NEETChemistryChemical Equilibrium
Match List-I with List-II. List-I (Reaction) List-II (Relation between K_P and K_C ) (A) H ₂( g ) + I ₂( g ) 2 HI ( g ) (I) K_P = K_C(RT)⁻² (B) N ₂( g ) + 3 H ₂( g ) 2 NH ₃( g ) (II) K_P = K_C (C) NH ₄ HS ( s ) NH ₃( g ) + H ₂ S ( g ) (III) K_P = K_C(RT)^2 (D) PCl ₅( g ) PCl ₃( g ) + Cl ₂( g ) (IV) K_P = K_C(RT) (V) K_P = K_C(RT)⁻¹ Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- B(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
- C(A)-(II), (B)-(I), (C)-(IV), (D)-(IV)
- D(A)-(II), (B)-(I), (C)-(III), (D)-(V)
Correct answer
A. (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Step-by-step solution
The relationship between K_P and K_C is K_P = K_C(RT)^ n_g , where n_g = moles of gaseous products - moles of gaseous reactants . (A) H ₂( g ) + I ₂( g ) 2 HI ( g ) n_g = 2 - (1 + 1) = 0 . Therefore, K_P = K_C . (Matches II) (B) N ₂( g ) + 3 H ₂( g ) 2 NH ₃( g ) n_g = 2 - (1 + 3) = -2 . Therefore, K_P = K_C(RT)⁻² . (Matches I) (C) NH ₄ HS ( s ) NH ₃( g ) + H ₂ S ( g ) Solid species are ignored in n_g calculation. n_g = (1 + 1) - 0 = 2 . Therefore, K_P = K_C(RT)^2 . (Matches III) (D) PCl ₅( g ) PCl ₃( g ) + Cl ₂( g