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

`[N_2]_(eq)=[H_2]_(eq)=0.09899M`

`[NO]_(eq)=0.00202M`

Step-by-step 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=([N_2][O_2])/([NO]^2)`

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

`2.4x10^3=(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.