Do you want to know if a reaction is possible or not? It's simple. For a reaction to happen, the change in enthalpy (∆H) must decrease, and there must be an increase in the change in entropy (∆S). But what if both ∆H and ∆S decrease? Is the reaction still possible? Don't worry, we have the answer.
Enthalpy (H) is the internal energy (E) plus the product of the pressure (P) and the volume (V) of a thermodynamic system. A decrease in enthalpy means the reaction released heat. Entropy (S) is the measure of the disorder of a thermodynamic system. An increase in entropy means the system became more disordered, such as a solid turning into a liquid. Now, let's talk about free energy. We use the equation for free energy to predict if a reaction is feasible or not. You'll also learn how the temperature affects reaction feasibility. We'll show you how to calculate ∆G using ∆Hº and ∆Sº values, as well as ∆Gº values. You'll find out how to determine the temperature at which a reaction becomes feasible. In conclusion, with the right tools and knowledge, you can predict if a reaction is possible or not. Keep reading to learn more about free energy and how it can
To determine whether a reaction is possible or not, we can't solely rely on the changes in enthalpy (∆H) or entropy (∆S). Instead, we use Gibbs free energy (∆G) to predict the feasibility of a reaction by considering both of these changes. Gibbs free energy is the amount of energy that is available within a system to do work.
The equation for Gibbs free energy is:
∆G = ∆H - T∆S
Where ∆H is the enthalpy change, T is the temperature, and ∆S is the entropy change.
Temperature plays a crucial role in the feasibility of a reaction. For example, combustion is a spontaneous process, but it requires an increase in temperature to occur. As the temperature increases, the feasibility of the reaction increases.
The equation for Gibbs free energy allows us to determine the feasibility of a reaction based on its enthalpy change, entropy change, and temperature. If ∆G is negative, the reaction is feasible. If ∆G is positive, the reaction is not feasible.
To summarize, the equation for Gibbs free energy is essential in determining the feasibility of a reaction. By considering the enthalpy change, entropy change, and temperature, we can predict whether a reaction is possible or not.
There are two ways you can calculate free energy. The first way is by using standard change in enthalpy (∆Hº) and entropy (∆Sº) values. The second is from the free energy (∆Gº) values of the substances in the system.
Before you learn these two methods to calculate free energy, you must remember that Gibbs free energy, like enthalpy and entropy, is a state function and has standard values for compounds. Most of the time, though, we calculate free energy under non-standard conditions. So you often see ∆G without the symbol for standard conditions (º).
The units for the equation ∆Gº = ∆Hº - T∆Sº :
∆G in ∆H in T in K∆S in
Be careful with the units of entropy when doing ∆G calculations. Remember to change them to by dividing by 1000.
∆Sº here represents the entropy change of a reaction and not total entropy.
1. Calculate ∆Gº for the combustion of methane. ∆Sº = -242.2 ∆Hº = -890.4 Change the value for entropy into by dividing by 1000 -242.2 ÷ 1000 = -0.2422 Write the equation for Gibbs free energy. ∆G = ∆H - T∆S Since we are asked for standard values, T = 298K. Fill in the given values:∆G = -890 - 298(-0.2422) ∆G = -817.6
2. Use the given values for to find the of the following reaction
When we know the ∆G of the substances in the reaction, we can use the equation
Fill in the equation using the given values:
∆Gº = [ 2(-394) + 3(-229) ] - [ (-175) + 3(0) ]
∆Gº = -1300
When we talk about spontaneity and feasibility we often mean the same thing. But so we’re on the same page:
When we say ‘feasible’, we mean the reaction is energetically favourable. In other words, the reaction should take place. When we say ‘spontaneous’, we usually mean the reaction happens on its own, without outside interference. Spontaneous reactions can be very, very slow!
