Question

Consider the equilibrium system: N2O4 (g) = 2 NO2 (g) for which the Kp = 0.1134 at 25 C and deltaH rx is 58.03 kJ/mol. Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5 L contains. What will be the equilibrium value of [N204]?

A) 0.358 M
B) 0.042 M
C) 0.0822 M
D) 0.928 M
E) 0.379 M

Answer 1

Answer: The equilibrium value of
`N_2O_4` is 0.379 M

Step-by-step explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

Using ideal gas equation :
`PV=nRT`

P = pressure of gas

V = volume of gas

n = no of moles

R = gas constant

T = Temperature

pressure of
`N_2O_4` =
`(1* 0.0821Latm/Kmol* 298)/(5L)=5atm`

pressure of
`NO_2` =
`(2* 0.0821Latm/Kmol* 298)/(5L)=10atm`

`N_2O_4(g)\rightleftharpoons 2NO_2(g)`

at t= 0 5 atm 10 atm

at eqm (5-x) atm (10+2x) atm

`K_p=([p_NO_2]^2)/([p_N_2O_4])`

`0.1134=((10+2x)^2)/((5-x))`

`x=-4.48`

pressure of
`N_2O_4` at equilibrium = (5-(-4.48))= 9.48 atm

pressure of
`N_2O_4` =
`(n* 0.0821Latm/Kmol* 298)/(V)`

9.48 =
`{M* 0.0821Latm/Kmol* 298}`

`M=0.379`

Thus the equilibrium value of
`N_2O_4` is 0.379 M