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:

$K=\frac{x^2}{(2-2x)}=3.2\cdot10^{-34}$

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:

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

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