Question

2. Hydrogen chloride (HCI (g)) decomposes in hydrogen gas and chlorine gas. The equilibrium constant, K, is 3.2 x 10^-34 at 25°C. Calculate the equilibrium concentrations of all entities if 2.00 mol HCl(9) is initially placed in a closed 1.00-L vessel. All the entities are in gaseous form. HCl H2 + Cl2

Answer 1

The chemical equation:

`2\text{HCl}\to H_2+Cl_2.`

And the concentrations of hydrogen chloride:

`M_(HCl)=\frac{2\text{ mol HCl}}{1\text{ L}}=2M_(HCl).`

We do an ICE chart:

`K=(x^2)/((2-2x))=3.2\cdot10^(-34)`

Doing the calculations in a software, we obtain that x is:

`x=\pm2.5982\cdot10^(-17).`

Using the positive result, we obtain that:

`\lbrack HCl\rbrack^2=2-2x=2-2(2.5982\cdot10^(-17))\text{.}`
`\lbrack HCl\rbrack=\sqrt[]{2}\approx1.41\text{ M}`
`\lbrack H_2\rbrack=\lbrack Cl_2\rbrack=x=2.5982\cdot10^(-17).`

These results is tellinig us that the direction of the equilibrium in the reaction goes like this:

`K=(\lbrack HCl\rbrack^2)/(\lbrack H_2\rbrack\lbrack Cl_2\rbrack)=\frac{2}{(\text{2}.5982\cdot10^(-17))^2}\approx2.963\cdot10^(33).`

When K > 0, the direction goes to the right: to the products.