Tutorial-Hess's law and its use

Note: Since I cannot type greek letter 'Delta' here, therefore I will be using D instead. Thus deltaH will be written as DH

What is Hess's law?

Hess's law states that the enthalpy change (DH) for any chemical or physical process is independent of the path and the number of steps required to complete the process. In other words, the DH value depends only on the initial and final states, and not on how the initial state is changed to the final state.
Hess's law is based on the principle of conservation of energy and the path independence of energy changes.

Why is it named as Hess's law?
This law is called as Hess's law after the name of Germain Hess, a Swiss-born Russian chemist and doctor, who gave this law.

What is the use of Hess's law?
Often we need to know the DH value for a reaction but it is difficult to measure it experimentally. But we know DH values of some other reactions and the other reactions are such that we can combine their chemical equations to get the chemical equation of the desired reaction. Then Hess's law allows us to calculate the DH of the desired equation by performing arithmetic operations on the chemical equations with known DH values.
If it is not clear, then do not worry; continue to read further and it will be clear.

You can perform the following arithmetic operations on chemical equations:-
1. Reverse an equation
When an equation is reversed, then the sign of DH is also reversed. Thus if DH is positive, then it becomes negative and if DH is negative, then it becomes positive.
Example: Consider the equation

2H2 (g) + O2 (g) --> 2H2O (l) DH = -571.6 kJ

On reversing the equation we get

2H2O(l) --> 2H2(g) + O2(g) DH = +571.6 kJ

Notice that the sign of DH changed.

2. Multiply or divide by a number
When a chemical equation is multiplied or divided by a number, then the number of each molecule gets multiplied or divided by that number. Also, the DH value for the equation is also multiplied or divided by that number.
Example: Consider the equation

2 H2 (g) + O2 (g) --> 2 H2O (l) DH = -571.6 kJ

On dividing the equation by 2, we get

H2(g) + 1/2 O2(g) --> H2O(l) DH = -285.8 kJ

Notice that the we divided the value of DH i.e. -571.6 kJ and got -285.8 kJ

3. Chemical equations can be added or subtracted
When chemical equations are added or subtracted, then the DH values of those equations should also be added or subtracted. Thus we get another chemical equation whose DH value we have calculated.
Example: Consider the following two equations:

C(s) + O2(g) --> CO2(g) DH = -393.5 kJ
CO2(g) --'> CO(g) + 1/2 O2(g) DH = +283.0 kJ

When we add these two equations, we get

We can simplify further by bringing 1/2 O2(g) to left. Then we will get

C(s) + 1/2 O2(g) '> CO(g) DH = -110.5 kJ

Next we consider an example of subtracting one chemical equation from another
Consider the equations

C(graphite) + O2(g) '> CO2(g) DH = -393.5 kJ
C(diamond) + O2(g) '> CO2(g) DH = -395.4 kJ

If from first equation we subtract the second equation then we will get

C(graphite) - C(diamond) --> DH = +1.9 kJ

When we subtract, then on the left hand side, O2(g) and O2(g) get cancelled and on the right hand side, CO2(g) and CO2(g) get cancelled.
There is minus sign before C(diamond). Therefore let us move it from left to right:-

In problems based on Hess's law, you will be given some chemical equations whose DH values will be given. You have to combine them to find the DH value of a desired chemical equation.

You have to use the arithmetic operations described above. However, the question is how you will know how to combine. In the next post, I will post some tips. In the mean time, let me know if that will really be useful for you.

tht was a gr8 tutorial
relly helpful ..i studied hess's law this year and last exam i lostfor not changing the sign ..while reversing the equation and whole problem went rong ..i was confused til now wat was the mistake in my answer bt now i get it 😃

thanchyuu again🤗

You will be given some chemical equations together with their DH values. You will be asked to calculate the DH value of another chemical equation. You can do that by performing arithmetic operations on the equations with given DH values in such a way that you get the desired equation.

However, it is not always easy to determine what airthmetic operations and in what sequence will help. The tips below will work in many cases. Let me know if there are problems in this topic which you are not able to solve using these tips and I will explained how to solve those.

Tips
1. See the first molecule in the desired equation.
2. Find which of given equations has this molecule. Consider only those equations, which you have not already considered for another molecule of the desired equation. If you find more than one such equation, then consider the next molecule in the desired equation. If there is no further molecule, then depending upon the given problem, you have to make a choice(sometimes a little trial and error may also be needed).
3. See if the molecule in the desired equation and in the given equation are in the same direction of the reaction arrow. If they are in different directions, then reverse the given equation.
4. See if the number of moles for that molecule is the same in both the given equation and the desired equation. If different, then multiply or divide the given equation with a number such that the number of moles of that molecule in the given equation becomes equal to that in the desired equation.
5. See if there is any more molecule in the desired equation. If yes, then go to step 2.
6. Add the current equation (after reversing and/or multiplying/dividing by a number) to the previously obtained equation.
7. If the equation that you get has some extra molecules as compared to the desired equation, then take one of those unwanted molecules. Repeat from step 2, but this time, rather than adding, you have to subtract a given equation because your aim is to get rid of this unwanted molecule. In many problems, you will get the desired equation without explicitly trying to eliminate all unwanted molecules because when you eliminate some unwanted molecules, then you will find that some more unwanted molecules also get eliminated.