For the following set of balanced reactions
$N_{2} O_{5} + O_{2} ⟶ 2 {NO}_{2} + O_{3} Δ H = 200 kJ$
${NO}_{2} + O_{2} ⟶ NO + O_{3} Δ H = - 20 k \cdot J$
10 moles of $N_{2} O_{5}$ and 30 moles of $O_{2}$ were taken in a chamber to cause complete conversion of $N_{2} O_{5}$ to ${NO}_{2}$ . ${NO}_{2}$ partially reacts with remaining oxygen such that volume percentage of $O_{3}$ is 50%. Calculate overall enthalpy change $kJ$

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answer is 1700.

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Detailed Solution

$N_{2} O_{5} + O_{2} ⟶ 2 {NO}_{2} + O_{3}$

$10 30 - -$

$- 20 20 10$

${NO}_{2} + O_{2} ⟶ NO + O_{3} 20 20 - - 20 - x 20 - x x x$

$\frac{(10 + x)}{[(20 - x) + (20 - x) + x + (10 + x)]} \times 100 = 50$

$\Rightarrow x = 15$

$Δ H = 10 \times 200 - 15 \times 20 = 1700 kJ$

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