Industrial Chemistry

Hess’s law

Index

Hess’s Law and its experimental verification

Hess’s Law calculations, 4 examples.

Hess’s Law

Hess’s Law and calculationsHess’s law states that

“enthalpy change is independent of the route

taken”

Verification of Hess’s Law

The conversion of solid NaOH to NaCl solution can be achieved by two possible routes. Route 1 is a single-step process, (adding HCl (aq) directlyto the solid NaOH) and Route 2 is a two-step process (dissolve the solid NaOH in water, then adding the HCl(aq)) All steps are exothermic.

If Hess’s Law applies, the enthalpy change for route 1 must be the same as the overall change for route 2.

H = enthalpy change

H 1 = H 2 H 3+

NaOH (s) NaCl (aq)Route 1H 1

NaOH (s)

NaOH (aq)

NaCl (aq)Route 2

H 2 H 3

Route 2 H 2 + H 3

50 ml H2O 50 ml HCl

then

2.50g of NaOH added to a dry, insulatedbeaker. Before adding the acid, its temperatureis recorded. The final temperature after adding the acid is also recorded.

Knowing the specific heat capacity for water, it is then possible to calculate the Enthaply change for this reaction.

1. 2.50g of NaOH added to a dry, insulatedbeaker. 2. Before adding the water, its temperatureis recorded. The final temperature riseafter adding the water is also recorded. 3. Now add the acid, again, recording the finaltemperature. Use the equation below to calculate H2 and H 3

H 2

H 1 = H 2 + H 3 will verify Hess’s Law

Route 1 H 1

50 ml 1mol l-1 HCl

H 1 = c m TH = c m T

H 3

Verification of Hess’s Law

Hess’s Law CalculationsHess’s Law can be used to calculate enthalpy changes that cannot be directly measured by experiment.

Route 1 Route 1 cannot be carried outin a lab, as carbon and hydrogen will not combine directly.

The enthalpy of combustion reactions can act as a stepping stone which enables a link with carbon and hydrogen (the reactants) with propene (the product)

Route 2a involves the combustion of both carbon and hydrogen

3C (s) + 3O2 (g) 3CO2 (g)

Route 2b involves the reverse combustion of propane

3CO2 (g) + 3H2O(l) C3H6 (g) + 4½O2 (g)

3H2 (g) + 1½O2 (g) 3H2O (l)and

3C (s) + 3H2 (g) C3H6 (g)

3CO2 (g) + 3H2O (l)

Route 2a Route 2b

H 1

3C (s) + 3H2 (g) C3H6 (g)Route 1

3CO2 (g) + 3H2O(l)

Route 2a Route 2b

H 1 = H 2a H 2b+

H1

Route 2a H c C= -394 kJ mol –1 H c H = -286 kJ mol –1

H 2a = -2040 kJ

H 2a= -( x 394) = -1182 kJ mol -1 -( x 286) = -858 kJ mol -1+3 3

3 3

Route 2b

H 2b = + 2058.5 kJ (note the reverse sign)

H c Propene = -2058.5 kJ mol –1

= -2040 kJ +( 2058.5) = + 18.5 kJ mol -1

Example 1

Alternative approach for example 1

3C(s) + 3H2 (g) C3H6(g)

C(s) + O2 (g) CO2(g) ΔHo298 = -394 kJ mol-1

H2(g) + ½O2(g) H2O(g) ΔHo298 = -286 kJ mol-1

C3H6(g) + 4½O2(g) 3H2O(g) + 3CO2(g)ΔHo298 = -2058.5 kJ mol-1

ΔHf = ?

Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also.3C(s) + 3O2 (g) 3CO2(g) ΔHc = 3 x -394 kJ

3H2(g) + 1½O2(g) 3H2O(g) ΔHc = 3 x -286 kJ

3H2O(g) + 3CO2(g) C3H6(g) + 4½O2(g) ΔHc = +2058.5 kJ

Equation has been reversed; (enthalpy now has opposite sign)

3C(graphite) + 3O2 (g) 3CO2(l) ΔHc = 3 x -394

3H2(g) + 1½O2(g) 3H2O(l) ΔHc = 3 x -286

3H2O(g) + 3CO2(g) C3H6(g) + 4½O2(l) ΔHc = +2058.5

Now add the equations and also the corresponding enthalpy values 3C(s) + 3H2(g) C3H6(g)

ΔHf = (3 x -394) + (3 x -286) + (+2058.5)

ΔHf = +18.5 kJ mol-1

Calculate the enthalpy change for the reaction:

The products of combustion act as a stepping stone which enables a link to be made with benzene and hydrogen (the reactants) with cyclohexane (the product).

