Question

Mercury(II) oxide decomposes to form mercury and oxygen, like this: (s)(l)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of mercury(II) oxide, mercury, and oxygen at equilibrium has the following composition:

Answer 1

The question is incomplete, here is the complete question:

Mercury(II) oxide decomposes to form mercury and oxygen, like this:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

At a certain temperature, a chemist finds that a 9.8 L reaction vessel containing a mixture of mercury(II) oxide, mercury, and oxygen at equilibrium has the following composition:

Compound                   Amount

HgO                              24.0 g

Hg                                 23.6 g

$O_2$               22.7 g

Calculate the value of the equilibrium constant $K_c$ for this reaction. Round your answer to 2 significant digits.

Answer: The value of $K_c$ for the given chemical reaction is $7.2\times 10^{-2}$

Explanation:

To calculate the molarity or concentration, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For oxygen gas:

Mass of oxygen gas = 22.7 g

Molar mass of oxygen gas = 32 g/mol

Volume of the solution = 9.8 L

Putting values in above equation, we get:

$\text{Molarity of }O_2=\frac{22.7}{32\times 9.8}=0.0724M$

For the given chemical equation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

The expression of $K_c$ for above equation follows:

$K_c=[O_2]$

The concentration of pure solids and liquids are taken as 1

So,

$K_c=0.0724=7.2\times 10^{-2}$

Hence, the value of $K_c$ for the given chemical reaction is $7.2\times 10^{-2}$