Gibbs Free Energy Calculations - Online Tutor, Practice Problems & Exam Prep

Understanding Gibbs free energy ($\mathrm{\Delta G}$) is crucial for predicting the spontaneity of chemical reactions. Under standard conditions (1 atmosphere pressure and 25 °C), the standard change in Gibbs free energy (${\mathrm{\Delta G}}^{\circ }$) is calculated using the formula:

${\mathrm{\Delta G}}^{\circ }={\mathrm{\Delta H}}^{\circ }-T{\mathrm{\Delta S}}^{\circ }$

where ${\mathrm{\Delta H}}^{\circ }$ is the change in enthalpy, $T$ is the temperature in Kelvin, and ${\mathrm{\Delta S}}^{\circ }$ is the change in standard entropy. For non-standard conditions, $\mathrm{\Delta G}$ varies and is not accompanied by the not sign. The standard Gibbs free energy of formation (${{\mathrm{\Delta G}}^{\circ }}_f$) for elements in their natural state is zero, and the ${\mathrm{\Delta G}}^{\circ }$ of a reaction can be determined by summing the Gibbs free energies of formation for all products and reactants, accounting for their respective moles. These values are typically provided in a chart, as they are unique for different compounds. This formula is a powerful tool for approximating whether reactions will be spontaneous at certain temperatures, guiding us through the thermodynamic landscape of chemical processes.

The Gibbs Free Energy Formula

Here we're going to see that the Gibbs free energy formula allows us to calculate the value of delta G typically in units of kilojoules by using the change in the standard enthalpy, change in our standard entropy and temperature in Kelvin values.

Now here we need to note that the change in the standard Gibbs free energy change is under standard conditions, while delta G without the little not sign is under non standard conditions. Here when we're talking about standard conditions, we're going to say that

ΔG ° = ΔH ° - T ΔS °

And here we're going to say that under standard conditions we're talking about a pressure of one atmosphere, a temperature of 25°C.

Gibbs Free Energy Calculations Example

Here it says for a particular reaction, the change in its standard enthalpy equals negative 111.4 kilojoules and a standard change in entropy equals -25 joules over Kelvin. Here we need to calculate the standard change in our Gibbs free energy for this reaction at 298 Kelvin and it says what can be said about the spontaneity the reaction at this temperature.

So to calculate our change in a standard Gibbs free energy, we're going to say ΔG=ΔH-TΔS. Here we need to make sure the units are the same. One's in kilojoules, but one's in joules. Typically we have ΔG in kilojoules, so remember that one KJ is equal to 1000 joules, so this is equal to negative 0.025 kilojoules per Kelvin.

Here we plug these numbers in, so that's going to be -111.4 kilojoules -298 Kelvin times -.025 kilojoules over Kelvin. Kelvin's here cancel out, so our answer at the end will be in kilojoules. When we do that, we get -103.95 kilojoules, so the standard change in our Gibbs free energy here it is a value less than 0.

That means that it is spontaneous as written. Remember, if your ΔG is less than 0, it's going to be a spontaneous reaction.

∆Gº and Temperature Conditions

Now when dealing with temperature conditions, we say here that the change in our standard Gibbs free energy formula can be used to approximate temperature at which reactions are spontaneous or non spontaneous.

Here we're going to say when our change in the standard Gibbs free energy values is unknown. We can utilize some basic techniques that help us to determine if a reaction will be always spontaneous, always non spontaneous or there can be spontaneous at high or low temperatures.

Now click on the next video and let's take a look at a typical question.

Gibbs Free Energy Calculations Example

Here it says for the reaction where one mole of nitrogen gas reacts with three moles of hydrogen gas to produce 2 moles of ammonia gas, our ΔH value is -92.4 kilojoules. ΔS₀ equals -198 Joules per Kelvin. Is the reaction spontaneous under standard conditions? If not, at which temperature will it be spontaneous?

All right, so here we can follow the following steps in order to find our answer. For step one we say using the Gibbs free energy formula, set ΔG₀ equal to 0. So here we're going to set it equal to 0. Plug in given values of ΔH and ΔS and solve for temperature. The foul temperature corresponds to equilibrium.