Some chemists like to describe the combustion of carbon as a spontaneous reaction. However, a piece of carbon left out in the open will not 'spontaneously' combust no matter how long you leave it there, unless you first apply some heat! The reaction is possible, it just has a high activation barrier. That's why some chemists prefer to use the term 'feasible' instead of 'spontaneous'.
Free energy helps us determine the feasibility of a reaction.
When ∆G is negative the reaction is feasible. When ∆G is positive the reaction is not feasible.
By rearranging the equation for free energy we can determine at what temperature a non-spontaneous reaction becomes feasible. A reaction becomes feasible at the point where ∆G = 0. In other words when ∆H = T∆S.
0 = ∆H - T∆S
.˙. ∆H = T∆S
.˙. T = ∆H / ∆S
At what temperature will the reaction between carbon oxide and carbon dioxide become spontaneous?
∆H = -178
∆S = -161
Change units for entropy to
161 / 1000 = 0.161 kJK-1mol-1
T = ∆H / ∆S
T = 178 / 0.161
T = 1105.59 K
To conclude, let's take a moment to discuss why we call it ‘free’ energy? What's so 'free' about it? Well, we used to call free energy by another name: available energy. In other words, free energy is the energy in a system we have available (or free) to do work after a reaction. Think about it: when a chemical reaction takes place, the enthalpy change is the difference between the energy used to break bonds and make new ones. Entropy change is the ‘energy cost’ of the reaction. The energy left over is what we know as free energy!
To explain further, have a look at the following reaction.
If you were to calculate ∆H for this reaction you would find that its value is negative (i.e., the reaction is exothermic. It gives out energy). Again, If you were to calculate ∆S (system) you would find that its value is also negative (since energy is less spread out in a solid than in a gas). Why then does the reaction take place? Because of the overall increase in the total entropy change for the system and its surroundings.
Remember the second law of thermodynamics:
In a spontaneous process, the total entropy change for a system and its surroundings is positive." -The second law
This means that the reaction is feasible because the heat energy given out into the surroundings (∆H) increases the entropy of the surroundings () enough to make up for the decrease in entropy in the system ().
In thermodynamics, a system is a substance or a collection of substances and energy. Everything that is not in the system, we call the surroundings. If a reaction takes place in a jar, then the jar is the system. Everything outside the jar is the surroundings.
Here are the actual ∆H and ∆S values for the above reaction:
∆H = -176
∆S = -284
We can conclude that the change in entropy of the surroundings () must be equal to or greater than +284 (0.284 ). The ∆H of -176 is more than enough to cover this! The value that interests us the most is the amount of the enthalpy change (∆H) we need to make the reaction feasible.
To find this out, we use the relationship between the entropy change of the surroundings and the change in enthalpy of a reaction:
If we rearrange this expression we get:
So at room temperature, 298K:
∆H = -(298)(0.284)
∆H = -85
This means, of the -175 of energy produced by the making and breaking of bonds, 86 of it must be released into the surroundings as heat for the reaction to be feasible. The rest of it is 'free' energy, and is available to be used as other forms of energy!
Great summary! Here's a slightly revised version:
Key takeaways on Free Energy:
What is free energy?
Free energy is the energy available in a system to do work. After a reaction is complete and a change in enthalpy and entropy has taken place, the energy left over is known as free energy. It is available to be used as other forms of energy.
How do you calculate Gibbs free energy?
We calculate free energy with the equation ∆G = ∆H - T∆S. Where ∆G is free energy, ∆H is the enthalpy change of the reaction, T is the given temperature and ∆S is the entropy change of the system.
What is an example of free energy?
1. Calculate ∆Gº for the combustion of methane. ∆Sº = -242.2 ∆Hº = -890.4 Change the value for entropy into by dividing by 1000 -242.2 ÷ 1000 = -0.2422 Write the equation for Gibbs free energy:∆G = ∆H - T∆S Since we are asked for standard values, T = 298K. Fill in the given values:∆G = -890 - 298(-0.2422) ∆G = -817.6
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