Thermochemistry summary

Question

Use the table of bond enthalpies below to calculate ΔH for the following reactions.

BOND ENTHALPY(kJ mol^-1)

H - H 436

Cl - Cl 242

H - Cl 431

C = C 612

C - C 348

C - H 412

Br - Br 193

C - Br 276

A. H₂ (g) + Cl₂ (g) ----> 2 HCl (g)

B. C₂H₄ (g) + Br₂ (g) ----> C₂H₄Br₂ (g)

Answer 1

Remember that it TAKES energy to break bonds and energy is GIVEN back when bonds are formed.

A) Find energy to break the H-H bond and the Cl-Cl bond in the reactants:

436 kJ/mol + 242 kJ/mol = 678 kJ added to the system

Find energy released when 2 mols HCl are formed:

2 mols H-Cl bonds x 431 kJ/mol = -862 kJ and since this is given off it receives a negative sign (-)

Add the reaction and the products:

∆H reaction = 431 - 862 kJ = -431kJ (exothermic) more heat is released than is added

B) Same procedure as above:

reactants:

H₂C=CH₂ has 1 C=C bond = 612 kJ

H₂C=CH₂ has 4 C-H bonds = 4 x 412 kJ = 1648 kJ

Br₂ has Br-Br = 193 kJ

TOTAL for the reactants = 612 + 1648 + 193 = 2453 kJ added to the system to break bonds

products:

C₂H₄Br₂ has 1 C-C bond = 348 kJ

C₂H₄Br₂ has 4 C-H bonds = 4 x 412 kJ = 1648 kJ

C₂H₄Br₂ has 2 C-Br bonds = 2 x 276 kJ = 552 kJ

TOTAL for the products = 348 + 1648 + 552 = -2548 kJ liberated when forming bonds in the product

∆H reaction = 2454 kJ - 2548 kJ = -94 kJ (exothermic) more heat is released than added