Answer:
The partial pressure of NO₂ is 0.958 bar and the partial pressure of N₂O₄ is 0.211 bar
Explanation:
The reaction is:
2NO₂ = N₂O₄
The equilibrium constant is:
[tex]K_{p} =\frac{P_{N_{2}O_{4} } }{P_{NO_{2} }^{2} } =\frac{0.58}{1.6^{2} } =0.227[/tex]
When the volume is double, the pressure will be halved, and will be:
PNO₂ = 0.8 bar
PN₂O₄ = 0.29 bar
The ICE table for this exercise is:
2NO₂ = N₂O₄
I......... 0.8..........0.29
C....... -2x.......... +x
E-----0.8-2x....... 0.29+x
The equilibrium constant is:
[tex]K_{p} =\frac{P_{N_{2}O_{4} } }{P_{NO_{2} }^{2} } \\0.23=\frac{0.29+x}{(0.8-2x)^{2} } \\x=-0.079[/tex]
The partial pressure of NO₂ is:
[tex]p_{NO_{2} } =0.8-2x=0.8-(2*(-0.079))=0.958bar[/tex]
The partial pressure of N₂O₄ is:
[tex]p_{N_{2}O_{4} } 0.29+x=0.29-0.079=0.211bar[/tex]