All right, so we're going to do what it says here. We're going to say here that we have ΔG = ΔH - TΔS. We set ΔG equal to 0 here. We're going to plug in the values here, so we're going to have -92.4 kilojoules for ΔH. We're solving for temperatures with temperatures unknown South. Entropy is in joules. Dividing that by 1000 will get us kilojoules. So now we have that filled in.

We're going to add 92.4 kilojoules to both sides, so when I do that I'm going to get 92.4 kilojoules equals. Now we have a negative times a negative. We're going to have equal to T * 198 kilojoules per Kelvin. Divide both sides here by the ΔS value and we'll isolate our temperature. So here kilojoules will cancel out and we'll be left with Kelvin. So here our temperature equals 466.67 Kelvin.

Now this is just the temperature to make ΔG equal to 0, which means this is the temperature for us to be at equilibrium. So found temperature corresponds equilibrium. So what do we do? Well, we go to Step 2. Here we say predict spontaneity by using the signs of ΔH and ΔS and that's where our Punnett square comes into play. Here on the left side we have +ΔH -ΔH. At the top we have +ΔS -ΔS.

The way we fill this in corresponds to our original equation, and it's set in a way where we're trying to manipulate temperature to help give us a ΔG that's less than 0 to make it spontaneous. Here, if ΔH and ΔS are both positive, then we'd be spontaneous at high temperatures, and the higher the temperature goes, the more spontaneous will become. So if spontaneous at high temperatures, reaction will become spontaneous above our calculated temperature.

If ΔH is positive and ΔS is negative, there will always be non spontaneous. So it doesn't matter what temperature we have. When ΔS is negative and ΔS is positive, then you're spontaneous. No matter the temperature, your Gibbs free energy will be less than 0. And then here when they're both negative, we're going to say we're only spontaneous at low temperatures is spontaneous at low temperatures, reaction will be spontaneous below our calculated temperature.

So let's go back to the original question. Here We have ΔS as negative and we have ΔS as negative, which means we fall. Here we're spontaneous at low temperatures O. That means here's our calculated temperature. It would mean that it's spontaneous below this temperature O any temperature below 466.67 Kelvin would give us a spontaneous reaction. So that would be the answer to this question.

Calculate ∆G° for the following reaction: P₄ (s) + 5 O₂ (g) → P₄O₁₀ (s), ∆H° = −2940 kJ/mol, 25 °C.

Does the reaction favor reactants or products?

                                                    

• A. 1419.3 kJ
• B. −2653 kJ
• C. −140.7 kJ
• D. 598.5 kJ

Determine if reaction is spontaneous under standard conditions, if not at what temperature will it be spontaneous?

3 A (g) + 5 B (s) → 3 AB (s) + B₂ (g)       ∆H° = 112.7 kJ, ∆S° = 78.3 J/K.

• A. 1439.3 K
• B. 274.6 K
• C. 1.439 K
• D. 719.65 K

Nickel has ∆H_vap = 370.4 kJ/mol and ∆S_vap = 123.3 J/mol•K. Will nickel boil at 2700 °C and 1 atm?

• A. Yes, nickel will boil at 2700 °C and 1 atm.
• B. No, the temperature is not high enough for nickel to boil.

∆Gº of Reactions

Now when it comes to Gibbs free energy of a reaction, we're going to say it's similar to the change in the standard entropy of a reaction formula. Here we're going to say the change in our standard Gibbs free energy of reaction can be calculated with free energy of formations, so ΔG⁰F these values will always be provided. That's because so many different compounds and elements have their own unique value. It'd be impossible for you to memorize all of them. So we'll be given to you in a chart within the question in some way.

Now elements in natural state tablet gives free energy of formation equal to 0, much like the change in the standard enthalpy of reactions when they also are in their natural states they're equal to 0. Now here we're going to say Gibbs free energy of a reaction formula is the change in the standard Gibbs free energy of a reaction equals the summation. So here this is Gibbs free energy of our reaction in kilojoules. Summation here is just Sigma or the sum of all your products, all your reactants and is the moles of your substance.

So it would be moles of the standard Gibbs for energy reformation of products minus the summation of the moles of Gibbs free energy of reactants products minus reactants. Here again, we're going to say that standard Gibbs free energy formation is in units of kilojoules per mole. So we're going to be utilizing this formula to help us figure out the change in the standard and the standard Gibbs free energy of a reaction, right? So we've seen this before within other variables. Now we're basically applying it to Gibbs free energy.

