;
$Mg + \frac{1}{2} O_2 \to MgO$,$\Delta H = -140.2 \ kJ$.
What is the $\Delta H$ of the reaction $3Mg + Fe_2O_3 \to 3MgO + 2Fe$ in $kJ$?
$Mg + \frac{1}{2} O_2 \to MgO$,$\Delta H = -140.2 \ kJ$.
What is the $\Delta H$ of the reaction $3Mg + Fe_2O_3 \to 3MgO + 2Fe$ in $kJ$?
Difficult
View SolutionGiven: $S + O_2 \rightarrow SO_2 : \Delta H_1 = -298.2 \ kJ$,$SO_2 + 1/2 \ O_2 \rightarrow SO_3 : \Delta H_2 = -98.7 \ kJ$,$SO_3 + H_2O \rightarrow H_2SO_4 : \Delta H_3 = -130.2 \ kJ$,and $H_2 + 1/2 \ O_2 \rightarrow H_2O : \Delta H_4 = -287.3 \ kJ$. Calculate the heat of formation of $H_2SO_4$ in $kJ$.
Medium
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