For the reaction: #N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)#, #K_c# - #8.3x10^-10# at 25 °C. What is the concentration of #N_2# gas at equilibrium when the concentration of #NO_2# is five times the concentration of #O_2# gas?

1 Answer
Mar 6, 2017

Answer:

#N_2=30120481927.710842mol#

Explanation:

#Kc = [NO_2]^2/ ([N2][O2]^2) #

The Kc is measured at the equilibrum

#8.3 * 10^-10 = x^2/([N_2][x]^2)#

Conc of #N_2 "when" NO_2 = O_2# = 1204819277.108434mol

#8.3 * 10^-10 = (5x^2)/(N_2 * x^2)#

Let's solve for #N_2#.

#8.3(10^−10)= (5x)^2 /(x^2N_2)#

Step 1: Multiply both sides by #N_2#

#0N_2=25#

Step 2: Divide both sides by 0.

#(0N_2)/0 = 25/0#

#N_2=30120481927.710842mol#

Answer:
#N_2=30120481927.710842mol#