Question

At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?

Answer 1

Answer:

$[N_2]_{eq}=[H_2]_{eq}=0.09899M$

$[NO]_{eq}=0.00202M$

Explanation:

Hello,

In this case, for the given chemical reaction:

$2NO\rightleftharpoons N_2+O_2$

We know the equilibrium constant and equilibrium expression:

$Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}$

That in terms of the reaction extent $x$ (ICE procedure) we can write:

$2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}$

In such a way, solving for $x$ by using a quadratic equation or solver, we obtain:

$x_1=0.09899M\\x_2=0.1010M$

Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:

$[N_2]_{eq}=[H_2]_{eq}=x=0.09899M$

$[NO]_{eq}=0.2M-2*0.09899M=0.00202M$

Regards.