# Hess’s Law: Definition, Formulas, and Example Problems

Understanding Hess’s Law – Enthalpy is a state function, which only depends on the initial and final state of each reaction and the reaction products without regard to the path from changing the substance of each reaction to the reaction product. In the previous enthalpy calculations, the enthalpy can be determined by calculating the heat of reaction at constant pressure.

However, not all reactions can be clearly known by the amount of heat of reaction at that time. In 1840, a chemist from Russia, named Germain Henri Hess, manipulated thermochemical equations in order to calculate the ΔH contained in a law called Hess’s law or the law of addition of heat. Gerrmain Henry Hess states that:

“If a reaction takes place in two or more reaction steps, then the enthalpy change for the reaction will be equal to the sum of the enthalpy changes for all the stages that occur.”

## Hess’s Law sounds

“The enthalpy of a reaction does not depend on the course of the reaction, but on the end result of the reaction.”

From  Hess’s Law  which was mentioned earlier, it is possible to calculate the enthalpy change for a reaction based on the enthalpy change for another reaction whose value is known. This can be done so that researchers do not have to do experiments every time.

Hess’s law can be described in a schematic way, namely as follows.

Given the Hess diagram for the reaction A → C

 Diagram Hess.

The change from A to C can take place through 2 stages, namely:

Stage I (directly)

A → C → ∆H1

Stage II (indirectly)

Based on Hess’s Law, the price ∆H 1  = ∆H 2  + ∆H 3

 A → B ∆H2 B → C ∆H3 A → C ∆H 2  + H 3

Many reactions can take place based on two or more steps.

Example:

The reaction of carbon and oxygen needed to form CO 2  can take place in one step (direct way) or can also take place in two stages (indirect way).

 1) One level: C(s) + O2(g) → CO2(g) ∆H = –394 kJ 2) Two levels: C(s) + ½ O 2 (g) → CO(g) ∆H = –110 kJ CO(g) + O 2 (g) → CO 2 (g) ∆H = –284 kJ C(s) + O2(g) → CO2(g) ∆H = –394 kJ

Hess’s law can also be expressed in the form of cycle diagrams or energy level diagrams. The cycle diagram for the carbon burning reaction is based on the previous example as follows:

 Carbon burning reaction cycle diagram.

Based on the reaction cycle previously described, carbon burning can go through two pathways, namely path-1 which can directly form CO 2 , while pathway- 2 , initially forming CO, then becoming CO 2 . So, ∆H 1  = ∆H 2  + ∆H 3

Energy level diagram:

 Energy level diagram for the reaction of carbon using oxygen and forming CO 2  according to two paths.

Hess’s law can be used to determine the heat based on a reaction which cannot be known in a direct way. Consider the following example.

Example:

 ½ N2(g) + O2(g) → NO2(g) ΔH1 = x kJ = + 33,85 kJ/mol 1 level ½ N2(g) + ½ O2(g) → NO(g) ΔH2 = y kJ = + 90,37 kJ/mol + 2 levels NO(g) + ½ O2(g) → NO2(g) ΔH3 = z kJ = – 56,52 kJ/mol ½ N2(g) + O2(g) → NO2(g) ΔH 1  = ΔH 2 + ΔH 3 x = y + z

According to Hess’s Law:

ΔH 1  = ΔH 2 + ΔH 3  or x = y + z

 The change that occurs from N 2  g) and O 2  g) to NO(g) is accompanied by an enthalpy change (ΔH 1 ) of +33.85 kJ/mol, even though the reaction has been defined as one step or two steps, ΔH 1  = ΔH 2 + ΔH 3 .

## Problems example

Problem 1

Determine the enthalpy of burning charcoal into a gas of carbon dioxide and also water vapor.

The reaction of burning charcoal can be written as follows.

C(s) + O 2 (g) → CO(g)

Not all changes that occur in enthalpy can be determined by conducting experiments. For example, in a combustion reaction of carbon (graphite) which turns into carbon monoxide. The enthalpy change for the reaction of burning carbon to pure CO tends to be more difficult to do because CO is flammable. What if we react carbon with excess oxygen, CO will immediately burn and turn into CO2 . Meanwhile, in a limited amount of oxygen, a mixture of CO and CO2 will be formed .

However, the enthalpy change that occurs in the formation of CO can be determined based on the reaction enthalpy change which can easily occur. The reaction that occurs more easily is a reaction of burning carbon and turning into carbon dioxide and burning of carbon monoxide which turns into carbon dioxide.

The enthalpy value of this reaction is unknown. The enthalpy value of the charcoal burning reaction can be determined by using a reaction whose enthalpy value is known beforehand. It is known that the enthalpy of formation of CO 2  = –393.5 kJ mol –1  and the enthalpy of combustion of CO = –283 kJ mol –1 .

