Hess’ Law Starter: Why is it easy to measure enthalpies of combustion but often difficult to measure enthalpies of formation?
Oct 07, 2014
Hess’ Law
Starter: Why is it easy to measure
enthalpies of combustion but often
difficult to measure enthalpies of
formation?
Problem
• How would you measure the enthalpy
change for the formation of methane from
graphite and hydrogen gas?
– problem: graphite doesn’t react with
hydrogen gas to form methane
• C(graphite) + 2H2(g) CH4(g)
Hess’s Law
StartFinish
Both lines accomplished the same result,
they went from start to finish.
Net result = same.
Hess’s Law
• When reactants are converted to products,
the change in enthalpy is the same
whether the reaction takes place in one
step or a series of steps.
Remember
• P4(s) + 3O2(g) P4O6(s) ΔH = -1640.1 kJ
• Transform the equation so that P4O6(s) is on the
reactant side.
• P4O6(s) P4(s) + 3O2(g) -ΔH = 1640.1 kJ
• Note: if the equation switches, the sign of ΔH
does as well.
Example
• What is the change in enthalpy for the
formation of methane from graphite and
hydrogen gas?
– problem: graphite doesn’t react with
hydrogen gas to form methane
• C(graphite) + 2H2(g) CH4(g)
Solution
• Find reactions that do occur that link
graphite to methane through a series of
steps.
If you formed the products from their elements you should need the same
amounts of every substance as if you formed the reactants from their elements.
Enthalpy of formation tends to be an exothermic process
Enthalpy of reaction from enthalpies of formation
Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
DH2 1 x C=C bond @ 611 = 611 kJ
4 x C-H bonds @ 413 = 1652 kJ
1 x H-H bond @ 436 = 436 kJ
Total energy to break bonds of reactants = 2699 kJ
Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
DH2 1 x C=C bond @ 611 = 611 kJ
4 x C-H bonds @ 413 = 1652 kJ
1 x H-H bond @ 436 = 436 kJ
Total energy to break bonds of reactants = 2699 kJ
DH3 1 x C-C bond @ 346 = 346 kJ
6 x C-H bonds @ 413 = 2478 kJ
Total energy to break bonds of products = 2824 kJ
Applying Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ
Sample calculation
Calculate the standard enthalpy change for the following reaction, given that the
standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,
+33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element
2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
PRODUCTS REACTANTS
[ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ]
DH°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0
ANSWER = - 252 kJ
Enthalpy of reaction from enthalpies of formation
DH = DHf of products – DHf of reactants
Enthalpy of reaction from enthalpies of combustion
If you burned all the products you should get the same amounts of oxidation products
such a CO2 and H2O as if you burned the reactants.
Enthalpy of combustion is an exothermic process
Enthalpy of reaction from enthalpies of combustion
Sample calculation
Calculate the standard enthalpy of formation of methane; the standard enthalpies of
combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .
C(graphite) + 2H2(g) ———> CH4(g)
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
REACTANTS PRODUCTS
[ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4]
DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890)
ANSWER = - 76 kJ mol-1
DH = DHc of reactants – DHc of products
Determine the heat of reaction for the reaction:
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) 2NO(g) DH = 180.6 kJ
N2(g) + 3H2(g) 2NH3(g) DH = -91.8 kJ
2H2(g) + O2(g) 2H2O(g) DH = -483.7 kJ
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g) C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) DH = -1401 kJ
C2H6(g) + 3.5 O2(g) 2CO2(g) + 3H2O(l) DH = -1550 kJ
H2(g) + 0.5 O2(g) H2O(l) DH = -286 kJ
Consult your neighbor if necessary.