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JEE Main 202118 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2021 · 18 March · Morning Shift · Chemistry · Thermodynamics (C)

For the reaction C _ 2 H _ 6 → C _ 2 H _ 4 + H _ 2 the reaction...

For the reaction
C2H6→C2H4+H2{C}_{2}{H}_{6} \to {C}_{2}{H}_{4} + {H}_{2}
the reaction enthalpy ∆rHin{∆}_{r}Hin kJmol−1{kJmol}^{- 1} is
(Round off to the Nearest Integer).
[Given : Bond enthalpies in kJmol−1:C−C:{kJmol}^{- 1}: C - C:
347,C=C:611;C−H:414,H−H:436]347, C = C: 611; C - H: 414, H - H: 436 ]

Correct answer

0

Step-by-step solution

ΔrH=[ϵC−C+2ϵC−H]−[ϵC=C+ϵH−H]{\Delta}_{r}H = \left[{ϵ}_{C - C} + 2{ϵ}_{C - H}\right] - \left[{ϵ}_{C = C} + {ϵ}_{H - H}\right]
=[347+2×414]−[611+436]= [ 347 + 2 \times 414 ] - [ 611 + 436 ]
=128= 128
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