Working Content > Thermodynamics and Statistical Physics > The 1st Law of Thermodynamics

Prerequisites:

• Chemical bonding
  
• Enthalpy

In expanding our idea of the concept of energy to include work done, we introduced the concept of enthalpy, a quantity that includes the amount of work that has to be done in order to maintain the system at a constant pressure. The difference between energy and enthalpy is particularly important in the case of chemical reactions between gases. To see how this works, let's work through a sample problem.

Consider the reaction corresponding to the splitting of water into hydrogen and oxygen gas in air: 2H2O  → 2H2 + O2. The binding energies of each (single or double) bond is shown in the figure at the right. Let's begin with a consideration of the chemical energies.   1. For the reaction of a single pair of water molecules shown, does energy have to be added to make the reaction go or will energy be produced when it goes? How much energy needs to be added or will be produced by the changes in the chemical bonding in the reaction?   2. If 1 mole of water molecules undergoes this reaction, how much energy (Joules/mole) will be required or produced?

3. If the reaction of 1 mole of water vapor molecules takes place in a cylinder at ~STP what will be the enthalpy change as a result of the reaction?   Initially the cylinder only contains water vapor molecules. Let's specifically say that the pressure p0 = 100 kPa and the temperature T is maintained at 300 K.

1. Considering only the chemical energy and ignoring thermal, the initial chemical energy is equal to the sum of the bond energies:

U_chemical = 4 x (-4.77 eV) = -19.08 eV

The chemical energies are negative because we are taking the 0 of PE to be when all the atoms are separated and at rest. To get these 6 atoms separated, we would have to put in +19.08 eV of energy. Therefore we had to start out at a negative value.

The final chemical energy is equal to the sum of the new bond energies:

U_chemical = -2 x (-4.52 eV) + (-5.03 eV) = -14.07 eV.

This is less negative than we started with, so we have to put a positive amount of energy in:

ΔU_chemical = U^f_chemical - Uⁱ_chemical = -14.07 eV - (-19.08 eV) = (+19.08 - 14.07) eV = 5.01 eV

This amount of energy needs to be added to split the water into hydrogen and oxygen. This agrees with our intuition, since we know that hydrogen gas burns explosively in air, combining with oxygen to give water vapor. Therefore, running the reaction the other way requires an input of energy.

2. To scale this up to a mole of water molecules, we need Avogadro's number of molecules: 6 x 10²³. Since we need 2 water molecules for each reaction, one mole of water molecules will have 3 x 10²³ reactions, and will require an energy input of

Q = (5.01 eV/molecule) x (3 x 10²³ molecules/mole) = 15 x 10²³ eV/mole.

Since 1 eV = 1.6 x 10^-19 J, we can convert this to Joules by multiplying by 1 = (1.6 x 10^-19 J)/(1 eV) giving 

Q = (15 x 10²³ eV/mole) x (1.6 x 10^-19 J)/(1 eV) = 2.4 x 10⁵ J = 240 kJ/mole.

(One kJ = 1 kiloJoules = 10³ J) 

3. To figure out the change in enthalpy we need to know the change in volume. Since we are starting out with pure water vapor at 1 atmosphere, we can use the ideal gas law to figure out the initial volume. We recall from chemistry that 1 mole of any gas occupies 22.4 liters (= 22.4 x 10³ cm³). 

After the reaction has taken place, we have the same number of molecules of hydrogen as we had of water, so 1 mole. We have half as many molecules of oxygen as we had of water, so 1/2 mole. The result of the reaction is to increase the number of gas molecules by 50% (from 1 mole to 1.5 moles). If the pressure is the same and the temperature is the same, since from pV = Nk_BT we get

V = Nk_BT/p

T and p stay the same but N increases by a factor of 1.5, so V must increase by a factor of 1.5. The change in volume is therefore 1/2 x 22.4 liters = 11.2 liters. 

The enthalpy change is

ΔH = ΔU + pΔV = 240 kJ/mole + (10⁵ N/m²) x (11.2 x 10³ cm³/mole)

The cm doesn't cancel the meters so we have to multiply by (1 m)/(10² cm) = 1 three times. This gives

ΔH = ΔU + pΔV 

     = 240 kJ/mole + (10⁵ N/m²) x (11.2 x 10³ cm³/mole) x (1 m/10² cm)³ 

     = 240 kJ/mole + 11.2 x 10^(5+3-6) N⋅m/mole 

     = 240 kJ/mole + 1120 J/mole

since a N⋅m equals a Joule. Converting both to kJ we get

ΔH = ΔU + pΔV  = (240 + 1.1) kJ/mole

It's clear that for this example, most of the enthalpy change comes from the change in chemical energy. The work term is effectively negligible.

Joe Redish 1/26/16