Calculate bond enthalpy (in kJ mol- 1 ) of C-Cl in CCl4(g) from the following Data: Δ vapHCCl4° =30.5 kJ mol Δ f H(CCl4 , l)° =-135.5 kJ/mole Δ aH° (C)=715 kJ/mol (i.e. enthalpy of atomisation) Δ aHCl2° =242 kJ/mole

Question

Calculate bond enthalpy (in kJ mol- 1 ) of C-Cl in CCl4(g) from the following Data: Δ vapHCCl4° =30.5 kJ mol Δ f H(CCl4 , l)° =-135.5 kJ/mole Δ aH° (C)=715 kJ/mol (i.e. enthalpy of atomisation) Δ aHCl2° =242 kJ/mole (A) 326 (B) 333.625 (C) 356.5 (D) 447

Tardigrade Admin · Accepted Answer

C(s) +2 Cl 2 (g) arrow CCl 4 (ℓ) Δ H =-135.5 KJ / mole CCl 4 (ℓ) arrow CCl 4 (g) Δ H =30.5 KJ / mole C(s) +2 Cl 2 (g) longrightarrow CCl 4 (g) Δ H =-105 KJ / mole 715+2 × 242-4 × x=-105 x =326 KJ / mole i.e. x = B . E . of C - Cl in CCl 4 (g)

$326$

$333.625$

$356.5$

$447$

Q. Calculate bond enthalpy (in kJmol−1 ) of C−Cl in CCl4​(g) from the following Data: Δvap​HCCl4​∘​=30.5kJmolΔf​H(CCl4​,l)∘​=−135.5kJ/mole Δa​H∘(C)=715kJ/mol (i.e. enthalpy of atomisation) Δa​HCl2​∘​=242kJ/mole

326

333.625

356.5

447

C(s)+2Cl2​(g)→CCl4​(ℓ)ΔH=−135.5KJ/mole CCl4​(ℓ)→CCl4​(g)ΔH=30.5KJ/ mole C(s)+2Cl2​(g)⟶CCl4​(g)ΔH=−105KJ/ mole 715+2×242−4×x=−105 x=326KJ/ mole i.e. x=B.E. of C−Cl in CCl4​(g)

$326$

$333.625$

$356.5$

$447$