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enthalpy - what is it?
enthalpy - what is it?
All reactions involve energy - enthalpy is the amount of heat energy there is a chemical reaction. The change in enthalpy can be can be negative or positive in value, and is shown by ΔH°r
All reactions involve energy - enthalpy is the amount of heat energy there is a chemical reaction. The change in enthalpy can be can be negative or positive in value, and is shown by ΔH°r
Exothermic or Endothermic reactions?
Exothermic or Endothermic reactions?
In a thermochemical equation the enthalpy change is given:
In a thermochemical equation the enthalpy change is given:
It is important to realise that changing states of matter are thermochemical processes, not reactions, and involve release or absorption of heat energy.
It is important to realise that changing states of matter are thermochemical processes, not reactions, and involve release or absorption of heat energy.
This leads onto the next point - that of bond enthalpies.
This leads onto the next point - that of bond enthalpies.
Bond enthalpies - what are they?
Bond enthalpies - what are they?
Bond strength = bond energy = bond enthalpy, with enthalpy being the preferred choice.
Bond strength = bond energy = bond enthalpy, with enthalpy being the preferred choice.
THE BOND DISSOCIATION ENTHALPY IS THE ENERGY REQUIRED TO BREAK APART ONE MOLE OF ATOMS IN THE GASEOUS STATE INTO SEPARATE ATOMS.
THE BOND DISSOCIATION ENTHALPY IS THE ENERGY REQUIRED TO BREAK APART ONE MOLE OF ATOMS IN THE GASEOUS STATE INTO SEPARATE ATOMS.
This is obviously all well and good for diatomic molecules, such as O₂ or Cl₂ in which there is only 1 bond between 2 identical atoms. However for most compounds there is more than one type of atom and multiple bonds, eg H₂SO₄ or CH₄, where it takes differing amounts of energy to remove the identical bonds.
This is obviously all well and good for diatomic molecules, such as O₂ or Cl₂ in which there is only 1 bond between 2 identical atoms. However for most compounds there is more than one type of atom and multiple bonds, eg H₂SO₄ or CH₄, where it takes differing amounts of energy to remove the identical bonds.
In methane although there are 4 C-H bonds when one is broken apart the energy required to break apart the next C-H bond is different - bond strengths are affected by the environment they are in, including the amount of bonds and what other atoms are bonded.
In methane although there are 4 C-H bonds when one is broken apart the energy required to break apart the next C-H bond is different - bond strengths are affected by the environment they are in, including the amount of bonds and what other atoms are bonded.
FOR THESE REASONS BOND ENTHALPIES ARE GIVEN AS AVERAGE VALUES.
FOR THESE REASONS BOND ENTHALPIES ARE GIVEN AS AVERAGE VALUES.
Calculations involving enthalpy changes
Calculations involving enthalpy changes
Here is another example calculating ΔH°r
Here is another example calculating ΔH°r
In this example the ΔH°r is known and the bond enthalpy for a particular bond has to be calculated.
In this example the ΔH°r is known and the bond enthalpy for a particular bond has to be calculated.
Calculations - thermochemical equations
Calculations - thermochemical equations
Let us look at the following thermochemical equation:
Let us look at the following thermochemical equation:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ∆H = -890.3 kJ mol -¹
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ∆H = -890.3 kJ mol -¹
From this it can be determined that the combustion of 1 mole of methane releases 890.3 kJ mol -¹ . 1 mole equates to 16 grams.
From this it can be determined that the combustion of 1 mole of methane releases 890.3 kJ mol -¹ . 1 mole equates to 16 grams.
If we wanted to know what the total heat released is when 2.5 moles of methane are burnt we would use the first equation in the image above, and rearrange it to find Q (total energy released):
If we wanted to know what the total heat released is when 2.5 moles of methane are burnt we would use the first equation in the image above, and rearrange it to find Q (total energy released):
Q = ΔHr x n = -890.3 x 2.5 = 2,225.75 KJ
Q = ΔHr x n = -890.3 x 2.5 = 2,225.75 KJ
As we know the relationship between moles and mass (equation 2 in the above image) it is straight forward to work out how much methane this in grams. We just need to convert moles to mass:
As we know the relationship between moles and mass (equation 2 in the above image) it is straight forward to work out how much methane this in grams. We just need to convert moles to mass:
m = n x M = 2.5 x 16 = 40g
m = n x M = 2.5 x 16 = 40g
From the above thermochemical equation it can also be determined that 890.3 KJ of energy is released when 2 moles of oxygen combust with the 1 mole of methane. From this we can determine the amount of energy released if 5 moles of oxygen is used to combust the methane:
From the above thermochemical equation it can also be determined that 890.3 KJ of energy is released when 2 moles of oxygen combust with the 1 mole of methane. From this we can determine the amount of energy released if 5 moles of oxygen is used to combust the methane:
2 moles of O₂ : 890.3 KJ released
2 moles of O₂ : 890.3 KJ released
1 mole of O₂ = 445.15 KJ released
1 mole of O₂ = 445.15 KJ released
5 moles = 445.15 x 5 = 2225.75 KJ
5 moles = 445.15 x 5 = 2225.75 KJ
Provided the ΔH and Q (energy released is known ) we can also determine the number of moles of a chemical species that was used. In this case we would rearrange the equation so that n = Q / ΔHr Equally if we know the moles we can determine the mass using m = n x M) and it we know the mass we can determine the moles (n = m/M) - this is often the first step as moles are much more useful . Remember the unit for ΔHr is kJ mol -¹.
Provided the ΔH and Q (energy released is known ) we can also determine the number of moles of a chemical species that was used. In this case we would rearrange the equation so that n = Q / ΔHr Equally if we know the moles we can determine the mass using m = n x M) and it we know the mass we can determine the moles (n = m/M) - this is often the first step as moles are much more useful . Remember the unit for ΔHr is kJ mol -¹.
Lets assume 735 KJ of energy is released in the reaction above and we want to determine the mass of H₂O produced:
Lets assume 735 KJ of energy is released in the reaction above and we want to determine the mass of H₂O produced:
n (H₂O) = Q / ΔHr
n (H₂O) = Q / ΔHr
n (H₂O) = 735 / 890.3 = 0.826
n (H₂O) = 735 / 890.3 = 0.826
As there are 2 moles (H₂O) in this reaction then:
As there are 2 moles (H₂O) in this reaction then:
n(H₂O) = 0.826 x 2 = 1.65
n(H₂O) = 0.826 x 2 = 1.65
Therefore the mass (H₂O) produced is n x M
Therefore the mass (H₂O) produced is n x M
mass = 1.65 x 18 = 29.7 grams
mass = 1.65 x 18 = 29.7 grams
WORKED EXAMPLES
WORKED EXAMPLES
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