Saved Bookmarks
| 1. |
(i) Derive the relation between K_p and K_c for general homogeneous gaseous reaction. (ii) How do you measure heat changes of a constant pressure ? |
|
Answer» Solution :Consider a gerneral reaction in which all reactants and products are ideal gases. `xA + y B hArr lC + mD` The equilibrium costant `K_(c)` is `K_(c)=([C]^l[D])/([A]^X[B]^y)` `K_p = (p^lC xx p_D^m)/(p_A^x xx p_B^y)` The ideal gas equation is `PV = nRT or P= (n)/(V)RT` Since, Active mass=molar concentration `= m/v` P= Active mass `xx` RT Based on the above expressio, the partial pressure of the reacants and products can be expressred as, `p_(A)^x =[A]^x.[RT]^x , p_B^y=[B]^y.[RT]^y` `p_C^l=[C]^l.[RT]^l , P_D^m = [D]^m.[RT]^m` On substiutiono in equation (2), `K_p = ([C]^l[RT]^l[D]^m[RT]^m)/([A]^x[RT]^x[B]^y[RT]^y)` `K_(p)= ([C]^l[D]^m)/([A]^x[B]^y) xx RT^((l+m)-(x+y))` By comparing equation (1) and (4) ,we get `K_(p)=K_(c)(RT)^(Delta ng)` where `Delta n_g` is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase. (i) If `Deltan_g =0, K_p = K_c(RT)^0 ` `K_p = K_c` Example : `H_(2)(g) + I_(2)(g) hArr 2HI(g)` (ii) when `Delta n_g = + ve` `K_p = K_c (RT)^(+ve) RARR K_(p)= K_(c)` Example `2NH_(3)(g) hArr N_2(g)+3H_(2)(g)` (iii) When `Delta n_g = -ve` `K_p = K_c(RT)^(-ve)` `K_p lt K_c` Example : `2SO_2(g) + O_(2)(g) hArr 2 SO_3 (g)` 1. Measurement of heat change at constant pressure can be DONE in a coffee cup calorimeter. 2 . We know that `Delta H=q_p` (at constant P) and therefore , heat absorbed or evoloved `q_r` constant pressure is also called tha heat of reaction or enthalpy of reaction `Delta H_r` 3. In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. Therefore `q_p` will be NEGATIVE and `Delta H_r` will also be negative 4. Similarly in an endothermic reaction, heat is absorbed `q_p` is postive and `Delta H_r` will also be positive. 
|
|