ENERGETICS1.HESS’S LAW

2. BORN HABER CYCLE

ΔHθ – The amount of heat absorbed/evolved when the molar quantities of reactants as stated in the equation react together under std conditions.

Std conditions : P = 1 atm

T = 298KSubstances in their normal physical states

Definition of std enthalpy change

ΔH f θ - The heat change required to produce 1

mol of the substance from its elements under standard conditions.

Ag(s) + ½ Cl2 (g) AgCl (s) ; ΔH f θ = -127

kJmol-1

ΔH c θ - The heat energy evolved when 1 mol of

the substance is completely burnt in excess oxygen under standard conditions.

C2 H4(g) + 3O2 (g) 2CO2 (g) +2 H2O (l) ; ΔH c θ

= -1411 kJmol-1

Definitions

Hess’s law states that overall heat change in a chemical reaction is constant and not dependent on the route taken.

X

A + B C + D (initial) (final)

Y ZΔHθ Route 1 = ΔHθ Route 2

X = Y + Z

Hess’s law

C(s) + O2 (g) CO2 (g) ; ΔH f θ = -406 kJmol-1

H2(g) + ½O2 (g) H2O (l) ; ΔH f θ = -285 kJmol-1

C2 H5OH(l) + 3O2 (g) 2CO2 (g) +3H2O (l) ; ΔH c θ = -

1423 kJmol-1

Find ΔH f θ C2 H5OH (l).

2C(s) + 3H2(g)+ ½O2 (g) C2H5OH(l) Working, ΔH f

θ = 2 ΔH1 + 3ΔH2 – ΔH3

= 2(-406) + 3(-285) + 1423 = -244 kJmol-1

Use algebraic method

ΔH 4

2C + 3H2 C2 H5OH

ΔH1 ΔH2 ΔH3

2CO2 + 3H2O

By Hess’s law ΔH1 + ΔH2 = ΔH4 + ΔH3

ΔH4 = ΔH1+ ΔH2- ΔH3

= 2(-406) + 3(-285) –(-1423)

= -244 kJmol-1

Use energy cycle

Def : The enthalpy change when 1 mol of an ionic compound is broken apart into its constituent gaseous ions under standard conditions.

NaCl(s) Na+(g) +Cl-(g) ; ΔH latt

θ =+771 kJmol-1

Lattice enthalpy

E.g.1

Draw a Born Haber cycle for RbI. Hence use the data

below to calculate the ΔHf

of RbI.

ΔH a Rb = +86 kJmol-1

ΔH a I = +107 kJmol-1

ΔH latt Rbl = +609 kJmol-1

IE Rb = +402 kJmol-1

I(g) + e I- (g) ; ΔH = -314 kJmol-1

Calculation using Born Haber cycle

Rb(s) + ½ l2(s) Rbl(s)◦ ΔH2 ΔH4 ΔH1

Rb(g) l(g) ΔH6

◦ ΔH3 ΔH5

Rb+ (g) I-(g) Route 2

Born Haber cycle

By Hess’s law,Heat change in route 1=Heat change in route 2ΔH1= ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6

= +86+402+107+(-314)+(-609)

= -328 kJmol-1

Hence ΔHf RbI = -328 kJmol-1

Find ΔH from Born Haber cycle

Let’s do the exercise !