Question: Antimony pentachloride decomposes according to this equation:

An equilibrium mixture in a 5.00-L flask at 448⁰C contains 3.85 g of \({\rm{SbC}}{{\rm{l}}_5}\),9.14 g of \({\rm{SbC}}{{\rm{l}}_3}\)and 2.84 g of \({\rm{C}}{{\rm{l}}_2}\).How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?

Given information:

-The volume of a flask is 5.00L

At equilibrium

The mass of\({\rm{SbC}}{{\rm{l}}_5}\)is 3.85g \(\left( {\frac{{3.85{\rm{g}}}}{{299,02{\rm{g}}/{\rm{mol}}}} = 1.29 \times {{10}^{ - 2}}{\rm{mol}}} \right)\)

\(\left( {\left( {SbC{l_5}} \right) = \frac{{1.29 \times {{10}^{ - 2}}{\rm{mol}}}}{{5.00{\rm{L}}}} = 2.58 \times {{10}^{ - 3}}{\rm{M}}} \right)\)

-the mass of \({\rm{SbC}}{{\rm{l}}_3}\)is 9.14g\(\left( {\frac{{9.14{\rm{g}}}}{{228,13{\rm{g}}/{\rm{mol}}}} = 4.01 \times {{10}^{ - 2}}{\rm{mol}}} \right)\)

\(\left( {\left( {SbC{l_3}} \right) = \frac{{4.01 \times {{10}^{ - 2}}{\rm{mol}}}}{{5.00{\rm{L}}}} = 8.02 \times {{10}^{ - 3}}{\rm{M}}} \right)\)

-the mass of \({\rm{C}}{{\rm{l}}_2}\)is 2.84g\(\left( {\frac{{2.84}}{{70.91{\rm{g}}/{\rm{mol}}}} = 4.01 \times {{10}^{ - 2}}{\rm{mol}}} \right)\)

\(\left( {\left( {C{l_2}} \right) = \frac{{4.01 \times {{10}^{ - 2}}{\rm{mol}}}}{{5.00{\rm{L}}}} = 8.02 \times {{10}^{ - 3}}{\rm{M}}} \right)\)

- The equilibrium constant for this equation is

\({K_c} = \frac{{\left( {SbC{l_3}} \right) \times \left( {C{l_2}} \right]}}{{\left( {SbC{l_5}} \right)}} = \frac{{8.02 \times {{10}^{ - 3}} \times 8.02 \times {{10}^{ - 3}}}}{{2.58 \times {{10}^{ - 3}}}} = 0.0249\)

We have to find the mass of each gas will be found if the mixture is transferred into a 2.00 L flask. (The temperature remains the same)

- The initial concentrations, under new conditions are

\(\begin{array}{*{20}{c}}{\left( {{\rm{SbC}}{{\rm{l}}_5}} \right) = \frac{{1.29 \times {{10}^{ - 2}}{\rm{mol}}}}{{2.00{\rm{L}}}} = 6.45 \times {{10}^{ - 3}}{\rm{M}}}\\{\left( {{\rm{SbC}}{{\rm{l}}_3}} \right) = \frac{{4.01 \times {{10}^{ - 2}}{\rm{mol}}}}{{2.00{\rm{L}}}} = 2.005 \times {{10}^{ - 2}}{\rm{M}}}\\{\left( {{\rm{C}}{{\rm{l}}_2}} \right) = \frac{{4.01 \times {{10}^{ - 2}}{\rm{mol}}}}{{2.00{\rm{L}}}} = 2.005 \times {{10}^{ - 2}}{\rm{M}}}\end{array}\)

Since volume is decreased, the pressure will increase, hence the equilibrium will move to the left (because of the fewer number of gas molecules).

Now we will find the value of x

\({K_c} = \frac{{\left( {{\rm{SbC}}{{\rm{l}}_3}} \right) \times \left( {{\rm{C}}{{\rm{l}}_2}} \right)}}{{\left. {{{({\rm{SbCl}})}_5}} \right)}}\)

\(0.0249 = \frac{{\left( {2.005 \times {{10}^{ - 2}} - x} \right) \times \left( {2.005 \times {{10}^{ - 2}} - x} \right)}}{{6.45 \times {{10}^{ - 3}} + x}}\)

\(\begin{array}{*{20}{c}}{0.0249 \times \left( {6.45 \times {{10}^{ - 3}} + x} \right) = {{\left( {2.005 \times {{10}^{ - 2}} - x} \right)}^2}}\\{0.16 \times {{10}^{ - 3}} + 0.0249x = 4.02 \times {{10}^{ - 4}} - 4.01 \times {{10}^{ - 2}}x + {x^2}}\\{0 = {x^2} - 0.065x + 2.42 \times {{10}^{ - 4}}}\\{x = 3.965 \times {{10}^{ - 3}}{\rm{M}}}\end{array}\)

The new equilibrium concentrations are [\({\rm{SbC}}{{\rm{l}}_5}\)] =\(6.45 \times {10^{ - 3}} + x = 10.415 \times {10^{ - 3}}{\rm{M}}\)

[\({\rm{SbC}}{{\rm{l}}_3}\)]\( = 2.005 \times {10^{ - 2}} - x = 16.085 \times {10^{ - 3}}{\rm{M}}\)

[\({\rm{C}}{{\rm{l}}_2}\)]\( = 2.005 \times {10^{ - 2}} - x = 16.085 \times {10^{ - 3}}{\rm{M}}\)

Therefore, the mass of each gas that would be found if the mixture is transferred into a 2.00 L flask is

\({\rm{SbC}}{{\rm{l}}_5}\)\( = 10.415 \times {10^{ - 3}}{\rm{M}} \times 2L \times 299,02{\rm{g}}/{\rm{mol}} = 6.23{\rm{g}}\)

\({m_{SbCl}}\)\( = 16.085 \times {10^{ - 3}}M \times 2L \times 228,13{\rm{g}}/{\rm{mol}} = \)7.34g

\({m_{Cl}}\)\( = 16.085 \times {10^{ - 3}}M \times 2L \times 70.91{\rm{g}}/{\rm{mol}} = 2.28{\rm{g}}\)