#color(white)(xxxxxxxxxxxxxxx)N_2O_4 color(white)(xxxx)-> color(white)(xxxx)2NO_2#
#"Before equilibrium" color(white)(xxxx)x color(white)(xxxxxxxxxxxxxx)0#
#"Change" color(white)(xxxxxxxxxx)x-3.80 color(white)(xxxxxxx)2(x-3.80)#
#"At equilibrium" color(white)(xxxxxxx)3.80 color(white)(xxxxxxxxx)2(x-3.80)#
Recall;
#N_2O_4 rightleftharpoons 2NO_2#
#3.80 -> 2(x-3.80)#
Where #x# is the initial concentration of #N_2O_4# before equilibrium..
#K_c = [NO_2]^2/[N_2O_4]#
Where;
#K_c = "Equilibrium constant"#
#NO_2 = "Product"#
#N_2O_4 = "Reactant"#
#K_c = 0.523#
#NO_2 = 2(x-3.80)#
#N_2O_4 = 3.80#
#0.523 = [2(x-3.80)]^2/[3.80]#
#0.523 = [2x-7.6]^2/[3.80]#
#0.523 xx 3.80 = 4x^2-30.4x + 57.76#
#1.987 = 4x^2 - 30.4x + 57.76#
#4x^2-30.4x + 57.76 - 1.987 = 0#
#4x^2-30.4x + 55.77 = 0 -> "Quadratic Equation"#
Using quadratic formula to solve;
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#
Where;
#a = 4#
#b = -30.4#
#c = 55.77#
#x = (-(-30.4) +- sqrt((-30.4)^2 - 4(4)(55.77)))/(2(4)#
#x = (-(-30.4) +- sqrt(924.16 - 892.32))/8#
#x = (30.4 +- sqrt31.84)/8#
#x = (30.4 +- 5.643)/8#
#x = 4.505 or 3.095#
Recall: #x# is the initial concentration of #N_2O_4#
Therefore we take #x# as #4.505M# and ignore #3.095# because the initial concentration of #N_2O_4# must be greater than #3.80M#
Also recall;
#N_2O_4 rightleftharpoons 2NO_2#
#3.80 -> 2(x-3.80)#
Therefore, Concentration of Nitrogen dioxide #(NO_2)# at equilibrium is;
#rArr 2(x-3.80) = 2(4.505 - 3.80) = 2(0.705) = 1.41M#