Gibbs Free Energy Calculations Example

Heritage has determined the change in the standard gifts for energy reaction or when we have nitric acid reacting with ammonia to produce ammonium nitrate solid. All right, so here we're given the standard Gibbs for energy of formations for each of these compounds. So remember that's just Delta G reaction equals products minus reactants equals.

So everything's a one to one relationship. So we're going to say 1 mole of our product of ammonium nitrate has a delta Geo formation of -183.8 kilojoules per mole minus our reactants, which is 1 mole of nitric acid, which has a value of -73.5 kilojoules per mole plus one more of ammonia which has a value of negative ones of one 6.4 kilojoules per mole.

Here we see that moles cancel out O. Our answer at the end will be in kilojoules. So when we plug this in, we're going to get a Gibbs free energy of reaction. Standard Gibbs free energy of reaction equal to -93.9 kilojoules. So this would be our final answer.

Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g) is a redox reaction. What would be its Gibbs Free energy change under standard conditions? Is the reaction spontaneous at 25 °C?

                                                                                                                                     

• A. 513.6 kJ; No, the reaction is not spontaneous at 25 °C.
• B. 56.4 kJ; No, the reaction is not spontaneous at 25 °C.
• C. −513.6 kJ; Yes, the reaction is not spontaneous at 25 °C.
• D. 56.4 kJ; Yes, the reaction is spontaneous at 25 °C.

Here’s what students ask on this topic:

How do you calculate standard Gibbs free energy?

To calculate the standard Gibbs free energy change (ΔG°) for a chemical reaction, you can use the following formula:

$\mathrm{\Delta G°}=\mathrm{\Delta H°}-T\mathrm{\Delta S°}$

Where:

• ΔG° is the standard Gibbs free energy change,
• ΔH° is the standard enthalpy change of the reaction,
• T is the temperature in Kelvin (K),
• ΔS° is the standard entropy change of the reaction.

All these values are typically given in joules per mole (J/mol). The standard enthalpy (ΔH°) and standard entropy (ΔS°) values can be found in tables or calculated from other data.

Another way to calculate ΔG° is by using the standard free energies of formation (ΔG_f°) for each reactant and product:

$\mathrm{\Delta G°}=\sum \left(\mathrm{\Delta G_f°}\text{products}\right)-\sum (\mathrm{\Delta G_f°}$

What is the standard Gibbs free energy?

The standard Gibbs free energy, denoted as ΔG^°, is a thermodynamic quantity that indicates the amount of energy available to do work during a chemical reaction at standard conditions (usually 1 bar pressure and 298.15 K temperature). It combines the enthalpy (ΔH^°) and entropy (ΔS^°) changes of a reaction into a single value. The equation for standard Gibbs free energy change is:

ΔG^° = ΔH^° - TΔS^°

Where:

• ΔG^° is the standard Gibbs free energy change,
• ΔH^° is the standard enthalpy change,
• T is the temperature in Kelvin,
• ΔS^° is the standard entropy change.

A negative ΔG^° value indicates that the reaction is spontaneous under standard conditions, meaning it can occur without the input of additional energy. Conversely, a positive ΔG^° means the reaction is non-spontaneous and requires energy to proceed. If ΔG^° is zero, the system is at equilibrium, and no net reaction occurs. This concept is fundamental in predicting the direction and extent of chemical reactions.

Which of the following correctly expresses the standard Gibbs free energy change of a reaction in terms of the changes in enthalpy and entropy?

The standard Gibbs free energy change (ΔG^°) of a reaction can be correctly expressed in terms of the changes in enthalpy (ΔH^°) and entropy (ΔS^°) using the following equation:

ΔG^° = ΔH^° - TΔS^°

In this equation, T represents the absolute temperature in Kelvin (K). The ΔG^° value predicts the spontaneity of a reaction under standard conditions (1 bar pressure and the substances involved being in their standard states). If ΔG^° is negative, the reaction is spontaneous; if it's positive, the reaction is non-spontaneous; and if it's zero, the system is at equilibrium. This relationship is fundamental in thermodynamics and helps to understand how changes in enthalpy, entropy, and temperature influence the spontaneity of a chemical reaction.