Asked by slayrat
use bond enthalpy values to calculate the enthalpy change for the following reaction. Then, label if the reaction is endothermic or exothermic. CH4(g)+Br2(g) --> CH3Br(g) + HBr(g)
Answers
Answered by
Bot
Breaking bond enthalpies:
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
Forming bond enthalpies:
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Overall enthalpy change:
(4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol = -56 kJ/mol
The enthalpy change is negative, indicating an exothermic reaction.
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
Forming bond enthalpies:
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Overall enthalpy change:
(4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol = -56 kJ/mol
The enthalpy change is negative, indicating an exothermic reaction.
Answered by
Bot
Yes, that is correct.
Answered by
Bot
Sure, here is the full calculation with units:
Bond enthalpies:
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Enthalpy change:
ΔH = (4 x C-H) + Br-Br - C-Br - H-Br
ΔH = (4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1652 kJ/mol + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1202 kJ/mol - 642 kJ/mol
ΔH = -56 kJ/mol
The negative sign indicates an exothermic reaction.
Bond enthalpies:
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Enthalpy change:
ΔH = (4 x C-H) + Br-Br - C-Br - H-Br
ΔH = (4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1652 kJ/mol + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1202 kJ/mol - 642 kJ/mol
ΔH = -56 kJ/mol
The negative sign indicates an exothermic reaction.
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