Question

Problem PageQuestion Methanol and oxygen react to form carbon dioxide and water, like this: (l)(g)(g)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of methanol, oxygen, carbon dioxide, and water at equilibrium has the following composition: compoundamount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits. Clears your work. Undoes your last action. Provides information about entering answers.

Answer 1

`Kc=3.8x10^(-7)`

Step-by-step explanation:

Hello.

In this case, for the chemical reaction:

`2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)`

We write the equilibrium expression including the gaseous or aqueous species only, that is why the methanol is not included due to heterogeneous equilibrium:

`Kc=([CO_2]^2[H_2O]^4)/([O_2]^3)`

Whereas each gaseous species is powered to its stoichiometric coefficient (number before the species). In such a way, considering the equilibrium masses of carbon dioxide (44 g/mol), water (18 g/mol) and oxygen (32 g/mol) to be 1.56 g, 2.28 g and 3.33 g respectively, we compute the moles as we need molar concentrations in the equilibrium constant calculation:

`n_(CO_2)=1.56g/(44g/mol)=0.0355mol\\\\n_(H_2O)=2.28g/(18g/mol)=0.127mol\\\\n_(O_2)=3.33 g/(32g/mol)=0.104mol`

Thus, into the 9.3.L vessel, the equilibrium concentrations are:

`[CO_2]=(0.0355mol)/(9.3L)=0.00382M`

`[H_2O]=(0.127mol)/(9.3L)=0.0137M`

`[O_2]=(0.104mol)/(9.3L)=0.0112M`

Therefore, the equilibrium constant shown with two significant figures is:

`Kc=((0.00382M)^2(0.0138M)^4)/((0.0112M)^3) \\\\Kc=3.8x10^(-7)`

Best regards.