Question

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

TiCl3 → Ti(s)+ 2Cl2(g)


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

Compound Amount
TiCl4 4.18g
Ti 1.32g
Cl2 1.08g

Calculate the value of the equilibrium constant Kc for this reaction.

Answer 1

`Kc=1.17x10^(5)`

Step-by-step explanation:

Hello,

In this case, we are talking about the following reaction at equilibrium:

`Ti(s)+ 2Cl_2(g) \rightleftharpoons TiCl_4(l)`

Which has the following associated law of mass action:

`Kc=(1)/([Cl_2]^2)`

As just the gases or the aqueous solutions are said to have concentrations, that is why just chlorine participates in that equation. For that reason, we compute the concentration of chlorine at equilibrium in terms of its molarity:

`[Cl_2]_(eq)=(1.08gCl_2*(1molCl_2)/(70.9gCl_2))/(5.2L)=2.93x10^(-3)M`

Finally, the equilibrium constant results:

`Kc=(1)/((2.93x10^(-3))^2)\\\\Kc=1.17x10^(5)`

Best regards.