Respuesta :
Solution :
For the reaction given :
[tex]$\text{2Hg}_{(l)}+\text{O}_2_{(g)} \rightarrow \text{2HgO}_{(s)}$[/tex]
Thus we know that the equilibrium constant [tex]$K_c$[/tex] contains aqueous an dgas species only.
∴ [tex]$K_c=\frac{1}{[O_2]}$[/tex] ............(1)
Now at the equilibrium, an amount of the 13.4 g of oxygen was found in the vessel of 6.9 liters. For determining the concentration of the oxygen gas, we use :
[tex]$[O_2]= \frac{n_{O_2}}{V_{soln}}$[/tex] ................... (2)
Here, [tex]$n_{O_2}$[/tex] = no. of moles of oxygen gas (mol)
[tex]$V_{soln}$[/tex] = volume of solution (L)
Therefore the number of moles of the oxygen gas is calculated by directly using the molecular weight (31.9988 g/mol) as the conversion factor.
∴ [tex]$n_{O_2}= 13.4 \ g \times \frac{\text{1 mol}}{31.9988 \ g}$[/tex]
= 0.418 mol
Now substituting the known values in (2), we can find the equilibrium concentration of the oxygen gas :
[tex]$[O_2] =\frac{0.418 \ \text{mol}}{6.9 \ \text{L}}$[/tex]
= 0.0605 M
Therefore substituting the result in (1), the equilibrium constant for the reaction is :
[tex]$K_c=\frac{1}{0.0605}$[/tex]
= 16.52
