At high temperatures, both NO₂(g) and N₂O₃(g) decompose into N₂(g) and O₂(g).

At -23∘C, a mixture of NO₂(g) and N₂O₃(g) has a total pressure of 100 kPa. Upon heating to 727∘C, these two gases decompose completely, and the total pressure of the N₂ + O₂ products is 840 kPa.

Determine the partial pressures, in kPa, of NO₂ and N₂O₃ in the original mixture at -23∘C.

A. 0 kPa NO₂, 100 kPa N₂O₃
B. 20 kPa NO₂, 80 kPa N₂O₃
C. 40 kPa NO₂, 60 kPa N₂O₃
D. 60 kPa NO₂, 40 kPa N₂O₃
E. 80 kPa NO₂, 20 kPa N₂O₃
F. 100 kPa NO₂, 0 kPa N₂O₃

Question

23-11-2023
Chemistry

High School

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At high temperatures, both NO₂(g) and N₂O₃(g) decompose into N₂(g) and O₂(g).

At -23∘C, a mixture of NO₂(g) and N₂O₃(g) has a total pressure of 100 kPa. Upon heating to 727∘C, these two gases decompose completely, and the total pressure of the N₂ + O₂ products is 840 kPa.

Determine the partial pressures, in kPa, of NO₂ and N₂O₃ in the original mixture at -23∘C.

A. 0 kPa NO₂, 100 kPa N₂O₃
B. 20 kPa NO₂, 80 kPa N₂O₃
C. 40 kPa NO₂, 60 kPa N₂O₃
D. 60 kPa NO₂, 40 kPa N₂O₃
E. 80 kPa NO₂, 20 kPa N₂O₃
F. 100 kPa NO₂, 0 kPa N₂O₃

Answer :

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CallumPhillips CallumPhillips · Answer 1 · 2024-08-20T01:05:58-04:00

Final answer:

The partial pressures of NO₂ and N₂O₃ in the original mixture at -23∘C are 40 kPa and 60 kPa respectively.

Explanation:

The partial pressures of NO₂ and N₂O₃ in the original mixture at -23∘C can be determined by using the given information and using the ideal gas law. The initial total pressure is 100 kPa and upon heating, the total pressure of N₂ + O₂ products is 840 kPa. By calculating the change in pressure for each gas and using the equation for the decomposition reaction, we can determine the partial pressures of NO₂ and N₂O₃. The correct answer is: O 40kPa NO₂, 60kPa N₂O₃.