Chemistry, 11.12.2019 19:31 m3lissa

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.

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Answer from: mp515

The value of equilibrium constant for the reaction is $8.5\times 10^{7}$

Explanation:

The given chemical equation follows:

$Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)$

The expression of $K_{eq}$ for the above reaction follows:

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

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of chlorine gas = 1.67 g

Molar mass of chlorine gas = 71 g/mol

Putting values in above equation, we get:

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

Volume of vessel = 7.0 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

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

Putting the value in equilibrium constant expression, we get:

$K_{eq}=\frac{1}{(3.43\times 10^{-3})^2}\\\\K_{eq}=8.5\times 10^{7}$

Hence, the value of equilibrium constant for the reaction is $8.5\times 10^{7}$

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Titanium and chlorine react to form titanium(iv) chloride, like this: ti(s) + 2 cl 2(g)-ticl 4( at...

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