C6H6(l) + 3H2 (g) C6H12(l) Route 1 ?

6H2O(l) + 6CO2(g)

Route 2a Route 2b

Route 2a involves the combustion of both benzene and hydrogen

C6H6(g) + 7½O2(g) 3H2O(l) + 6CO2(g) H c benzene = -3268 kJ mol –1

H2(g) + ½O2(g) H2O(l) H c hydrogen = -286 kJ mol –1

Route 2b involves the reverse combustion of cyclohexane

6CO2 (g) + 6H2O(l) => C6H12 (g) + 7½O2 (g)H c cyclohexane = -3924 kJ mol –1

H 1 = H 2a + H 2b = ( -3268 + ( x - 286)) + 3924 = 202 kJ mol -1 3

3

Example 2

C6H6(l) + 3H2(g) C6H12(l) ΔHf = ?

Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also.C6H6(l) + 7½O2 (g) 6CO2(g) + 3H2O(g) ΔHc = -3268 kJ

3H2(g) + 1½O2(g) 3H2O(l) ΔHc = 3 x -286 kJ

6H2O(g) + 6CO2(g) C6H12(g) + 9O2(g) ΔHc = +3924 kJ

Equation has been reversed; (enthalpy now has opposite sign)

C6H12(l) + 9O2 (g) 6H2O(l) + 6CO2(g) ΔH = -3924 kJmol-1

H2(g) + ½O2(g) H2O(l) ΔH = -286 kJmol-1

C6H6(l) + 7½O2(g) 3H2O(l) + 6CO2(g) ΔH = -3268 kJmol-1

Alternative approach for example 2

Now add the equations and also the corresponding enthalpy values

C6H6(l) + 3H2(g) C6H12(l)

ΔH f = -3268 + (3 x -286) + 3924

ΔH f = -202 kJ mol-1

C6H6(l) + 7½O2 (g) 6CO2(g) + 3H2O(g) ΔHc = -3268

3H2(g) + 1½O2(g) 3H2O(g) ΔHc = 3 x -286

6H2O(g) + 6CO2(g) C6H12(l) + 9O2(g) ΔHc = +3924

3. Use the enthalpy changes of combustion shown in the table to work out the enthalpy change of formation of ethyne, C2H2.

Substance C(graphite) H2(g) C2H2(g)

ΔHo(combustion) -394 kJmol-1 -286 kJmol-1 -1299 kJmol-1

“Using the Second method”

“Required” equation, 2C(s) + H2(g) C2H2(g) ΔHf = ?

• 2C(s) + 2O2(g) 2CO2(g) ΔHc = 2 x -394 kJ mol-1

•H2(g) + ½O2(g) H2O(g) ΔHc = -286 kJ mol-1

C2H2(g) + 2½O2(g) 2CO2(g) + H2O(g) ΔHc = -1299 kJ mol-1

•2CO2(g) + H2O(g) C2H2(g) + 2½O2(g) ΔHc = +1299 kJ mol-1

ΔHf = (2 x -394) + (-286) + 1299

ΔHf = +225 kJ mol-1

Adding “bulleted” equations gives us 2C(graphite) + H2(g) C2H2(g)

4. Using the following standard enthalpy changes of formation, ΔHo

f / kJmol-1 : CO2(g), -394; H2O(g), -286; C2H5OH(l), -278 calculate the standard enthalpy of combustion of ethanol i.e. the enthalpy change for the reaction C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔHc = ?

2C(s) + 3H2(g) + ½O2(g) C2H5OH(l) ΔHf = -278 kJ

● 2C(s) + 2O2(g) 2CO2(g) ΔHf = 2 x -394 kJ

● 3H2(g) + 1½O2(g) 3H2O (g)ΔHf = 3 x -286 kJ

● C2H5OH(l) 2C(s) + 3H2(g) + ½O2(g) ΔHf = +278 kJ

Add bulleted equationsC2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Solve equation for ΔHc ΔHc = + 278 + (2 x -394) + (3 x -286)

ΔHc = -1368 kJ mol-1

“Using the Second method”