STANDARD HEAT OF FORMATION

DEFINITION

• The change in enthalpy that accompanies the formation of one mole of the compound at its normal state from its constituent elements in their standard state.

• Standard state is 1atm & 25°C (298K)

• Every free element in its standard state is assigned a ΔHf of 0.0 kJ. Free elements include single elements and any diatomic molecule (H2O2N2Cl2Br2I2F2)

EQUATION

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)° °

**Σ is the Greek letter sigma. It means sum, or add. We are finding the sum (Σ) of the change in heat of formation (ΔHf) of the reactants and subtracting it from the sum of the change in heat of formation of the products.

Use the standard enthalpies of formation to calculate ∆Hrxn

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

∆Hf(CO2(g))= -394kJ

∆Hf(H2O(l))= -286kJ

∆Hf(CH4(g))= -75kJ

∆Hf(O2(g))= 0kJ

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)

= [-394+(2×-286)] – [-75+0]

= -966 + 75

=-891 kJ exothermic

reactants products

2HF(g) H2(g) + F2(g)

2(-271.1) 0 0

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)

= 0 − -271.1

= 542.2kJ endothermic

**look up ΔHf values in the table provided

reactants products

2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g)

2(-20.63) 0 2(-285.830) 2(-296.830)

sum=Σ= -41.26 sum=Σ= -1165.66

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)

= -1165.66 − -41.26

= -1124.06kJ exothermic

reactants products

4Fe(s) + 3O2(g) 2Fe2O3(s)

0 0 2(-824.2)

sum=Σ= 0 -1648.4

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)

= -1648.4 − 0

= -1648.4kJ exothermic

reactants products