Thermodynamics problem... calculate the bond enthalpy of C-Cl in CCl4
Abhishek Dogra , 6 Years ago
Grade 11
AIIMS MBBS> Thermodynamics problem... calculate the b...
1 Answers
Arun
Last Activity: 6 Years ago
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
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
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