Based on these two enthalpy data, using Hess’s law the enthalpy of burning carbon that turns into carbon monoxide can be calculated in the following way.

The thermochemical equation that results in the formation of carbon dioxide (CO 2 )

(1) C(s) + O2(g) → CO2(g) = -393,5 kJ/mol

The thermochemical equation that results in the combustion of carbon monoxide (CO)

(2) CO(g) + 1/2 O2(g) → CO2(g) AH = -283 kJ/mol

In order to obtain the equation for the reaction of burning carbon which turns into carbon monoxide, reaction (2) can be reversed and can then be added using a reaction (6).

 C(s) + O2(g) → CO2(g) ∆H = – 393,5 kJ CO 2 (g) → CO(g) + O 2 ∆H = + 283 kJ C(s) + O2 (g) + CO2 (g) → CO2 (g) + CO(g) + ½ O2(g) ∆H = – 110,5 kJ

Disposal of two substances that are the same on both sides will be able to produce a reaction equation, namely:

C(s) + ½ O2 (g) → CO(g)     ∆H = -110,5kJ

Based on the sum of the two stages of the reaction, the enthalpy change for the burning of carbon that turns into carbon monoxide can be determined in a way that tends to be easier, namely by adding up the enthalpy changes for the two stages of the reaction that have occurred.

The determination of the number of enthalpy changes in this way was discovered by a chemist from Russia, namely GH Hess (1840). Through a series of experiments he has done, Hess made a statement that the enthalpy change depends only on the initial and final states of a reaction and does not depend on the course of the reaction.

So, if a chemical reaction takes place through several reaction steps, the enthalpy change is determined by adding up the enthalpy change for each step. This statement from Hess later became known as Hess’ law (or also often referred to as the Law of Addition of Heat). The application of Hess’ law makes it easy to determine the enthalpy change for a reaction which is difficult to determine when conducting experiments.

The thermochemical equation is prepared in such a way that the result of the sum is the reaction whose enthalpy change will be determined. Often, several equations must be multiplied by suitable coefficients to obtain a required thermochemical equation.

This calculation method is in accordance with Hess’s law which states that the enthalpy of the reaction absorbed or released by a reaction does not depend on the course of a reaction. Some of the principles of calculating thermochemical equations according to Hess’s law that need to be considered are:

a. If a reaction equation must be reversed, then change the sign of ∆H. For example,

Reaction : H 2 (g) + O 2 (g) → H 2 O 2 (l) ∆H = –187.8 kJ

Reversed: H 2 O 2 (l) → H 2 (g) + O 2 (g) ∆H = +187,8 kJ

b. If in the addition of the reaction there is a substance that appears on both sides of the equation with the same substance phase, then the substance can be removed. Examples are:

 H2(g) + ½ O2(g) → H2O(g) ∆H = +241,80 kJ H2O(l) → H2(g) + ½ O2(g) ∆H = –285,85 kJ H 2 O( l ) → H 2 O(g) ∆H = –44,05 kJ

The calculation of the ∆H of the reaction  can also be carried out using the basic data of the standard heat of reaction of formation (∆H f °). The standard heat of formation is a heat of formation of compounds based on their elements. Look at the general equilibrium reaction equation below.

aA + bB → cC + dD

∆Hreaksi = (c × C + d × D) – (a × A + b × B)

= ∆H°  products  – ∆H° f reactants

So, in general the ∆H of the reaction  can be determined using the formula:

∆H reaction  = ∆H° f products  – ∆H° f reactants

Information :

∆H° f product  : is the total enthalpy of standard formation based on the product substances.
∆H° f reactants  : is the total enthalpy of standard formation based on the reactants.

Problem 2:

Determine the ∆H value of the reaction  for the decomposition reaction of SO 3  according to the following reaction equation.

SO3(g) → SO2(g) + ½ O2(g)

Solution :

From the table it is known:

∆H°f SO3 = –395,2 kJ mol-1,

∆H°f SO2 = –296,9 kJ mol-1

∆H reaction  = ∆H°  product  – ∆H°  reactant

= {1× (–296.9 kJ mol -1 ) + ½ × 0} – {1 × (–395.2 kJ mol -1 )}

= –296.6 kJ mol -1  + 395.2 kJ mol-1

= +98.6 kJ times -1

So, the decomposition of SO 3  is +98.6 kJ mol -1 .

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Question 3 :

Carbon forms two types: graphite and diamond. The enthalpy for burning graphite is –3939.5 kJ while the enthalpy for burning diamond is –395.4 kJ.

