Cindy Spangler
Enthalpy of Reaction
March 23, 2007
The enthalpy of reaction occurs as a result of reactants forming products. In this process, energy
is released or absorbed by the chemical process in order to make the products of interest. The
enthalpy change is measured in constant pressure environment and will provide the data
necessary to change easily obtainable and controllable experiments into data that is collected in
different conditions.
Model 1 A SIMPLE CHEMICAL PROCESS
When two hydrogen atoms form hydrogen gas, the rH for the reaction would be negative
because the process would be spontaneous and it would be at the minimum enthalpy. When the
reaction is reversed and the bond in the hydrogen gas is broken, this would have the same
magnitude of the enthalpy change but opposite sign.
Model 2: Hess’s Law
Making the assumption that the enthalpy of reaction is unaffected by the number of steps that
occurs in a reaction, we are able to calculate the unknown value for enthalpy of reaction for a
reaction if we know the enthalpies for the intermediate steps. Where
 H
r
products
  r H reac tan ts
eq (1)
This can be shown in the formation of nitrogen dioxide.
Example:
Reaction 1:
N2(g) + O2(g)  2 NO
rH1 = 180 kJ
Reaction 2:
2 NO(g) + O2(g)  2 NO2
rH2 = -112 kJ
Reaction 3:
N2(g) + 2 O2(g)  2 NO2
rH3 = 68 kJ
Reaction 1 and reaction 2 are summed together to find the enthalpy of reaction for reaction 3.
This gives us rH=-68 kJ for the formation of nitrogen oxide gas. It was obtained by
 r H1   r H 2   r H 3
eq(2)
This same method can be used to find the enthalpy of reaction 1 by equation 3.
 r H1   r H 3   r H 2
eq(3)
This can give us a guideline on unobtainable enthalpies if we know the steps that form the final
reaction.
Model 3: Some Chemical Reactions at 298 K.
 H
The enthalpy of formation, r f , is for the formation of one mole of products from its elements
in the stable state at 1 atm and 298 K. For a reaction equation to be valid all reactants should be
in the stable state at 1 atm and 298 K. Equation 4 is invalid because carbon is a gas.
Mg (s) + C (g) + 3/2 O2 (g)  MgCO3 (s)
eq(4)
Another violation is that more than one mole of products are formed, shown in equation 5.
N2 (g) + O2 (g)  2 NO (g)
eq(5)
And the reaction for NH4Cl (aq) is given by
½ N2 (g) + 2H2 (g) + ½ Cl2 (g) → NH4Cl (aq)
The  r H f of elements in their stable state is given by 0 kJ/mole because they are the definition
of enthalpy of formation. As a way to generalize the use of enthalpy of formation we can use the
following equation to find the enthalpy of reaction for a chemical reaction.
P


R

 r H    r H f ( products) * moles products    r H f ( reacants) * molesreacants
i

eq(6)
i
Equation 6 combines both Hess’s law and the enthalpies of formation to find the enthalpy of
reaction for an equation. The equation has made assumptions of constant pressure, temperature
and that the enthalpy of formation is not affected by the number of steps.
ChemActivity T5
Model: A reaction at different Temperatures
In many given tables for chemical constants, the enthalpy of reaction are at a given temperature
but often times it is necessary to find what the change in the change of enthalpy is for different
temperatures. When a substance is heated, there is a change in enthalpy change, given by
equation 7:

H  C p  r H T 2   r H T 1

eq(7)
This difference occurs from the differences in temperature because  r H is dependent on the
change of temperature and the heat capacity. With this knowledge and that enthalpy is a state
function, we are able to construct a cycle that will provide a method of determining the d r H .
H  C p ( products) T 
Reactants
/Products

p
H T1  p H T 2
 r HT1

 r HT 2
T1
T2
H  C p ( reacant) T 

p
H T1  p H T 2

Figure 1: Cycle for finding the change of the enthalpy change
Using figure 1, we have the equation
 C p T products   r H T 1  C p ( reacant) T   r H T 2  0
eq(8)
Reduces to:
d r H   r C p dT
eq(9)
This is showing that the change of the enthalpy at two different temperatures is equal to the
change in the heat capacity at constant pressure over the temperature change. When equation 9
is integrated,  r H is found.
 r H   r C p (T2  T1 )
eq(10)
EXERCISE 1
Using equation 10, we are able to find the  r H  for the freezing of water at a temperature lower

than 273 K. We know the heat capacities of ice and water and  r H 273
K . This is found to be
(-5649 J/mol) for the freezing of water at 10 below 273 K. There is almost a difference of 400
J/mol for the change in temperature of freezing; a significant change in the enthalpy.
EXERCISE 2
When Cp is a function of temperature, we have to integrate with respect to the heat capacity. We
start with equation 9 and equation 11, which is the how heat capacity varies with temperature.
C p  a  bT  cT 2
eq(11)
 r C p  a  bT  cT 2
H2
T2
T2
H1
T1
T1
  d r H    r C p dT   (a  bT  cT 2 )dT
eq(12)
T2
T2
T2
T1
T1
T1
  r H 2   r H 1  a  dT  b  TdT  c  T 2 dT
  r H 2   r H 1  a(T2  T1 ) 
1 1
b 2
T2  T12  c   
2
 T2 T1 


Using this general expression for finding the change in the enthalpy change from two
temperatures, we find that  r H 2 =2.26 kJ/mol for the formation of hydrogen gas and oxygen gas
from water vapor when the temperature is at 500 K.
From this lab, we were able to use the idea of Hess’s law in order to combine the heat capacities
of the steps in formation of a final product. By doing this, we are able to estimate the change of
the change in enthalpy of formation and apply this value to processes that are different from
perfect lab conditions. Enthalpy is also a temperature dependent factor that is a state function.
As shown in figure 1, enthalpy is zero if done in a cyclic process.