Question

In the reaction, \[PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}\], the amount of each\[PC{l_5}\], \[PC{l_3}\], \[C{l_2}\] is \[2\] mole at equilibrium and total pressure is \[3\] atmospheres. The value of \[{K_p}\] will be:
A.\[4\]
B.\[3\]
C.\[2\]
D.\[1\]

Answer 1

Hint: \[{K_p}\] is known as the equilibrium constant obtained from the partial pressures of the products and partial pressure of reactants. The partial pressure cam be calculated from total pressure and mole fraction. The mole fraction of a component can be obtained from the moles of solute to total moles present in the solution.
Formula used:
\[{K_p} = \dfrac{{{P_{PC{l_3}}} \times {P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}\]
Where \[{K_p}\] is equilibrium constant
\[{P_{PC{l_3}}}\] is partial pressure of \[PC{l_3}\]
\[{P_{C{l_2}}}\] is partial pressure of \[C{l_2}\]
\[{P_{PC{l_5}}}\] is partial pressure of \[PC{l_5}\]

Complete answer:
The mole fraction of each component can be obtained from the number of moles of a compound to the total moles in the solution.
The total number of moles in the solution are \[6\].
The mole fraction of \[PC{l_5}\] is \[\dfrac{2}{6} = 0.333\]
Partial pressure will be \[0.333 \times 3 = 1\]atm
The mole fraction of \[PC{l_3}\] is \[\dfrac{2}{6} = 0.333\]
Partial pressure will be \[0.333 \times 3 = 1\]atm
The mole fraction of \[C{l_2}\] is \[\dfrac{2}{6} = 0.333\]
Partial pressure will be \[0.333 \times 3 = 1\]atm
Thus, the equilibrium constant value obtained from the partial pressures can be obtained by substituting the above values in the equation
\[{K_p} = \dfrac{{1 \times 1}}{1} = 1\]
Thus, the value of \[{K_p}\] is \[1\]

Option D is the correct one.

Note:
The mole fraction has no units and the pressure has the units of atm and the partial pressure can be calculated from the total pressure and mole fraction. Given that each reactant and product has two moles. Thus, the total moles will be six. Mole fraction of each component are equal, as the same number of moles are the same for each component.