Question

C(g), Kp 0.0444 at 395°C. Assume that initially n moles of A(g) and no 8. For a reaction A(g)B(g) moles of B(g) and C(g) are present. What are the partial pressures of A(g) and B(g) at 1.0 bar total pressure? O PA 0.75 bar, PB = 0.125 bar O PA 0.846 bar, PB 0.077 bar 0.8 bar, PB O PA 0.1 bar O PA 0.658 bar, Pp = 0.171 bar

Answer 1

8. The given neaction is at 395€ Ko=0:0444 19) <Beg) + COO) Initial rmole o o & is the equilibrium moles advancement on the en el sentimentos na te entend of the grasectes Total numbers of moles in the system at equilibrium :(n-F.) + fet se ontfe . | Holefractions of the gaseous :"A noe s&B- Je ste se u species, & nthe eth" unteve he is the fraction of dissociation at equilibracion, de a se Now, the equilibrium Constant (kp) can be comitten as using the partial pressures of the individual gaseous Species. Koo PB(g) * Peco) 20.0444 at 395c Pacg)

- pantial presswees (P.) a mole fraction (X;)x total pessetalp? - PA TAP 3 Pg=&BP ; Peadap Now, Substituting there partial pressure teroms into i equation we get Ķpa dopo dop Apa "Blxtol Ps total pressure of daß the System 2 1.0 baro ... x B x de P Cl-Le) Citae) Now her skp > 0.6144

12 0.0444 om, 0.0444 of the root DAR on, I ve 2 alt 1 : 1:0444 0.0444 0.0444 on, le 20.0444 ... 1:0444.. ... des. 0.0444 0.206185 (onkey tve value 1104447 taken as m. parotial presscore of AG); Te cannot be PA ! -de pa (1.0.206185 negative) Gxlbar (!+0.206185) *** 30.658120 bano -0.658 baro and, the partial presscore of Begin Per Te P, 0.206185 xabaro 2-0.170939 baro it le 170 206185 20.17lbar · Last Option : PA = 0.658 bars , Pres > 0.171 bars Ae