Many chemical reactions occur at constant pressure rather than constant volume: any reaction in an open beaker will be under the constant 1 atmosphere of pressure. Gases expanding out of open reaction vessel are doing $p \Delta V$ work on the surroundings, but this work is lost from a practical point of view, since it is dissipated into the surroundings. When we do a calorimetry experiment at constant (atmospheric) pressure and measure the heat given off by the system, we don't know the $p \Delta V$ work done by the system on the surroundings.
$\Delta U = q + w = q - p \Delta V$
We want to measure $\Delta U$ for the reaction because it is a state function, with is history independent (unlike heat and work). To get around this problem we do a mathematical trick. We define a new state function called Enthalpy:
$H = U + p V$
Since internal energy, pressure, and volume are all state functions, then enthalpy, H, must also be a state function.
In a constant pressure experiment the change in enthaply is given by
$\Delta H = \Delta(U + p V) = \Delta U + p \Delta V $
Remember $p$ is constant. Thus, we obtain
$\Delta H = q_p$
Enthalpy is the perfect state function to use for measuring the energy flowing out of a reaction as heat at constant pressure. Also, enthalpy, like internal energy is an extensive parameter.
50.0 mL of 1.0 M HCl at 25°C is mixed with 50.0 mL of 1.0 M NaOH also at 25°C in a constant atmospheric pressure calorimeter. After the reactants are mixed the temperature increases to 31.9°C. What is ΔH for this reaction? Assume the density of H2O is 1.0 g/mL.
H+(aq) + OH-(aq) → H2O(l)
First let's calculate the heat evolved?
$q = - m C_s \Delta T$
We have a negative sign in this equation because heat transfer is out of the system. Next we calculate ΔH per mole of H2O(l) produced.
Thus we have
H+(aq) + OH-(aq) → H2O(l) + 58 kJ
There are 5.8 kJ of heat evolved at constant pressure. In other words, ΔH = - 58 kJ.
Because enthalpy is a state function we can always write for a chemical reaction that
ΔH = Hproducts - Hreactants
Some other characteristics of enthalpy are...
If the reaction is reversed then the sign of ΔH is also reversed. For example, for the reaction
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
the ΔH is -802 kJ, that is, it is exothermic.
In contrast, the reverse reaction
CO2(g) + 2 H2O(g) → CH4(g) + 2 H2O(g)
has a ΔH of +802 kJ, that is, it is now endothermic.
ΔH is extensive. If the coefficients in a balanced reaction are multiplied by an integer, then the value of ΔH is multiplied by the same integer. For example,
Xe(g) + 2 F2(g) → XeF4(g) ΔH = -251 kJ.
2 Xe(g) + 4 F2(g) → 2 XeF4(s) ΔH = -502 kJ.
Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is independent of the pathway. This leads us to Hess' Law...
For example, consider the following reaction
|N2(g) + 2O2(g) → 2 NO2(g)||ΔH = 68 kJ.|
This reaction can also be carried out in two distinct steps,
|N2(g) + O2(g) → 2 NO(g)||ΔH1 = 180 kJ.|
|2 NO(g) + O2(g) → 2 NO2(g)||ΔH2 = - 112 kJ.|
|N2(g) + 2O2(g) → 2 NO2(g)||ΔHtotal = ΔH1 + ΔH2 = 68 kJ.|
Determine ΔH for
|Cgraphite(s) → Cdiamond(s)|
|Cgraphite(s) + O2(g) → CO2(g)||ΔH = - 394 kJ|
|Cdiamond(s) + O2(g) → CO2(g)||ΔH = - 396 kJ|
To solve this problem we simply switch the direction of the 2nd reaction (the diamond reaction) and add the two reactions together. That is,
|Cgraphite(s) + O2(g) → CO2(g)||ΔH1 = - 394 kJ|
|CO2(s) → Cdiamond(s) + O2(g)||ΔH2= + 396 kJ|
|Cgraphite(s) → Cdiamond(s)||ΔHtotal = ΔH1 + ΔH2 = 2 kJ.|
Hess' Law is great because it means we can get the ΔH for a reaction even if we can't easily measure it with a calorimeter. All we need are reactions of known ΔH that will add up to our unknown reaction ΔH. Before we start making tables of all known reaction enthalpies we need to adopt a reference state for the enthalpies of substances.
Thus, we define
For example, the formation of CO2(g) from it's elements, carbon and oxygen, in their standard states is written
|Cgraphite(s) + O2(s) → CO2(s)||ΔH = - 393.5 kJ|
therefore, we define the ΔH for this reaction as the Standard Enthalpy of formation for CO2(g)
ΔHf° (CO2(g)) = -393.5 kJ/mole.
Another example: the formation of H2O(l) from it's elements, hydrogen and oxygen, in their standard states is written
|H2(g) + 1/2 O2(g) → H2O(l)||ΔH = - 286 kJ|
therefore, the Standard Enthalpy of formation for H2O(l) is
ΔHf° (H2O(l)) = - 286 kJ/mole.
We use the coefficient 1/2 in front of O2 because we interested in heat of formation for 1 mole of H2O(l).
Obviously, any element that is already in its standard state will have a ΔHf° of zero.
Here are some examples of ΔHf° values.
|CO2(g)||carbon dioxide||- 393.5|
Using all the ΔHf° for the reaction's reactants and products you can calculate ΔH° for a reaction using
ΔH°rxtn = Σ n ΔHf° (products) - Σ n ΔHf° (reactants)
where n and m are the number of moles of each product and reactant, respectivelty, involved in the reaction.
What is ΔH°rxtn for
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH°rxtn = [3ΔHf° (CO2(g)) + 4ΔHf° (H2O(g))] - [ΔHf° (C3H8(g)) + 5ΔHf° (O2(g))]
ΔH°rxtn = [3 (-393.5 kJ/mole) + 4 (-241.8 kJ/mole) ] - [ (-103.85 kJ/mole)]= -2220 kJ
The heat of formation of CO2(g) and H2O(l) are -394 kJ/mole and -285.8 kJ/mole, respectively. Using the data for the following combustion reaction, calculate the heat of formation of C3H4(g).
|C3H4(g) + 4 O2(g) → 3 CO2(g) + 2 H2O(l)||ΔH = -1939.1 kJ|
To solve this problem we can start by reversing the reaction above to make C3H4(g) a product and then add together the reactions for the heat of formation of CO2(g) and H2O(l)
|3 CO2(g) + 2 H2O(l) → C3H4(g) + 4 O2(g)||ΔH1 = + 1939.1 kJ|
|3 C(s) + 3 O2(g) → 3 CO2(g)||3 ΔHf = 3 (-394 kJ/mole)|
|2 H2(g) + 2 (1/2) O2(g) → 2 H2O(l)||2 ΔHf° = 2 (-285.8 kJ/mole)|
|3 C(s) + 2 H2(g) → C3H4(g)||ΔHf° = 185.5 kJ/mole|
Chemisty, The Central Science, 10th Ed.
5.31, 5.33, 5.35, 5.37, 5.39, 5.41, 5.43, 5.45, 5.53, 5.55, 5.57, 5.59, 5.61, 5.63, 5.65, 5.67, 5.69, 5.71, 5.73, 5.75, 5.77
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