NEETChemistryChemical Equilibrium
Match List-I with List-II. (A) N _ 2( g ) + 3 H _ 2( g ) 2 NH _ 3( g ) (I) K_p = K_c (B) CaCO _ 3( s ) CaO _ ( s ) + CO _ 2( g ) (II) K_p = K_c(RT) (C) H _ 2( g ) + Cl _ 2( g ) 2 HCl _ ( g ) (III) K_p = K_c(RT)⁻² (D) NH ₄ HS _ ( s ) NH _ 3( g ) + H ₂ S _ ( g ) (IV) K_p = K_c(RT)² Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- B(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
- C(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- D(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Correct answer
D. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Step-by-step solution
The relationship between K_p and K_c is given by K_p = K_c(RT)^ n_g , where n_g = ( moles of gaseous products ) - ( moles of gaseous reactants ) . For (A): n_g = 2 - (1 + 3) = -2 . Thus, K_p = K_c(RT)⁻² . For (B): Solids are not included in the calculation of n_g . n_g = 1 - 0 = 1 . Thus, K_p = K_c(RT) . For (C): n_g = 2 - (1 + 1) = 0 . Thus, K_p = K_c(RT)⁰ = K_c . For (D): Solids are not included. n_g = (1 + 1) - 0 = 2 . Thus, K_p = K_c(RT)² . Therefore, the correct match is (A)-(III), (B)-(II), (C)-(I), (D)-(IV).