Question

A reaction begins with 1.00 mol of HCl in a 4.00 liter container at a certain temperature.

2HCl(g) ⇋ H2(g) + Cl2(g) Kc = 4.00 What is the equilibrium concentration of HCl?

Answer 1

`[HCl]_(eq)=0.05M`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action at equilibrium results:

`Kc=([H_2]_(eq)[Cl_2]_(eq))/([HCl]^2_(eq))`

Next, in terms of the change
`x` due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:

`4.00=((x)(x))/((0.25-2x)^2)`

Thus, solving for
`x` via quadratic equation or solver, the following results are obtained:

`x_1=0.1M\\x_2=0.17M`

Clearly, the solution is
`x=0.1M` as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:

`[HCl]_(eq)=0.25M-2*0.1M`

`[HCl]_(eq)=0.05M`

Best regards.