Question

Calculate the enthalpy change for the process ccl4

Answer 1

Answer:

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