Question

13.8 g of N2O4 was placed to a 1L reaction vessel at 450 K and allowed to attain Equilibrium. The total pressure at equilibrium was found to be 9.15 bar. Calculate kc,kp and partial pressure of each gas at equilibrium. N2O4(g) = 2NO2(g)​

Answer 1

We know that, PV=nRT

molar mass of N2O4 =92g

number of moles = 13.8÷92

=0.15 of the gas

R=0.083 bar L mol

T=400K

So, PV=nRT p×1L

=0.15×0.083×400=4.98 bar

If N2O4 2NO2 initial pressure : 4.98 bar O at eqn.

: (4.98−λ) bar 2x bar

Hence

P Total =P N2O4 + P NO2

9.15=(4.98−x)+2x9.5=4.98+xx

=9.15−4.98=4.17 bar

Kp = (8.34)×2 ÷0.81

⇒Kp =85.87

Now,

Kp=Kc (RT)Δn

⇒85.87=Kc (0.083×400)

⇒Kc = 85.87 ÷ 0.083×400

⇒Kc =2.586

• I Hope It Will Help You •

If right mark it brainliest