NEET2008ChemistryChemical EquilibriumActual
The dissociation equilibrium of a gas AB ₂ can be represented as, 2 AB ₂( ~g ) 2 AB ( g )+ B ₂( ~g ) . The degree of dissociation is ' x ' and is small compared to 1. The expression relating the degree of dissociation ( x ) with equilibrium constant K_P and total pressure P is
Options
- A( K_p P )
- B( 2 K_P P )
- C( 2 K_P P )^ 1 / 3
- D( 2 K_P P )^ 1 / 2
Correct answer
C. ( 2 K_P P )^ 1 / 3
Step-by-step solution
aligned & 2 AB ₂ 2 AB + B ₂ & 1 & 00 & 1-x & x x / 2 & aligned Total mole at equi. =1+ x 2 K_P= ( x 1+x / 2 P )^2 ( x / 2 1+x / 2 P ) ( 1-x 1+x / 2 P )^2 (Here x is degree of dissociation) or K_P= x^3 P 2 or x^3= 2 K_P P or x= ( 2 K_P P )^ 1 / 3