Examples of ΔH calculations

Heat energy = mass (m) x specific heat capacity (c) x change in temperature (ΔT)

Specific heat capacity for water is 4.18kj kg-1

K-1

ΔH = m x c x ΔT

Note

State symbols

Moles

+ or -

Units K kj mol-1

or kJ kg (not grams)

Example 1

50 cm3 of 1.00 mol dm

-3 HCl(aq) added to 50 cm

3 of NaOH(aq).

Start temperature was 16.7 C

Maximum temperature was 23.5 oC

Calculate enthalpy change ΔH

Step 1 Equation

HCl(aq) + NaOH(aq)� NaCl(aq) + H2O(l)

Step 2 Moles of each

Moles = volume x morality

= 50/1000 X 1.00 = 5.00 x 10-2

mol

Step 3 heat evolved

2 x50cm3 = 100cm

3 = 100g = 0.100 kg

Temperature change is 6.8 K

Heat evolved = 0.100 X 4.18 x 6.8 = 2.84kj

ΔH 2.84 x ( 1/5.00 x10-2

) =-56.8 kj mol-1

Example 2:

0.690g ethanol (C2H5OH) was burned, it caused a temperature raise of 13.2 in 250g of water

Calculate ΔH

1. Equation

2. Moles of ethanol

3. Heat evolved and ΔH

Example 3

Four grams of sodium hydroxide pellets were dissolved in 100 cm3 of water in a simple

calorimeter. The temperature before adding the sodium hydroxide pellets was 25ºC. After

adding the pellets and stirring, it was 35ºC. Calculate the enthalpy of solution of sodium

hydroxide in kJ/mole.

(specific heat capacity of water = 4.2 J/K/g, Molar masses: Na=23, O=16, H=1))

Example 4

Bond enthalpy calculations

Hydrogenation of ethene

C2H4 + H2 � C2H6

Energy absorbed � Energy released to

To break bonds form bonds

C=C 612 C-C 348

C-H 412 x 4 C-H 412 x 6

H-H 436

2696kJ 2820kJ

ΔH = (2820-2696) = -124kJ mol-1

Example 5

N2H4 + =O2 � N2 + H2O

Example 6

Calculate the standard enthalpy change when 1 mole of methane is formed.

Enthalpies of combustion are:

C -393

H -286

CH4 -890

Step 1: equation

ΔHѲ

C(s) + 2H2(g) �CH4(g)

Step 2: energy cycle showing different routes

C + 2H2--------------�CH4

CO2 + 2H2O

Step 3: Use hess’ law to show the energy changes along the two routes:

ΔHѲ

C + 2 ΔHѲ

H2 = ΔHx + ΔH

Ѳ CH4

Rearrange the equation to find ΔHx

ΔHx =

ΔH

Ѳ C + 2 ΔH

Ѳ H2 - ΔH

Ѳ CH4

-393 + (2 x -286) – (-890) = -75 kJ mol-1

O2 O2 2O2