Question

Question: \(\;A\;0.72 - mol\)sample of \(PC{l_5}\)is put into a \(1.00 - L\) vessel and heated. At equilibrium, the vessel contains \(0.40mol\) of \(PC{l_3}(g)\) and \(0.40mol\;of\;C{l_2}(g)\). Calculate the value of the equilibrium constant for the decomposition of \(PC{l_5}\;to\;\)\(PC{l_3}\;and\;C{l_2}\)at this temperature.

Short answer

The value of \(K = 0.5\)

Step-by-step solution

Define Equilibrium

Equilibrium is the state in which market supply and demand balance each other, and as a result prices become stable. Generally, an over-supply of goods or services causes prices to go down, which results in higher demand—while an under-supply or shortage causes prices to go up resulting in less demand. The balancing effect of supply and demand results in a state of equilibrium.

Stoichiometry coefficient

In this table we are writing concentrations. It says in the task that\({\rm{PC}}{{\rm{l}}_5}\)is added in the vessel in concentration of\(0.72{\rm{M}}\)and then heated. This is its initial concentration. Because we did not have anything else in that vessel, initial concentrations of the products,\({\rm{C}}{{\rm{l}}_2}\)and\({\rm{PC}}{{\rm{l}}_3}\)are zero. Change in reactant concentration is\( - x\)because its stoichiometry\(1\)(that's why it is\( - 1\)) and we do not know for how much does it change (hence it is marked as an). It is negative because it is reactant and it is being unitized in the reaction (it is disappearing). For reactants we write\( + x\)because they are being formed during the reaction\(( + )\), their stoichiometry is\(1( + 1x)\)and we do not know for how much it is changing (hence\(x\)).

\(c({\rm{\;equilibrium\;}}) = c({\rm{\;initial\;}}) + c({\rm{\;change\;}})\)

Determining equilibrium concentration

The value of equilibrium concentration of the\(PC{l_3}\)and\(C{l_2}\)

\(c{\left( {{\rm{C}}{{\rm{l}}_2}} \right)_{eq}} = c{\left( {{\rm{PC}}{{\rm{l}}_3}} \right)_{eq}} = 0.40\frac{{{\rm{mol}}}}{{\rm{L}}}\)

The equilibrium concentration from the table is\(x\)

\(\begin{array}{*{20}{c}}{x = c{{\left( {{\rm{C}}{{\rm{l}}_2}} \right)}_{eq}} = c{{\left( {{\rm{PC}}{{\rm{l}}_3}} \right)}_{eq}}}\\{x = 0.40\frac{{{\rm{mol}}}}{L}}\end{array}\)

Calculating the equilibrium

calculate equilibrium value for\({\rm{PC}}{{\rm{l}}_5}:\)

\(\begin{array}{*{20}{c}}{c{{\left( {{\rm{PC}}{{\rm{l}}_5}} \right)}_{eq}} = 0.72 - x}\\{c{{\left( {{\rm{PC}}{{\rm{l}}_5}} \right)}_{eq}} = (0.72 - 0.40)\frac{{{\rm{mol}}}}{{\rm{L}}}}\\{c{{\left( {PC{l_5}} \right)}_{eq}} = 0.32\frac{{{\rm{mol}}}}{{\rm{L}}}}\end{array}\)

Having all the data,

\(\begin{array}{*{20}{c}}{K = \frac{{c{{\left( {{\rm{PC}}{{\rm{l}}_3}} \right)}_{eq}} \cdot c{{\left( {{\rm{C}}{{\rm{l}}_2}} \right)}_{eq}}}}{{c{{\left( {{\rm{PC}}{{\rm{l}}_5}} \right)}_{eq}}}}}\\{K = \frac{{0.40 \cdot 0.40}}{{0.32}}}\\{K = 0.5}\end{array}\)

The value of \(K = 0.5\)