JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 18)
For the reaction C2H6 $$ \to $$ C2H4 + H2
the reaction enthalpy $$\Delta$$rH = __________ kJ mol$$-$$1. (Round off to the Nearest Integer).
[ Given : Bond enthalpies in kJ mol$$-$$1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
the reaction enthalpy $$\Delta$$rH = __________ kJ mol$$-$$1. (Round off to the Nearest Integer).
[ Given : Bond enthalpies in kJ mol$$-$$1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
Answer
128
Explanation
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$$\Delta H = {E_{C - C}} + 6{E_{C - H}} - {E_{C = C}} - 4{E_{C - H}} - {E_{H - H}}$$
$$ = 347 + 6(414) - 611 - 4 \times 414 - 436$$
$$ = 347 + 828 - 1047$$
$$ = 128$$ KJ/ mol
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