JEE Main202118 Mar 2021Morning ShiftChemistryThermodynamics (C)Actual
For the reaction C 2 H 6 → C 2 H 4 + H 2 the reaction enthalpy ∆ r H in kJmol - 1 is (Round off to the Nearest Integer). [Given : Bond enthalpies in kJmol - 1 : C - C : 347 , C = C : 611 ; C - H : 414 , H - H : 436 ]
Correct answer
0
Step-by-step solution
Δ r H = ϵ C - C + 2 ϵ C - H - ϵ C = C + ϵ H - H = [ 347 + 2 × 414 ] - [ 611 + 436 ] = 128