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Titanium and chlorine react to form titanium(IV) chloride, like this: Ti(s) + 2 Cl 2(g)-TiCl 4( 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 1.67 g Cl 2.93 g TiCI 2.02 g Ti Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.

Respuesta :

Answer: The value of equilibrium constant for the reaction is [tex]8.5\times 10^{7}[/tex]

Explanation:

The given chemical equation follows:

[tex]Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)[/tex]

The expression of [tex]K_{eq}[/tex] for the above reaction follows:

[tex]K_{eq}=\frac{1}{[Cl_2]^2}[/tex]

Concentration of pure solids and liquids are taken as 1 in the equilibrium constant expression.

Thus, titanium and titanium tetrachloride does not appear in the expression.

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

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of chlorine gas = 1.67 g

Molar mass of chlorine gas = 71 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of chlorine gas}=\frac{1.67g}{71g/mol}=0.024mol[/tex]

Volume of vessel = 7.0 L

Concentration of a substance is calculated by:

[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]

So, concentration of chlorine gas = [tex]\frac{0.024}{7.00}=3.23\times 10^{-3}M[/tex]

Putting the value in equilibrium constant expression, we get:

[tex]K_{eq}=\frac{1}{(3.43\times 10^{-3})^2}\\\\K_{eq}=8.5\times 10^{7}[/tex]

Hence, the value of equilibrium constant for the reaction is [tex]8.5\times 10^{7}[/tex]

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