Question

Question 6.15 Calculate the enthalpy change for the process

CCl4(g)→C(g) + 4Cl(g)

and calculate bond enthalpy of C–Cl in CCl4(g).

ΔvapHθ (CCl4) = 30.5 kJ mol–1.

ΔfHθ (CCl4) = –135.5 kJ mol–1.

ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy of atomisation

ΔaHθ (Cl2) = 242 kJ mol–1

Class XI Thermodynamics Page 183

Answer 1

CCl₄ ---------> 4Cl (g) ; ΔvapH° = +30.5 KJ/mol----------------(1)
C(s)  + 2Cl₂ (g)  ---------> ΔfH° = -135.5 KJ/mol ------------------(2)
C (s) ----------> C(g); ΔaH° = 715 KJ/mol --------------------------(3)
Cl₂ (g) ---------->2Cl (g) ;ΔaH° = 242 KJ/mol --------------------(4)
   multiplying with 2 in equation (4)
2Cl₂(g) ---------->4Cl(g) ; ΔaH° = 484 KJ/mol 
       now add equations (3) and (4) 
C(s) + 2Cl₂(g) ------>C(g) + 4Cl , ΔH₁ = 715 + 484 = 1199 KJ/mol -----(5)
       now add reverse of equation (1) 
CCl₄ (g) ------> CCl₄ (l), ΔH₂ = -30.5 KJ/mol-------------(6)
CCl₄ (l) ---------> C(s) + 2Cl₂(g) ,ΔH₃ = 135.5 KJ/mol---------(7)
     adding equations (5), (6) and (7) 
CCl₄ (g) --------> C(g) + 4Cl , ΔH = 135.5 - 30.5 + 1199 = 1304 KJ/mol
      now bond enthalpy  of C --- Cl is 1/4 of ΔH
                         Cl
                          |
      Cl------------ C ------------ Cl
                         |
                        Cl
 now, bond enthalpy = 1304/4 = 326 KJ/mol
 


   

Answer 2

Answer:

The chemical equations implying to the given values of enthalpies are:

(i) CCl4(l) → CCL4(g) ΔvapHθ = 30.5 kJ mol-1

(ii) C(s) → C(g) ΔaHθ = 715.0 kJ mol-1

(iii) Cl2(g) → 2Cl(g) ΔaHθ = 242 kJ mol-1

(iv) C(g) + 4Cl(g) → CCl4(g) ΔfH = -135.5 kJ mol-1

Enthalpy change for the given process C(g) + 4Cl(g) → CCl4(g) can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)

ΔH = ΔaHθ(C) + 2ΔaHθ (Cl2) - ΔvapHθ - ΔfH

= (715.0 kJ mol-1) + 2(242 kJ mol-1) - (30.5 kJ mol-1) - (-135.5 kJ mol-1)

∴ΔH = 1304 kJ mol-1

Bond enthalpy of C-Cl bond in CCl4(g) = 326 kJ mol-1