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Energy Profiles (Energy Diagrams) Chemistry Tutorial

Key Concepts

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Energy Profile for Exothermic Reactions

The synthesis of ammonia gas (NH3(g)) from nitrogen gas (N2(g)) and hydrogen gas (H2(g)) is an exothermic reaction.
92.4 kJ mol-1 (of N2(g)) is released.
Energy (heat) is a product of the reaction:

N2(g) + 3H2(g) → 2NH3(g) + 92.4 kJ mol-1

In order for energy to be conserved during the chemical reaction, the "energy of the reactants" must be greater than the "energy of the products".

energy of reactants = energy of products + energy released

energy of N2(g) and H2(g) = energy of NH3(g) + 92.4 kJ mol-1

Chemists refer to the "energy of the reactants" as their enthalpy, Hreactants.
Enthalpy of products, Hproducts, is the "energy of the products".

So, we can write:

enthalpy of reactants = enthalpy of products + energy released

H(N2(g) and H2(g)) = H(NH3(g)) + 92.4 kJ mol-1

We know the enthalpy change for the reaction: ΔH = -92.4 kJ mol-1.
(Remember the minus sign (-) tells us energy is released, energy is a product of the reaction, the reaction is exothermic.)
If we assume the total enthalpy of the reactants is 192.4 kJ mol-1, then we calculate the enthalpy of the products:
enthalpy of products = enthalpy of reactants - 92.4 = 192.4 - 92.4 = 100 kJ mol-1.
We could sketch a diagram to show the relative enthalpies of reactants, H(N2(g) and H2(g)), and products, H(NH3(g)), and the enthalpy change for the reaction (ΔH), as shown below:

energy
kJ mol -1
Relative Enthalpy of Reactants and Products

reaction coordinate →
(reaction path →)

Note that the energy of the reactants is greater than the energy of the products by an amount equal to the energy that is released by the reaction (92.4 kJ mol-1).
The x-axis is labelled "reaction coordinate" or "reaction path". The reaction coordinate tells us about the energy of the system at a particular stage of the reaction.
Initially at stage 1, or the first coordinate, only the energy of the reactant molecules is being considered.
On the diagram above the final stage, or the final coordinate, of the reaction is when the energy of product molecules are considered but not reactant molecules.
The reaction coordinate (reaction path) is not the same as time.

But, we have a problem.
The air we breathe is made up of about 78% nitrogen gas (N2(g)) and a tiny amount (about 0.00005%) of hydrogen gas (H2(g)), and, no measurable ammonia on this scale.
If N2(g) and H2(g) easily react to form NH3(g), there shouldn't be any hydrogen gas in the atmosphere but we should be detecting ammonia gas instead of hydrogen gas!
There must be some "barrier" that prevents the nitrogen gas and hydrogren gas in the atmosphere reacting to form ammonia gas.
In order for reactants to react, they need to have a minimum amount of energy. If the reactant molecules have this minimum amount of energy, then, when the reactant molecules collide, they can react to form product molecules (which we call successful or fruitful collisions). If the reactant molecules do not have this minimum amount of energy, then collisions between reactant molecules will not be successful and product molecules will not be produced.
The amount of energy we need to supply in order for N2(g) and H2(g) molecules to collide successfully must be quite large, otherwise the nitrogen and hydrogen molecules in our atmosphere would successfully collide with each other to form ammonia gas in the atmosphere.
We can refer to this "extra energy" we need to supply as an "energy barrier".
Our energy diagram needs to be ammended to show the reactant molecules absorbing some energy before the product molecules can be made.

Let's assume the "energy barrier" is 100 kJ mol-1, that is, the reactant molecules must absorb 100 kJ mol-1 of energy before they have sufficient energy to allow for successful collisions between nitrogen molecules and hydrogen molecules.
Our sketch of the relative enthalpy of reactants and products needs to include a new stage, or coordinate, representing this absorbed energy.
The new diagram now looks like the one shown below:

energy
kJ mol -1
Relative Enthalpy of Reactants and Products

reaction coordinate →
(reaction path →)

Chemists call this "energy barrier" the "activation energy" for the chemical reaction.
Activation energy is usually given the symbol Ea.
Activation energy represents the minimum amount of energy that must be absorbed by the reactant molecules before they can collide successfully and produce product molecules.

Once reactant molecules have sufficient energy they collide and form a high-energy intermediate product known as the activated complex.
This activated complex is unstable, as soon as it forms it breaks apart into the molecules that make up the products of the reaction, releasing energy in the process.
This activated complex stage of the reaction must be very short.
Therefore our sketch of the relative energies of reactants and products for our reaction, needs to show the highest energy achieved as a point, not a line, on the energy diagram.
The ammended diagram, which we now refer to as an "energy profile" is shown below:

energy
kJ mol -1
Energy Profile

reaction coordinate →
(reaction path →)

The energy profile shows us the:

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Energy Profile for Endothermic Reactions

We saw above that the synthesis of ammonia gas from nitrogen gas and hydrogen gas was an exothermic process:

N2(g) + 3H2(g) → 2NH3(g) + 92.4 kJ mol-1

and we constructed an Energy Profile to show the relative enthalpies of reactants and products.

