Question

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

2HgO(s)--------Hg(l) + O2(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:
compound amount
HgO 24.0g
Hg 23.6g
O2 22.7g
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

`Kc=0.71=7.1x10^(-1)`

Step-by-step explanation:

Hello,

In this case, since the equilibrium constant is written including gaseous and aqueous species only, for this reaction, it is:

`Kc=[O_2]`

Thus, since the volume is missing, we are going to assume 1 L, but you can change it based on the one you are given, thus, the concentration of oxygen at equilibrium is:

`[O_2]=(22.7g*(1mol)/(32g) )/(1L) =0.709M`

It means, that the equilibrium constant, with two significant figures is:

`Kc=0.71=7.1x10^(-1)`

Best regards.