Question

The initial concentration for the compounds involved in the reaction displayed were determined to be [NO2(g)] = 1.243 mol/L, [N2O4(g)] = 0.2578 mol/L. Calculate the value of the equilibrium constant (Kc) at 500.0 K if the equilibrium concentration of NO2(g) was 0.7629 mol/L.


2NO2(g) = N2O4(g)

Answer 1

..............2NO2 ==> N2O4
initial......1.243...0.2578
change.......-2x........+x
equil.......0.7629......0.2578+x

So 2x = 1.243-0.7829 = ?
and x = ?
(NO2) = 0.7629
(N2O4) = 0.2578+x
Substitute in Kc expression and solve for Kc. Post your work if you get stuck.