Question

Question: The density of trifluoroacetic acid vapor was determined at 118.1 °C and 468.5 torr, and found to be 2.784 g/L.

CalculateKcfor the association of the acid.

Short answer

The value for the association constant of the acid is \(26.5\)

Step-by-step solution

Definition

Chemical equilibrium is defined as a dynamic state where the concentration of all reactants remains constant. They may not be equal but they are not changing. In a chemical reaction, a double arrow indicates an equilibrium situation

The association of trifluoroacetic acid

The Temperature\(T = {118.1^ \circ }{\rm{C}}((118.1 + 273){\rm{K}} = 391.1{\rm{K}})\)

The Pressure\(P = 468.5{\rm{\;torr\;}}\left( {468.5{\rm{\;torr\;}} \times \frac{{1atm}}{{760{\rm{\;torr\;}}}} = 0.616{\rm{atm}}} \right)\)

The Denisty\(d = 2.784{\rm{g}}/{\rm{L}}\)

The Molar mass of the TFA-monomer is\(114,02{\rm{g}}/{\rm{mol}}\)

The Molar mass of the TFA-dimer is\(228.04{\rm{g}}/{\rm{mol}}\)

We will assume that the volume of the vessel is 1L

We have to calculate \({{\rm{K}}_c}\)for the association of the trifluoroacetic acid

Total Number of moles

We will determine the total number of moles (n of monomer + n of dimer)

\(\begin{array}{*{20}{c}}{PV = nRT}\\{n = \frac{{PV}}{{RT}}}\\{ = \frac{{0.616{\rm{\;atm\;}} \times 1L}}{{0.08206L{\rm{\;atm\;}}/{\rm{molK}} \times 391.1K}}}\\{ = 0.0192{\rm{mol}}}\end{array}\)

Now, we will find the mass of both monomer and dimer

\(\begin{array}{*{20}{c}}{m = d \times V}\\{ = 2.784{\rm{g}}/L \times 1L}\\{ = 2.784{\rm{g}}}\end{array}\)

Concentration of the monomer and dimer

\(\begin{array}{*{20}{c}}{n = {n_{{\rm{monomer\;}}}} + {n_{{\rm{dimer\;}}}}}\\{{n_{{\rm{dimer\;}}}} = n - {n_{{\rm{monomer\;}}}}}\\{ = 0.0192{\rm{\;mol\;}} - {n_{{\rm{monomer\;}}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{m = {m_{{\rm{monomer\;}}}} + {m_{{\rm{dimer\;}}}}}\\{2.784g = {n_{{\rm{monomer\;}}}} \times {M_{{\rm{monomer\;}}}} + {n_{{\rm{dimer\;}}}} \times {M_{{\rm{dimer\;}}}}}\\{2.784g = {n_{{\rm{monomer\;}}}} \times {M_{{\rm{monomer\;}}}} + \left( {0.0192{\rm{\;mol\;}} - {n_{{\rm{monomer\;}}}}} \right) \times {M_{{\rm{dimer\;}}}}}\\{2.784g = {n_{{\rm{monomer\;}}}} \times {M_{{\rm{monomer\;}}}} + \left( {0.0192{\rm{\;mol\;}} - {n_{{\rm{monomer\;}}}}} \right) \times {M_{{\rm{dimer\;}}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{2.784g = {n_{{\rm{monomer\;}}}} \times 114,02g/mol + \left( {0.0192m{\rm{\;mol\;}} - {n_{{\rm{monomer\;}}}}} \right) \times 228.04{\rm{g}}/{\rm{mol}}}\\{2.784 = 114.02{n_{{\rm{monomer\;}}}} + 4.378 - 228.04{n_{{\rm{monomer\;}}}}}\\{{n_{{\rm{monomer\;}}}} = 0.0140{\rm{mol}}}\end{array}\)

Since, We assumed that the volume is 1L, the concentration of monomer would be\(0.0140{\rm{M}}\)

And the concentration of the dimer would be\(0.0052{\rm{M}}\)

\({\rm{\;And the value of\;}}{{\rm{K}}_c}{\rm{\;is\;}}\)

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{({\rm{\;dimer\;}})}}{{{{({\rm{\;monomer\;}})}^2}}}}\\{ = \frac{{0.0052}}{{{{(0.0140)}^2}}}}\\{ = 26.5}\end{array}\)