Question

Titanium(IV) chloride decomposes to form titanium and chlorine, like this: (l)(s)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of titanium(IV) chloride, titanium, and chlorine at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

The question is incomplete ,the complete question is:

Titanium(IV) chloride decomposes to form titanium and chlorine, like this:

`TiCl_4\rightleftharpoons Ti+2Cl_2`

At a certain temperature, a chemist finds that a 5.2 L reaction vessel containing a mixture of titanium(IV) chloride, titanium, and chlorine at equilibrium has the following composition:

Compound amount

`TiCl_4` 4.18 g

Ti 1.32 g

`Cl_2` 1.08g

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

Answer:

`8.6* 10^(-6)` the value of the equilibrium constant for this reaction.

Step-by-step explanation:

`Concentration = (Moles)/(Volume (L))`

We have : Volume of vessel = 5.2 L

Concentration of Titanium(IV) chloride at equilibrium =:

`[TiCl_4]=(4.18 g)/(190 g/mol* 5.2 L)=0.004231 mol/L`

Concentration of Titanium at equilibrium =:

`[Ti]=(1.32 g)/(48 g/mol* 5.2 L)=0.005288 mol/L`

Concentration of chloride at equilibrium =:

`[Cl_2]=(1.08 g)/(71 g/mol* 5.2 L)=0.002925 mol/L`

`TiCl_4\rightleftharpoons Ti+2Cl_2`

The equilibrium expression will be given as:

`K_c=[Cl_2]^2`

The concentration of the solids and liquid is taken as unity.

`=(0.002925 mol/L)^2}`

`K_c=8.6* 10^(-6)`

`8.6* 10^(-6)` the value of the equilibrium constant for this reaction.