Finding partial pressure at equilibrium

[haha0p1]

In the coursebook the question says:
The reaction below was carried out at a pressure of 10×10⁴ Pa and at constant temperature.
N2 + O2 ⇌ 2NO
the partial pressures of Nitrogen and Oxygen are both 4.85×10⁴ pa
 Ccalculate the partial pressure of the nitrogen(ll) oxide, NO(g) at equilibrium.
In this question the partial pressure of nitrogen oxide is given at equilibrium or it is initial partial pressure. Also will the answer be 10.00×10⁴ - 4.85×10⁴ ?

 

[DrClaude]

I would assume that all pressures are at equilibrium.

Also will the answer be 10.00×10⁴ - 4.85×10⁴ ?

No. How is the total pressure decomposed into partial pressures?

 

[haha0p1]

I would assume that all pressures are at equilibrium.


No. How is the total pressure decomposed into partial pressures?

Ohkk. Then will the question be resolved in this way:
Partial pressure of nitrogen and Oxygen at equilibrium: 10×10⁴-4.85×10⁴= 5.15×10⁴
Partial pressure of Nitrogen oxide= ??

 

[Mayhem]

Ohkk. Then will the question be resolved in this way:
Partial pressure of nitrogen and Oxygen at equilibrium: 10×10⁴-4.85×10⁴= 5.15×10⁴
Partial pressure of Nitrogen oxide= ??

Take a look at Dalton's law

 

[haha0p1]

I have understood the question now. 10×10⁴-(4.85×10⁴+4.85×10⁴) = Partial pressure of NO. right ?

 

[DrClaude]

I have understood the question now. 10×10⁴-(4.85×10⁴+4.85×10⁴) = Partial pressure of NO. right ?

Correct.