For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.00 l rigid container are [h2] = 0.0500 m, [f2] = 0.0100 m, and [hf] = 0.400 m. if 0.345 mol of f2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished in moles/liter. h2(g) + f2(g) < > 2 hf(g)

Explanation:

Hello,

At the first condition, one computes the equilibrium constant based on the law of mass action:

Now, we compute the final molarity of fluorine after the mentioned addition of 0.345 mol:

Now, one proposes the law of mass action in therms of the affection of the equilibrium concentrations considering there is no change in the equilibrium constant:

Finally, one selects 0.0345M since the other value gives negative concentrations, thus, the new concentrations are:

Best regards.

explanation:

0.07 g/cm³

1.37 g/cm³