Respuesta :
Explanation:
Since, the given reaction is as follows.
[tex]2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)[/tex]
Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of [tex]O_{2}[/tex] at equilibrium will be calculated as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
[tex]2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}[/tex]
x = 18.1 atm
Thus, we can conclude that partial pressure of [tex]O_{2}[/tex] at equilibrium is 18.1 atm.
Answer:
partial pressure O2 = 17.867 atm
Explanation:
Step 1: Data given
Temperature = 200 °C
Kp = 2.40 *10^3
Pressure NO = 36.1 atm
Step 2: The balanced equation
2 NO ⇔ N2 + O2
Step 3: The initial pressure
pNO = 36.1 atm
pN2 = 0 atm
pO2 = 0 atm
Step 4: the pressure at the equilibrium
For 2 moles NO we'll have 1 mol N2 and 1 mol O2
pNO = 36.1 - 2X atm
pN2 = X atm
pO2 = X atm
Step 5: Calculate partial pressures
Kp = pN2 * pO2 / (pNO)²
2.40*10³ = x²/(36.1 - 2x)²
48.99 = x/(36.1-2x)
x = 1768.5 -97.98x
x = 17.867
x = partial pressure O2 = 17.867 atm
