Question

\[{K_c}\] for \[PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)\]​, is 0.04 at 250$^{\circ}$C. How many moles of \[PC{l_5}\] must be added to a 3 litre flask to obtain a \[C{l_2}\] concentration of 0.15 M?
A. 2.8
B. 2.1
C. 5.4
D. 3.0

Answer 1

Hint:Equilibrium refers to a condition when the rate of forward reaction is equal to the rate of reverse reaction. The equilibrium constant, denoted by K, expresses the relationship between reactants and products of a reaction at an equilibrium condition with respect to a specific unit.

Complete answer:
For a generalised chemical reaction taking place in a solution:
\[aA + bB \rightleftharpoons cC + dD\;\]
The equilibrium constant can be expressed as follows:
$K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
where [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, D respectively at equilibrium. The coefficients like a, b, c, and d in the generalised chemical equation become exponents as seen in the above expression. The given reaction in the question is:
\[PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)\]
\[x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]initial concentration on chemical species (Assumption)
\[x\left( {1 - \alpha } \right)\;\;\;\;\;\;\;\;x\alpha \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\alpha\;\;\;\;\;\;\;\;\;\;\;\; \] final concentration of chemical species
To avoid any ambiguities, the initial and final concentration values of the chemical species are mentioned in the below table as well.

Chemical species	Initial concentration	Final concentration
\[PC{l_5}\]	\[x\]	\[x\left( {1 - \alpha } \right)\]
\[PC{l_3}\]	\[0\]	\[x\alpha \]
\[C{l_2}\]	\[0\]	\[x\alpha \]

Equilibrium constant for this reaction is \[{K_c}\] = 0.04 (Given)
Concentration of \[C{l_2}\] i.e. \[x\alpha \] = 0.15 (Given)
Now, substitute the values in the aforementioned formula of equilibrium constant.
\[
\Rightarrow {K_c} = 0.04 = \dfrac{{{{(x\alpha )}^2}}}{{x(1 - \alpha )}} \\
\Rightarrow 0.04 = \dfrac{{{{(0.15)}^2}}}{{x - 0.15}} \\
\Rightarrow x = 0.7125
\]
Therefore, the concentration of \[PC{l_5}\] that was added initially was $0.7125$ M.
$\rm{Concentration \,of\, PCl_5 = \dfrac{No.\,of\,moles\,of\,PCl_5}{Volume\, in\,Litres}}$
$\Rightarrow \rm{No.\,of\,moles\,of\,PCl_5=Concentration \,of\, PCl_5 \times Volume\, in\,Litres}$
$\Rightarrow \rm{No.\,of\,moles= 0.7125\times3}$
$\Rightarrow \rm{No.\,of\,moles= 2.1375\simeq 2.1}$
Hence $2.1 \,moles$ of \[PC{l_5}\] must be added to a 3 litre flask to obtain a \[C{l_2}\] concentration of 0.15 M.

Therefore,the correct option is B. 2.1 L

Note:
While calculating the value of K, always remember few points such as
(i) K is constant for a particular reaction at a particular temperature. K changes on changing the temperature,
(ii) Pure solids or pure liquids are not generally included in the expression of equilibrium and
(iii) The chemical reaction must always be balanced including the coefficients (lowest possible integer values) to attain the correct value for K.