law

1. Chemistry 27/9/2022
Hess’s Low

2. Hess’s law
states that ‘the total enthalpy change in a
chemical reaction is independent of the route
by which the chemical reaction takes place as
long as the initial and final conditions are the
same’. The states of the reactants and products
must also be the same whichever route is
followed

3. Explaining Hess's Law:
Hess's Law is saying that if you convert
reactants A into products B, the overall enthalpy
change will be exactly the same whether you do
it in one step or two steps or however many
steps. If you look at the change on an enthalpy
diagram, that is actually fairly obvious.
A
C
B
H1
H2
H3
H3
H1+ H2
=

7. CH4 (g) + Cl2 (g) CH3Cl3(g) + HCl (g)
Hr

CH4 (g) Cl2 (g) CH3Cl3(g) + HCl (g)
KJmol -1 -74 o -83.7 -92.3
Hf

Enthalpy change of reaction from enthalpy changes of formation

8. CH4 (g) + Cl2 (g) CH3Cl3(g) + HCl (g)
C(s) + 2H2(g) + Cl2 (g)
-74.9
Hr

Hf
= -83.7
Hf
 =-92.3
The total enthalpy change of formationn for right hand
side = -83.7+-92.3 = -176 KJmol-1
standard enthalpy change of reaction=
+ 74.9 kjmol-1 + (-176)= -101.1 KJmol-1

9. Calculate the standard enthalpy change for the reaction:
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)
Solution
The relevant enthalpy changes of formation are:
[NaHCO3(s)] = −950.8 kJ mol−1
[Na2CO3(s)] = −1130.7 kJ mol−1
[CO2(g)] = −393.5 kJ mol−1
[H2O(l)] = −285.8 kJ mol−1
The enthalpy cycle is shown in Figure 6.9.
.
Hf
Hf
Hf
Hf

10. Figure 6.9: The enthalpy cycle for the decomposition of sodium hydrogencarbonate.
The dashed line shows the twostep
route

13. The reaction between ethene and hydrogen chloride
C2H4+HCL C2H5CL

15. +52.2 - 92.3 + ΔH = -109
Rearranging and solving:
ΔH = -52.2 + 92.3 - 109
ΔH = -68.9 kJ mol-1

16. 1.3Mg(S) +2AlCl3 (s) 3MgCl2 +2Al
HF

Mg ALCl
2
MgCl
2
Al(s)
Kjmol-1 -394 -286 -2877 0
Homework
Left side
3x394+2x286=1754
Right side
3x-2877=-8631
H =1754+_8631=_6877

17. Mg AlCl3 MgCl2 Al(s)
H KJmol 0 -705.6 -641 0
Calculate the enthalpy change of the following reaction
Mg AlCl3 MgCl2 Al(s)
HKJmol 0 -705.6 -641 0
3Mg(S) +2 AlCl3 3Mg Cl2 +3AL
Left side
H= 2X-705=1411
Right side
3x-641=-1924=-513.6 Kjmol-1

18. C2H5OH C2H4 H2O
HKJmol -277.1 52.2 -285
C2H5OH C2H4 +H2O
Left side
52.5+-285= -233.3
Right side
-277.1
277.1+-233.3= 43.8 Kjmol-1

19. Use the values for standard enthalpy of formation
to calculate standard enthalpy change for the
reactions
NH3(g) +HCl(g) NH4Cl(s)
ΔHf(NH3(g)) = -46.1 kJ mol-1
ΔHf(HCl(g)) = -92.3 kJ mol-1
ΔHf(NH4Cl(s)) = -314.4 kJ mol

20. Left side
46.1+92.3=138.4
Right side
-314.4
H =138.4 +-314.4=-176