Question

titanium(iv) chloride decomposes to form titanium and chlorine, like this: ticl₄(l)→ti(s)+2cl₂(g) at a certain temperature, a chemist finds that a 8.6 l reaction vessel containing a mixture of titanium(iv) chloride, titanium, and chlorine at equilibrium has the following composition: calculate the value of the equilibrium constant kₑ for this reaction. round your answer to 2 significant digits. kₑ =

Explanation:

Step1: Calculate the molarity of \(TiCl_4\)

The molar mass of \(TiCl_4\) is \(M_{TiCl_4}=47.87+(4\times35.45)=189.67\space g/mol\).
The number of moles of \(TiCl_4\), \(n_{TiCl_4}=\frac{m}{M}=\frac{3.13\space g}{189.67\space g/mol}\approx0.0165\space mol\).
The molarity of \(TiCl_4\), \(C_{TiCl_4}=\frac{n}{V}=\frac{0.0165\space mol}{8.6\space L}\approx1.92\times10^{-3}\space M\).

Step2: Calculate the molarity of \(Cl_2\)

The molar mass of \(Cl_2\) is \(M_{Cl_2}=2\times35.45 = 70.9\space g/mol\).
The number of moles of \(Cl_2\), \(n_{Cl_2}=\frac{m}{M}=\frac{1.01\space g}{70.9\space g/mol}\approx0.0142\space mol\).
The molarity of \(Cl_2\), \(C_{Cl_2}=\frac{n}{V}=\frac{0.0142\space mol}{8.6\space L}\approx1.65\times10^{-3}\space M\).

Step3: Write the equilibrium - constant expression

For the reaction \(TiCl_4(l)
ightleftharpoons Ti(s)+2Cl_2(g)\), the equilibrium - constant expression \(K_c=\frac{[Cl_2]^2}{[TiCl_4]}\) (since \(Ti\) is a solid and its concentration is not included in the \(K_c\) expression).

Step4: Substitute the values into the \(K_c\) expression

Substitute \(C_{Cl_2}\approx1.65\times 10^{-3}\space M\) and \(C_{TiCl_4}\approx1.92\times 10^{-3}\space M\) into the \(K_c\) expression.
\(K_c=\frac{(1.65\times 10^{-3})^2}{1.92\times 10^{-3}}\)
\(K_c=\frac{2.72\times10^{-6}}{1.92\times10^{-3}}\)
\(K_c = 1.42\times10^{-3}\approx1.4\times10^{-3}\)

Answer:

\(1.4\times 10^{-3}\)