Respuesta :
Answer: The equilibrium partial pressure of NO is 0.0034 atm
Explanation:
For the given chemical equation:
[tex]NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)[/tex]
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{(p_{NOBr})^2}{p_{NO}\times p_{Br_2}}[/tex]
We are given:
Equilibrium partial pressure of [tex]Br_2[/tex] = 0.0159 atm
Equilibrium partial pressure of NOBr = 0.0768 atm
[tex]K_p=109[/tex]
Putting values in above equation, we get:
[tex]109=\frac{(0.0768)^2}{p_{NO}\times 0.0159}\\\\p_{NO}=0.0034atm[/tex]
Hence, the equilibrium partial pressure of NO is 0.0034 atm
The equilibrium partial pressure of NO is 0.0034 atm
Calculation of the partial pressure of NO at equilibrium:
Since The following reaction has Kp = 109 at 25°C. 2 NO(g) + Br2(g) equilibrium reaction arrow 2 NOBr(g). There is equilibrium partial pressure of Br2 is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm
Now the partial pressure is
100 = 0.0768^2 /(Partial pressure * 0.0159)
Partial pressure = 0.0034 atm
Learn more about pressure here; https://brainly.com/question/18999457