Now consider the decomposition of ammonia gas (NH3(g)) to produce hydrogen gas (H2(g)) and nitrogen gas (N2(g)).
This reaction will be the reverse of the ammonia synthesis reaction above, that is, the chemical equation for the decomposition of ammonia gas is:

2NH3(g) + 92.4 kJ mol-1 → N2(g) + 3H2(g)

and the energy profile for the decomposition reaction will also be the "reverse" of that for the synthesis reaction:

energy
kJ mol -1
Energy Profile

reaction coordinate →
(reaction path →)

Note that the reactant (NH3(g)) molecules must now absorb 92.4 + 100 = 192.4 kJ mol-1 of energy in order to give them sufficient energy for successful (or fruitful) collisions to occur resulting in product molecules.
The activation energy for this reaction is 192.4 kJ mol-1.
Ea = 192.4 kJ mol-1

Once the reactant molecules have absorbed this amount of energy (the activation energy, Ea), the high-energy intermediate product known as the activated complex will form.
As soon as the activated complex forms, it breaks apart, releasing energy and forming the products of the reaction.
From our energy profile diagram we see that 192.4 kJ mol-1 of energy was absorbed by the reactant molecules, but only 100 kJ mol-1 was released as the activated complex broke apart to make the product molecules.
Overall, the system absorbed a net amount of energy of 192.4 - 100 = 92.4 kJ mol-1.
In other words, the difference in the enthalpy of the products and reactants is 92.4 kJ mol-1.
Because the reaction is endothermic, energy is absorbed by the system, the value for the enthalpy change, ΔH, is positive (+), ΔH = +92.4 kJ mol-1.

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Effect of a Catalyst on an Energy Profile

A catalyst can be used to increase the rate of a reaction.
A catalyst DOES NOT change:

A catalyst DOES lower the activation energy required for the reaction to proceed.
The catalyst provides an alternative, lower-energy, pathway for the reaction to follow, using a lower-energy intermediate product (lower-energy activated complex).
If the catalyst is a solid, it can do this by providing a surface on which the reactant molecules can "stick" in the correct orientation, increasing the rate at which successful collisions occur.

A number of solid catalysts are available for increasing the rate of commercial ammonia gas production (see the Haber Process tutorial).
Let's consider a catalyst that is capable of reducing the activation energy for the synthesis of ammonia gas by 50%.
That is, instead of requiring an activation energy of 100 kJ mol-1, the activation energy for the reaction is decreased to just 50 kJ mol-1.
The energy profile for the reaction would now look like the one below:

energy
kJ mol -1
Energy Profile

reaction coordinate →
(reaction path →)

Note that the catalyst lowers the activation energy for both the forward and reverse reactions.
For the forward reaction,


N2(g) + 3H2(g) → 2NH3(g)     ΔH = -92.4 kJ mol-1
Ea(catalysed) = 50 kJ mol-1
Ea(uncatalysed) = 100 kJ mol-1

For the reverse reaction,

2NH3(g) → N2(g) + 3H2(g)     ΔH = +92.4 kJ mol-1
Ea(catalysed) = 50 + 92.4 = 142.4 kJ mol-1
Ea(uncatalysed) = 100 + 92.4 = 192.4 kJ mol-1

So, the rate of the forward reaction will increase for the catalysed reaction, and, the rate of the reverse reaction will also increase for the catalysed reaction.

Note that you could find a substance that slows down the rate of the forward and reverse reactions by increasing the activation energy for the reaction.
This kind of substance has the opposite effect to a catalyst, so it is sometimes known as a negative catalyst, but is more often known as an inhibitor because it inhibits the reaction.

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Summary of Energy Profiles (Energy Diagrams)

The table below provides a summary of the energy profiles (energy diagrams) for fast and slow exothermic and endothermic reactions with or without the use of a catalyst:

Type of Reaction Faster Reaction
Lower activation energy
Slower Reaction
Higher activation energy
Effect of a Catalyst
A catalyst lowers the activation energy
Exothermic
Reaction
Endothermic
Reaction

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Worked Example of Finding Enthalpy Change for a Reaction given the Energy Profile

Question 1. The energy profile below is for the decomposition of N2O4(g) into NO2(g).
energy
kJ mol -1
Energy Profile

reaction coordinate →
(reaction path →)

Determine the enthalpy change for the reaction:

N2O4(g) → 2NO2(g)

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the enthalpy change for the forward reaction:
    ΔH = ? kJ mol-1

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    Read off the energy profile:

    enthalpy of reactants: H(N2O4(g)) = 250 kJ mol-1

    enthalpy of products: H(2NO2(g)) = 50 kJ mol-1

  3. What is the relationship between what you know and what you need to find out?
    ΔH(reaction) = H(products) - H(reactants)

    ΔH(reaction) = H(2NO2(g)) - H(N2O4(g))

  4. Substitute the data into the equation to calculate the enthalpy change

    ΔH(reaction) = H(2NO2(g)) - H(N2O4(g))

    ΔH(reaction) = 50 - 250 = -200 kJ mol-1

  5. Is your answer plausible?

    The energy profile clearly shows that the energy of the products is much lower than the energy of the reactants:
    reactants → energy + products
    Therefore the reaction releases energy, it is exothermic, so the enthalpy change for the reaction (ΔH) must be negative.

    We can work backwards, using the value for the enthalpy of reactants (250 kJ mol-1) and the enthalpy change for the reaction (-200 kJ mol-1) to calculate the enthalpy of the products:
    ΔH = H(products) - H(reactants)
    -200 = H(products) - 250
    -200 + 250 = H(products)
    +50 = H(products)

    Since this value for H(products) agrees with what we can read off the energy profile, we are reasonably confident that our value for ΔH is plausible.

  6. State your solution to the problem "enthalpy change for the reaction":

    ΔH = -200 kJ mol-1

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