ENTHALPY (H)Accounts for heat flow in chemical reactions

happening at constant pressure when work is performed due to P-V change.

The enthalpy of a substance is its internal energy plus a term that takes into account the pressure & volume of the substance.

H = U + PV

We cannot calculate the actual value of H. Instead, we can calculate the change in enthalpy.

ΔH = ΔU + PΔV (since W = - PΔV)

(Equation 1)

ΔH = ΔU – W

In a process carried out at constant pressure:

ΔU = qp + W or (Equation 2)

(Equation 1) = (Equation 2) ΔH = qp (at constant pressure)

qp= ΔU - W

ONLY at constant Pressure:

ΔH = Qp

For this reason, the term “ heat of reaction” and “change in enthalpy” are used interchangeably for rxns studied at constant pressure.

In summaryAt constant volume At constant pressure

ΔU measures heat lost/gained.

ΔH measures heat lost/gained.

The difference between ΔU & ΔH is very small. ΔH is generally satisfactory to use.

ExerciseUnder standard conditions, 1 mol CO is burnt in a sealed flask w/ constant volume and 281.75 kJ energy is released. The same amount of CO is burnt under the same conditions in a flask w/ frictionless movable piston, 283kJ of energy is released. What are the values of ΔH, ΔU, and w for both conditions?

Solution:1st condition--constant volume, w=0ΔU=qv = -281.75 kJ

ΔH=ΔU-w (w=0)- ΔH= ΔU=-281.75 kJ, w=0

2nd condition--constant pressure, ΔH= qp= -283kJ

ΔU=-281.75 kJ,ΔH=ΔU-w, w=ΔU-ΔH, w= - 281.75 +

283kJ= + 1.25 kJ

For a chemical rxn:

ΔH = Hproducts –Hreactants

When Hproducts > Hreactants

-ΔH is (+).-Heat will be absorbed by the system.-Reactants are more stable than the products.-reactants have stronger bonds than the products.

When Hproducts < Hreactants

-ΔH is (-).-Heat will be released by the system.-Reactants are less stable than the products.-Products have stronger bonds than the reactants.

ENTHALPIES OF RXNS1) Enthalpy is an extensive property.ΔH is directly proportional to the amounts of

reactants consumed in chemical rxns.

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ

2CH4(g) + 4O2(g) --->2 CO2(g) + 4H2O(l) ΔH = -1780 kJ

(890x2)

ExerciseWhen 1 mole of methane is burned at

constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure.

Solution:CH4: 16 g/mol

When 16g of CH4 burns 890 kJ energy is released

5.8 g of CH4 ?

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? = - 320 kJ of energy

2) ΔH for a rxn is equal in magnitude but

opposite in sign, to ΔH for the reverse rxn.

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ

CO2(g) + 2H2O(l) ---> CH4(g) + 2O2(g) ΔH = + 890 kJ

3) ΔH depends on the state (gas, liquid, solid, crystalline structure)of the reactants and products.

- Therefore, we should write the states of the substances in the rxn equation.

- I. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH1 = -890 kJ

- II. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g) ΔH2 = ? kJ

- Which one is greater in amount?

ΔH1 ( -890 kJ)> ΔH2 (- 802 kJ):

H2O(l) H2O(g) ΔH= + 88 kJ

4) ΔH depends on temperature & pressure of

the rxn medium.- We will generally assume that the

reactants & products are both at the same temperature, 25 ° C, unless otherwise stated.

5) ΔH is a state function. Therefore, it

doesn’t depend on the rxn pathway.