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:

$8.6\times 10^{-6}$ the value of the equilibrium constant for this reaction.

Explanation:

$Concentration = \frac{Moles}{Volume (L)}$

We have : Volume of vessel = 5.2 L

Concentration of Titanium(IV) chloride at equilibrium =:

$[TiCl_4]=\frac{4.18 g}{190 g/mol\times 5.2 L}=0.004231 mol/L$

Concentration of Titanium at equilibrium =:

$[Ti]=\frac{1.32 g}{48 g/mol\times 5.2 L}=0.005288 mol/L$

Concentration of chloride at equilibrium =:

$[Cl_2]=\frac{1.08 g}{71 g/mol\times 5.2 L}=0.002925 mol/L$

$TiCl_4\rightleftharpoons Ti+2Cl_2$

The equilibrium expression will be given as:

$K_c=[Cl_2]^2$

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

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

$K_c=8.6\times 10^{-6}$

$8.6\times 10^{-6}$ the value of the equilibrium constant for this reaction.