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* 10^(-2)`

Step-by-step 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}* \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=(22.7)/(32* 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* 10^(-2)`

Hence, the value of
`K_c` for the given chemical reaction is
`7.2* 10^(-2)`