N2+O2→2NO ΔH=+Q1N2+2O2→N2O4 ΔH=−Q2NO+1/2O2→NO2 ΔH=−Q32NO2→N2O4 ΔH=−Q4then

−Q2=Q1−2Q3−Q4 N2+O2→2NO Q12(NO+12O2)→2NO2 −2Q3 2NO2→N2O4 −Q4________________________ N2+2O2→N2O4=−Q2

N2+O2→2NO ΔH=+Q1N2+2O2→N2O4 ΔH=−Q2NO+1/2O2→NO2 ΔH=−Q32NO2→N2O4 ΔH=−Q4then

−Q2=Q1−2Q3−Q4 N2+O2→2NO Q12(NO+12O2)→2NO2 −2Q3 2NO2→N2O4 −Q4________________________ N2+2O2→N2O4=−Q2

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$N_{2} + O_{2} \to 2 NO ΔH = + Q_{1}$
$N_{2} + 2 O_{2} \to N_{2} O_{4} ΔH = - Q_{2}$
$NO + 1 / 2 O_{2} \to {NO}_{2} ΔH = - Q_{3}$
$2 {NO}_{2} \to N_{2} O_{4} ΔH = - Q_{4}$ then

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$Q_{2} = Q_{1} (- Q_{3}) + (- Q_{4})$

$- Q_{2} = Q_{1} - 2 Q_{3} - Q_{4}$

$Q_{1} + Q_{2} = (Q_{3} + Q_{4})$

$Q_{1} - Q_{4} = Q_{2} - Q_{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} - Q_{2} = Q_{1} - 2 Q_{3} - Q_{4} \\ N_{2} + O_{2} \to 2 NO Q_{1} \\ 2 (NO + \frac{1}{2} O_{2}) \to 2 {NO}_{2} - 2 Q_{3} \\ 2 {NO}_{2} \to N_{2} O_{4} - Q_{4} \\ ________________________ \\ N_{2} + 2 O_{2} \to N_{2} O_{4} = - Q_{2} \end{array}$