Use bond enthalpy values to calculate the enthalpy change for the following reactions.
CH4 (g) + Br2 (g) CH3Br (g) + HBr (g)
C2H4 (g) + HBr (g) C2H5Br (g)
C4H8 (g) + H2 (g) C4H10 (g)
C3H6 (g) + I2 (g) C3H6I2 (g)
Then, label each reaction as either endothermic or exothermic based on the enthalpy value.
CH4 (g) + Br2 (g) CH3Br (g) + HBr (g)
C2H4 (g) + HBr (g) C2H5Br (g)
C4H8 (g) + H2 (g) C4H10 (g)
C3H6 (g) + I2 (g) C3H6I2 (g)
To calculate the enthalpy change for a reaction, you need to use the bond enthalpies of the bonds broken and formed in the reaction.
CH4 (g) + Br2 (g) CH3Br (g) + HBr (g)
Bond enthalpies:
CH4 (g) : C-H: 413 kJ/mol, H-H: 436 kJ/mol
CH3Br (g) : C-H: 413 kJ/mol, C-Br: 290 kJ/mol, H-Br: 366 kJ/mol
HBr (g) : H-Br: 366 kJ/mol
ΔH = (1413 + 3436) + (1413 + 1290 + 1366) - (1413 + 3436) - (2366) = -104 kJ/mol
C2H4 (g) + HBr (g) C2H5Br (g)
Bond enthalpies:
C2H4 (g) : C-H: 413 kJ/mol, C=C: 614 kJ/mol, H-H: 436 kJ/mol
C2H5Br (g) : C-H: 413 kJ/mol, C=C: 614 kJ/mol, C-Br: 290 kJ/mol, H-Br: 366 kJ/mol
HBr (g) : H-Br: 366 kJ/mol
ΔH = (2413 + 1614 + 2436) + (2413 + 1614 + 1290 + 1366) - (2413 + 1614 + 2436) - (1*366) = -28 kJ/mol
C4H8 (g) + H2 (g) C4H10 (g)
Bond enthalpies:
C4H8 (g) : C-H: 413 kJ/mol, H-H: 436 kJ/mol
C4H10 (g) : C-H: 413 kJ/mol, H-H: 436 kJ/mol
ΔH = (4413 + 8436) + (4413 + 10436) - (4413 + 8436) - (2*436) = -28 kJ/mol
C3H6 (g) + I2 (g) C3H6I2 (g)
Bond enthalpies:
C3H6 (g) : C-H: 413 kJ/mol, C=C: 614 kJ/mol, H-H: 436 kJ/mol
C3H6I2 (g) : C-H: 413 kJ/mol, C=C: 614 kJ/mol, C-I: 298 kJ/mol, I-I: 150 kJ/mol
I2 (g) : I-I: 150 kJ/mol
ΔH = (3413 + 2614 + 6436) + (3413 + 2614 + 2298 + 1150) - (3413 + 2614 + 6436) - (2*150) = -96 kJ/mol
Based on the value of ΔH, the reactions are:
CH4 (g) + Br2