Q: At 92.2 C, the Kp for the following reaction is 0.2000 atm^-1. If you were to place exactly 0.200 atm of N2O4 (g) into a 1.00 L vessel, what would the partial pressure of NO2 (g) be once equilibrium was established?
2NO2 (g) --><-- N2O4 (g)
I'm having trouble understanding the part about partial pressure. The solution says that half of 0.200 atm of N2O4 shifts over then the partial pressure of NO2 is 0.200 atm.
I understand the calculation part but I would like to understand the process TBR used without calculating. I'm not understanding how the partial pressure of NO2 is 0.200 atm. Can someone please elaborate on this part? Thanks!
2NO2 (g) --><-- N2O4 (g)
I'm having trouble understanding the part about partial pressure. The solution says that half of 0.200 atm of N2O4 shifts over then the partial pressure of NO2 is 0.200 atm.
I understand the calculation part but I would like to understand the process TBR used without calculating. I'm not understanding how the partial pressure of NO2 is 0.200 atm. Can someone please elaborate on this part? Thanks!