5.4 Hess’s Law

Hess’s Law

Hess was a Russian chemist. This law was named after him in 1840. Hess’s law confirms that ΔH^o is state function which does not depend on the path of the chemical reaction. Hess’s law states that regardless of the multiple steps that the reaction can take, the total enthalpy change for the reaction is the sum of all changes.

The videos below illustrate the concept of Hess’s Law and some calculations examples:

Hess Law Chemistry Problems – Enthalpy Change – Constant Heat of Summation

Hess’s Law and Heats of Formation

Hess’s Law

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Examples:

1. Calculate the enthalpy for this reaction ΔH°:

Given the following thermochemical equations:

Step 1:

Determine what we must do to the three given equations to get our target equation:

a) First equation: flip it so as to put C₂H₂ on the product side
b) Second equation: multiply it by two to get 2C
c) Third equation: do nothing. We need one H₂ on the reactant side and that’s what we have.

Step 2:

Rewrite all three equations with changes applied:

Notice that the ΔH values changed as well.

Step 3:

Checking for what can be canceled:

2CO₂ ⟶ first & second equation

H₂O ⟶ first & third equation

⁵⁄₂O₂ ⟶ first & sum of second and third equation

Step 4:

Add up ΔH^o values for our answer:

ΔH°  =  +1299.5 kJ + (−787 kJ) + (−285.8 kJ) =+226.7 kJ

2. Calculate the enthalpy of the following chemical reaction:

CS₂(l) + 3O₂(g) ⟶ CO₂(g) + 2SO₂(g)

Given:

What to do to the data equations:

a) Leave equation 1 untouched (want CO₂ as a product)
b) Multiply second equation by 2 (want to cancel 2S, also want 2SO₂ on product side)
c) Flip 3rd equation (want CS₂ as a reactant)

Step 2:

The result:

Step 3:

Add the three revised equations. C and 2S will cancel.

ΔH^o = [-393.5 kJ/mole – 593.6 kJ/mole -87.9 kJ/mole] = 1075.0 kJ/mole

3. Given the following data:

Find the ΔH of the following reaction:

C(s, gr) + O₂(g) ⟶ CO₂(g)

Step 1:

Analyze what must happen to each equation:

a) First equation ⟶ flip it (this put the CO₂ on the right-hand side, where we want it)

b) Second equation⟶ do not flip it, divide through by two (no flip because we need to cancel the SrO, divide by two because we only need to cancel one SrO)

c) Third equation ⟶ flip it (to put the SrCO₃ on the other side so we can cancel it), divide by two (since we need to cancel only one SrCO₃)

Notice that what we did to the third equation also sets up the Sr to be cancelled. Why not also multiply first equation by two (to get 2SrO for canceling)? Because we only want one CO₂ in the final answer, not two. Notice also that I ignored the oxygen. If everything is right, the oxygen will take care of itself.

Step 2:

Apply all the above changes (notice what happens to the ΔH values):

Here is a list of what gets eliminated when everything is added:

SrCO₃, SrO, Sr, ¹⁄₂O₂

The last one comes from ³⁄₂O₂ on the left in the third equation and ¹⁄₂O₂ on the right in the second equation.

Step 4:

Add the equations and the ΔH values:

+234 + (+592) + (−1220) = −394

4. Given the following information:

Calculate the enthalpy change for the reaction below:

Step 1:

Analyze what must happen to each equation:

a) First equation ⟶ flip; multiply by ³⁄₂ (this gives 3NO₂ as well as the 3NO which will be necessary to get one NO in the final answer)

b) Second equation ⟶ divide by 2 (gives two nitric acid in the final answer)

c) Third equation ⟶ flip (cancels 2NO as well as nitrogen)

Step 2:

Notes on the oxygen atoms:

a) step 1a above puts ³⁄₂O₂ on the right
b) step 1b puts ⁵⁄₂O₂ on the left
c) step 1c puts ²⁄₂O₂ on the right

In addition, a and c give ⁵⁄₂O₂ on the right to cancel out the ⁵⁄₂O₂ on the left.

Step 3:

Apply all the changes listed above:

Step 4:

Add the equations and the ΔH values:

+174 + (−128) + (−183) = −137 kJ

Use of Enthalpy Diagram using Hess’s Law:

Enthalpy diagram

Comparison and interpretation of enthalpy changes is materially aided by a graphical construction in which the relative enthalpies of various substances are represented by horizontal lines on a vertical energy scale. The zero of the scale can be placed anywhere, since energies are always arbitrary; it is generally most useful to locate the elements at zero energy, which reflects the convention that their standard enthlapies of formation are zero.

This very simple enthalpy diagram for carbon and O2

Figure 5.26 Enthalpy Diagram (1)

oxygen and its two stable oxides shows the changes in enthalpy associated with the various reactions this system can undergo. Notice how Hess’ law is implicit in this diagram; we can calculate the enthalpy change for the combustion of carbon monoxide to carbon dioxide, for example, by subtraction of the appropriate arrow lengths without writing out the thermochemical equations in a formal way.

The zero-enthalpy reference states refer to graphite, the most stable form of carbon, and gaseous oxygen. All temperatures are 298 K.

Figure 5.27 Enthalpy Diagram (2)

This enthalpy diagram for the hydrogen-oxygen system shows the known stable configurations of these two elements. Reaction of gaseous H2 and O2 to yield one mole of liquid water releases 285 kJ of heat (1).

If the H2O is formed in the gaseous state, the energy release will be smaller.

Notice also that…

The heat of vaporization of water (2) (an endothermic process) is clearly found from the diagram.

Hydrogen peroxide H2O2, which spontaneously decomposes into O₂ and H₂O, releases some heat (3) in this process. Ordinarily this reaction is so slow that the heat is not noticed. But the use of an appropriate catalyst can make the reaction so fast that it has been used to fuel a racing car.

Figure 5.28 Race Cars

Ref: commonswikimedia.org

Why you can’t run your car on water ?

You may have heard the venerable urban legend, probably by now over 80 years old, that some obscure inventor discovered a process to do this, but the invention was secretly bought up by the oil companies in order to preserve their monopoly. The enthalpy diagram for the hydrogen-oxygen system shows why this cannot be true — there is simply no known compound of H and O that resides at a lower enthalpy level.

Figure 5.29 Enthalpy Diagram (3)

Enthalpy diagrams are especially useful for comparing groups of substances having some common feature. This one shows the molar enthalpies of species relating to two hydrogen halides, with respect to those of the elements. From this diagram we can see at a glance that the formation of HF from the elements is considerably more exothermic than the corresponding formation of HCl. The upper part of this diagram shows the gaseous atoms at positive enthalpies with respect to the elements. The endothermic processes in which the H2 and the dihalogen are dissociated into atoms can be imagined as taking place in two stages, also shown. From the enthalpy change associated with the dissociation of H₂ (218 kJ mol–1), the dissociation enthalpies of F₂ and Cl₂ can be calculated and placed on the diagram.