JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

2 moles of \[PC{{l}_{5}}\] were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of \[PC{{l}_{5}}\] is dissociated into \[PC{{l}_{3}}\] and \[C{{l}_{2}}\]. The value of equilibrium constant is

A) 0.266

B) 0.53

C) 2.66

D) 5.3

Correct Answer: A

Solution :

[a] \[\underset{2}{\mathop{PC{{l}_{5}}}}\,\]\[\rightleftharpoons \] \[\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\]

\[\frac{2\times 60}{100}\] \[\frac{2\times 40}{100}\] \[\frac{2\times 40}{100}\]

Volume of container = 2 litre.

\[{{K}_{c}}=\frac{\frac{2\times 40}{100\times 2}\times \frac{2\times 40}{100\times 2}}{\frac{2\times 60}{100\times 2}}=0.266\].