NEETChemistryChemical Equilibrium
Match List I with List II for the expression of K _ p in terms of forward rate constant ( k _ f ), backward rate constant ( k _ b ), R , and T . List I (Reaction) List II ( K _ p expression) (A) H ₂( g ) + I ₂( g ) 2 HI ( g ) (I) ( k _ f k _ b )( RT )⁻² (B) N ₂ O ₄( g ) 2 NO ₂( g ) (II) ( k _ f k _ b ) (C) N ₂( g ) + 3 H ₂( g ) 2 NH ₃( g ) (III) ( k _ f k _ b )( RT )⁻¹ (D) 2 SO ₂( g ) + O ₂( g ) 2 SO ₃( g ) (IV) ( k
Options
- A(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- B(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- C(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
- D(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Correct answer
A. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Step-by-step solution
For any reaction, the equilibrium constant in terms of concentration is K _ c = k _ f k _ b . The relationship between K _ p and K _ c is given by K _ p = K _ c ( RT )^ n _ g = ( k _ f k _ b )( RT )^ n _ g . (A) H ₂( g ) + I ₂( g ) 2 HI ( g ) : n _ g = 2 - (1+1) = 0 . Thus, K _ p = ( k _ f k _ b ) . (Matches II) (B) N ₂ O ₄( g ) 2 NO ₂( g ) : n _ g = 2 - 1 = 1 . Thus, K _ p = ( k _ f k _ b )( RT )¹ . (Matches IV) (C) N ₂( g ) + 3 H ₂( g ) 2 NH ₃( g ) : n _ g = 2 - (1+3) = -2 . Thus, K _ p = ( k _ f k _ b )( RT )⁻²