Calculate the heat of dissociation N2O4 into NO2 from the data :N2+O2→2NO; ΔH=43.10kcal . . . (1) N2+2O2→N2O4; ΔH=−1.870kcal . . . .(2)2NO2→2NO+O2;ΔH=+26.10kcal . . . .(3)

We have to find ΔH for N2O4→2NO2; ΔH=?Add Eqs. (2) and (3)N2+2O2→N2O4; ΔH=−1.87kcal2NO2→2NO+O2; ΔH=26.10kcal∴N2+2O2+2NO2→N2O4+2NO+O2; ΔH=+24.23kcal or N2+O2+2NO2→N2O4+2NO;ΔH=24.23kcalSubtract Eq. (l) from Eq. (4), we get.N2+O2⟶2NO;ΔH=+43.10kcalN2+O2⟶2NO2⟶N2O2+2NO;ΔH=24.23kcal--------------------2NO2⟶N2O4;ΔH=−18.87kcal or N2O4→2NO2; ΔH=+18.87kcal i.e., Heat of dissociation of N2O4=+18.87kcal

Calculate the heat of dissociation N2O4 into NO2 from the data :N2+O2→2NO; ΔH=43.10kcal . . . (1) N2+2O2→N2O4; ΔH=−1.870kcal . . . .(2)2NO2→2NO+O2;ΔH=+26.10kcal . . . .(3)

We have to find ΔH for N2O4→2NO2; ΔH=?Add Eqs. (2) and (3)N2+2O2→N2O4; ΔH=−1.87kcal2NO2→2NO+O2; ΔH=26.10kcal∴N2+2O2+2NO2→N2O4+2NO+O2; ΔH=+24.23kcal or N2+O2+2NO2→N2O4+2NO;ΔH=24.23kcalSubtract Eq. (l) from Eq. (4), we get.N2+O2⟶2NO;ΔH=+43.10kcalN2+O2⟶2NO2⟶N2O2+2NO;ΔH=24.23kcal--------------------2NO2⟶N2O4;ΔH=−18.87kcal or N2O4→2NO2; ΔH=+18.87kcal i.e., Heat of dissociation of N2O4=+18.87kcal

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Calculate the heat of dissociation N₂O₄ into NO₂ from the data :
$N_{2} + O_{2} \to 2 NO; Δ H = 43.10 kcal . . . (1) N_{2} + 2 O_{2} \to N_{2} O_{4}; Δ H = - 1.870 kcal . . . . (2)$
$2 {NO}_{2} \to 2 NO + O_{2}; ΔH = + 26.10 kcal . . . . (3)$

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+24.23

-18.87

+18.87

-24.23

answer is A.

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Detailed Solution

$We have to find Δ H for$

$N_{2} O_{4} \to 2 {NO}_{2}; Δ H = ?$

Add Eqs. (2) and (3)

$\begin{array}{l} N_{2} + 2 O_{2} \to N_{2} O_{4}; Δ H = - 1.87 kcal \\ 2 {NO}_{2} \to 2 NO + O_{2}; Δ H = 26.10 kcal \\ ∴ N_{2} + 2 O_{2} + 2 {NO}_{2} \to N_{2} O_{4} + 2 NO + O_{2}; Δ H = + 24.23 kcal \\ or N_{2} + O_{2} + 2 {NO}_{2} \to N_{2} O_{4} + 2 NO; Δ H = 24.23 kcal \end{array}$

Subtract Eq. (l) from Eq. (4), we get.

$\begin{array}{l} N_{2} + O_{2} ⟶ 2 NO; Δ H = + 43.10 kcal \\ N_{2} + O_{2} ⟶ 2 {NO}_{2} ⟶ N_{2} O_{2} + 2 NO; Δ H = 24.23 kcal \\ - - - - - - - - - - - - - - - - - - - - \\ 2 {NO}_{2} ⟶ N_{2} O_{4}; Δ H = - 18.87 kcal \end{array}$