Calculate the enthalpy change for the process: CCl 4 (g) → C(g) + 4Cl(g) and calculate bond enthalpy of C–Cl in CCl 4 (g). ∆ vap H ⊖ (CCl 4 ) = 30.5 kJ mol –1 ; ∆ f H ⊖ (CCl 4 ) = –135.5 kJ mol –1 . ∆ a H ⊖ (C) = 715.0 kJ mol –1 , where ∆ a H is enthalpy of atomisation ∆ a H ⊖ (Cl 2 ) = 242 kJ mol –1

Question

Calculate the enthalpy change for the process: CCl 4 (g) → C(g) + 4Cl(g) and calculate bond enthalpy of C–Cl in CCl 4 (g). ∆ vap H ⊖ (CCl 4 ) = 30.5 kJ mol –1 ; ∆ f H ⊖ (CCl 4 ) = –135.5 kJ mol –1 . ∆ a H ⊖ (C) = 715.0 kJ mol –1 , where ∆ a H is enthalpy of atomisation ∆ a H ⊖ (Cl 2 ) = 242 kJ mol –1, 412 kJ mol –1, 326 kJ mol –1, 369 kJ mol –1, 442 kJ mol –1

MARKS App · Accepted Answer

Correct Option is : (B) 326 kJ mol –1

Search any question & find its solution

Looking for more such questions to practice?