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Titanium and chlorine react to form titanium(IV) chloride, like this:
Ti(s) + 2 Cl_2(g) -------> TiCl_4(l)
At a certain temperature, a chemist finds that a 7.0 L reaction vessel containing a mixture of titanium, chlorine, and titanium(IV) chloride at equilibrium has the following composition:
compound amount
Ti 1.67 g
Cl_2 2.93 g
TiCI_4 2.02 g
1. Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Respuesta :

Olajidey

Answer:

the value of equilibrium constant for the reaction is 8.5 * 10⁷

Explanation:

Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)

equilibrium constant Kc = [tex]\frac{1}{[Cl_2]^2}[/tex]

Given that,

We are given:

Equilibrium amount of titanium = 2.93 g

Equilibrium amount of titanium tetrachloride = 2.02 g

Equilibrium amount of chlorine gas = 1.67 g

We calculate the No of mole = mass / molar mass

mass of chlorine gas = 1.67 g

Molar mass of chlorine gas = 71 g/mol

mole of chlorine = 1.67 / 71

= 7.0L

Concentration of chlorine is = no of mole / volume

= 0.024 / 7

= 3.43 * 10⁻³M

equilibrium constant Kc  = [tex]\frac{1}{[Cl_2]^2}[/tex]

= [tex]\frac{1}{[3.43 * 10^-^3]^2}[/tex]

= 8.5 * 10⁷