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

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 $N_2O_4$ = $\frac{1\times 0.0821Latm/Kmol\times 298}{5L}=5atm$

pressure of $NO_2$ = $\frac{2\times 0.0821Latm/Kmol\times 298}{5L}=10atm$

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

at t= 0    5 atm                                10 atm

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

$K_p=\frac{[p_NO_2]^2}{[p_N_2O_4]}$

$0.1134=\frac{(10+2x)^2}{(5-x)}$

$x=-4.48$

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

pressure of $N_2O_4$ = $\frac{n\times 0.0821Latm/Kmol\times 298}{V}$

9.48 = ${M\times 0.0821Latm/Kmol\times 298}$

$M=0.379$

Thus the equilibrium value of $N_2O_4$ is 0.379 M