Question #17305

1 Answer
Sep 16, 2015

Answer:

#K_c = 6.94 xx 10^(-3)#

Explanation:

#N_2O_4(g) -> 2NO_2(g)#

#K_c = ([NO_2(g)]^2)/([N_2O_4(g)])#

We need to find the equilibrium concentrations for #N_2O_4(g) " and " 2NO_2(g)#; therefore we can use #color(blue)("ICE table")# ( I don't know how to format it into a table!):

Remember, the volume is equal to 1.000 L and therefore,
#[N_2O_4(g)]_0 = n/V = (0.0300 mol)/(1.000 L) = 0.0300M#

at equilibrium: #[N_2O_4(g)] = n/V = (0.0236 mol)/(1.000 L) = 0.0236M#

#N_2O_4(g) " "->" " 2NO_2(g)#

Initial: #" " " " " ""0.0300M" " " " " " " " " " ""0M"#

Change: #" " " " " "color(blue)(- x)"M" " " " " " " " " " "color(blue)(+2x)"M"#

Equilibrium: #" " "0.0236M" " "" " " " " " " "+2x"M"#

Therefore, at equilibrium, #[N_2O_4(g)] = (0.0300 - x)M = 0.0236M#

#=> x= 0.0300 - 0.0236 = 0.0064M#

#[NO_2(g)] = +2xM = 2xx0.0064M = 0.0128M#

#K_c = ([NO_2(g)]^2)/([N_2O_4(g)]) = ((0.0128)^2)/0.0236 = 0.00694 =6.94xx10^(-3) #