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Question: If \[PC{l_5}\], \[PC{l_3}\] and \[C{l_2}\] are at equilibrium at \[500K\] in a closed containers and...

If PCl5PC{l_5}, PCl3PC{l_3} and Cl2C{l_2} are at equilibrium at 500K500K in a closed containers and their concentrations are 0.8×10−3molL−10.8 \times {10^{ - 3}}mol{L^{ - 1}}, 1.2×10−3molL−11.2 \times {10^{ - 3}}mol{L^{ - 1}} and 1.2×10−3molL−11.2 \times {10^{ - 3}}mol{L^{ - 1}} respectively.
The value of Kc{K_c} for the reaction:
PCl5(g)⇌PCl3(g)+Cl2(g)PC{l_5}(g)\rightleftharpoons PC{l_3}(g) + C{l_2}(g) will be:
A. 1.8×103molL−11.8 \times {10^3}mol{L^{ - 1}}
B. 1.8×10−3molL−11.8 \times {10^{ - 3}}mol{L^{ - 1}}
C. 1.8×10−3Lmol−11.8 \times {10^{ - 3}}Lmo{l^{ - 1}}
D. 0.55×104Lmol−10.55 \times {10^4}Lmo{l^{ - 1}}

Explanation

Solution

Equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction. The law of mass action states that the rate at which the reactant reacts is directly proportional to its active mass and the rate of a reaction is directly proportional to the product of the active mass of the reacting species. Le Chatelier's principle is one of the important laws in chemical equilibrium. It states that "If a system of equilibrium is subjected to any disturbance (or) stress, then the equilibrium will tend to shift in such a direction to nullify the effect of stress.

Complete step by step answer:
Equilibrium constant (Kc{K_c}) is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants.
Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine molecules. The reaction can be written as follows,
PCl5(g)⇌PCl3(g)+Cl2(g)PC{l_5}(g)\rightleftharpoons PC{l_3}(g) + C{l_2}(g)
The equilibrium constant for this reaction can be written as,
Kc=[PCl3][Cl2][PCl5]{K_c} = \dfrac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}}
The concentrations of PCl5PC{l_5}, PCl3PC{l_3} and Cl2C{l_2} are given as 0.8×10−3molL−10.8 \times {10^{ - 3}}mol{L^{ - 1}}, 1.2×10−3molL−11.2 \times {10^{ - 3}}mol{L^{ - 1}} and 1.2×10−3molL−11.2 \times {10^{ - 3}}mol{L^{ - 1}}.
⇒Kc=1.2×10−3molL−1×1.2×10−3molL−10.8×10−3molL−1\Rightarrow {K_c} = \dfrac{{1.2 \times {{10}^{ - 3}}mol{L^{ - 1}} \times 1.2 \times {{10}^{ - 3}}mol{L^{ - 1}}}}{{0.8 \times {{10}^{ - 3}}mol{L^{ - 1}}}}
⇒Kc=1.8×10−3molL−1\Rightarrow {K_c} = 1.8 \times {10^{ - 3}}mol{L^{ - 1}}
Thus, the value of Kc{K_c} for the reaction is 1.8×10−3molL−11.8 \times {10^{ - 3}}mol{L^{ - 1}}

So, the correct answer is Option B.

Note: When the concentration of species is written in terms of partial pressure, then the equilibrium constant can be written as Kp{K_p}. Degree of dissociation (△ng\vartriangle ng) is defined as the difference between the sum of products to the sum of reactants. The equilibrium constants Kp{K_p} and Kc{K_c} can be related as,
Kp=Kc×(RT)△ng{K_p} = {K_c} \times {(RT)^{\vartriangle ng}}
Where,
R-gas constant
T-temperature
Kp{K_p}-equilibrium constant in terms of partial pressure
Kc{K_c}-equilibrium constant in terms of molar concentration
△ng\vartriangle ng-degree of dissociation
When the degree of dissociation is zero i.e. the number of moles of product is equal to the number of moles of reactant, then Kp=Kc{K_p} = {K_c}. Kp{K_p} will have the units of pressure and Kc{K_c} will have the unit of molar concentration.