C(grafit) + O2(g) → CO2(g) ∆H = -393.5 kJ

C(intan) + O2(g) → CO2(g) ∆H = -395.4 kJ

Calculate the ∆H used to turn graphite into diamond.

Solution :

What we need is ∆H for the reaction:

C (graphite) → C (diamond)

C (grafit) + O(g) → CO(g) ∆H = -393.5 kJ

CO(g) → C (intan) + O(g) ∆H = +395.4 kJ

C (graphite) → C (diamond) ∆H = +1.9 kJ

By using the law of the conservation of energy, we can also use it through the form of energy diagrams in a reaction. An example of burning methane is useful to produce gaseous H 2 O and then condensing gaseous H 2 O to the solid state. Based on the energy diagram it looks as shown in the following.

 Diagram of the enthalpy change for the combustion reaction of methane.

So, to find out the reaction enthalpy is:

CH4(g) + O2(g) → CO2(g) + 2H2O(l)

The value will be equal to ∆H 1  = ∆H 2  + ∆H 3

To find the enthalpy of a reaction:

CH4(g) + O2(g) → CO2(g) + 2 H2O(g)

The value will be equal to ∆H 2  = ∆H 1  – ∆H

To find the enthalpy of a reaction:

2 H 2 O(g)) → 2 H 2 O( l )

The value will be equal to ∆H 3  = ∆H 1  – ∆H 2

Question 4 :

Known Hess cycle diagram

Determine the standard enthalpy for the formation of CO2 gas !

Solution :

According to Hess’s Law

∆H1 = ∆H2 + ∆H3 = –222 + (–566) kJ = –788 kJ

maca ∆H f ° gas CO 2  = – (788/2) = –394 kJmol –1

Question 5 :

The energy level diagram is known as follows.

Determine the standard enthalpy for the formation of CO2 gas !

According to Hess’s Law

∆H1 = ∆H2 + ∆H3 = –222 + (–566) kJ = –788 kJ

maca ∆H f ° gas CO 2  = – (788/2) = –394 kJmol –1

The value of ∆H for the reaction can be calculated using the data on the change in standard enthalpy of formation (∆H f °)

Formula :
∆H reaction  = ∆H product  – ∆H reactant

Problems example :

Is known :

∆Hf° CH4 = –79,3 kJ

∆Hf° CO2 = –393,52 kJ

∆H f ° H 2 O = –296.0 kJ

Determine ∆H c ° gas CH 4 !

Required reaction: CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O

∆H reaction  = ∆H f ° (CO 2  + 2 (H 2 O)) – ∆H f °(CH 4  + 2 (O 2 ))= (–393.52) + 2(–286) – (– 79.3) kJ
= (–965.52 + 79.3) kJ
= –886.22 kJ
Practicum for Determination of Enthalpy Change based on Hess’s Law
In this experiment, we will observe the ∆H of the reaction between solid NaOH and 0.5 M HCl solution using two ways.

Method 1:
Solid NaOH is first dissolved in water, then the NaOH solution will be reacted using HCl solution.

Method 2:
Solid NaOH will be immediately dissolved using HCl solution.

Work steps:

Method 1:
• Weigh 2 grams of NaOH, then store it in a closed container. Prepare 50 mL of water, and measure the temperature. Enter the NaOH into the water, then stir and record the maximum temperature. Calculate the ∆H of the reaction from dissolving NaOH (∆H 1 ).
• Prepare 50 mL of 1 M HCl and measure the temperature.
• Measure the temperature of 50 mL of NaOH solution that has been made before.
• React the NaOH solution using HCl solution, then record the maximum temperature. Calculate ∆H for the reaction (∆H 2 ).

Method 2:
• Weigh 2 grams of NaOH, then store it in a closed container.
• Prepare 100 mL of 0.5 M HCl solution, measure the temperature.
• React solid NaOH with HCl, record the maximum temperature
. Calculate ∆H for the reaction (∆H 3 ).

Questions:
1. Calculate ∆H 1 , ∆H 2 , and ∆H3 for each mole of NaOH!2. Write the equation of the thermochemical reaction in:
a. dissolving solid NaOH substance to become a solution of NaOH substance (aq),
b. neutralization reaction of NaOH (aq) with HCl (aq),
c. neutralization reaction of NaOH (s) with HCl (aq).

3. Make a reaction diagram in the previous experiment!
4. According to Hess’s Law, namely ∆H 1  + ∆H 2  = ∆H 3

Does your experimental data agree with Hess’s Law? If not, state three factors that cause it!
From the experiment above you will be able to find the result ∆H 1  = ∆H 2  + ∆H 3 .

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