Entropy Calculations - Online Tutor, Practice Problems & Exam Prep

Understanding the entropy of a system and its surroundings is crucial in predicting the spontaneity of a chemical reaction. Entropy, a measure of disorder or randomness, is denoted in units of joules per Kelvin (J/K). The total entropy change in the universe (ΔS_universe) is the sum of the entropy change of the system (ΔS_system) and the surroundings (ΔS_surroundings). A positive ΔS_universe indicates a spontaneous process, aligning with the second law of thermodynamics, which states that the entropy of the universe tends to increase. Conversely, a negative value suggests a non-spontaneous process, while a value of zero indicates equilibrium.

The entropy change of the surroundings can be calculated using the formula:

ΔS_surroundings=-ΔH_reaction/T where ΔH_reaction is in kilojoules and T is the temperature in Kelvin, assuming constant pressure and temperature conditions.

To calculate the entropy change of a system, the standard molar entropy (S°) of each substance is used. These values, which are always greater than zero, are provided due to the vast number of substances and their unique entropies. The standard entropy change of a reaction (ΔS°_reaction)

Total Entropy Formula

Here, when discussing entropy calculations, we need to take into account the entropy of the universe. Now here we're going to see the total entropy change in the universe is represented by the following equation or formula below. Here we're going to say that the total change in entropy, which is the same thing as the entropy of the universe, equals the change in entropy of my system, which is the same thing as my reaction, plus the change in entropy of my surroundings, which is everything else here.

Units of this change in entropy are typically in units of Joules per Kelvin. Now here are change in entropy. Total can be either a positive number, a negative number, or equal to 0. If it's positive, then it goes in line with the second law of thermodynamics, which says that the entropy of the universe, or total, is ever increasing. The second law of thermodynamics is talking about the natural increase in entropy in the universe, making it a natural process, making it spontaneous.

Now if we're negative, we're the exact opposite of that, so it would be non spontaneous. And then finally for Δ0, we're not spontaneous, we're not not spontaneous. What we are we're at equilibrium. So these are the three types of conditions that exist depending on what we calculate for the change in entropy total or change in entropy of the universe.

Entropy Calculations Example

Here it says calculate the total entropy change for reaction when the change in entropy of our surroundings equals 2.7 joules per Kelvin and the standard entropy change of our reaction equals negative -450 kilojoules per Kelvin, is this reaction spontaneous?

All right, so we need the total entropy. So delta S total here will equal delta S of my surroundings plus delta S standard of our reaction. Now here typically we'll have the change in our entropy being Joules per Kelvin. So I'll use this 2.7 joules per Kelvin.

Here I need to convert kilojoules to joules. Remember, 1K is 10 to the three, so that will come out to -450,000 joules per Kelvin. When we add them together we're going to get negative 449997.3 joules per Kelvin.

Because our change in total entropy is a negative value, that would mean that this is a non spontaneous reaction. OK, so here we'd say it's non spontaneous for this particular example.

Entropy of Surroundings Formula

When it comes to the entropy of our surroundings, we can say we can also calculate it if we know the temperature at which the reaction is taking place and we're going to say the conditions here will be under constant pressure and temperature.

Now if we take a look here we have the entropy of surroundings formula. That is, the change in the entropy of our surroundings is equal to negative change in the enthalpy of our reaction divided by temperature.

Now here we say that the enthalpy of our reaction is usually in units of kilojoules, and we say here that temperature as always, will be in Kelvin. So again, we can calculate the real value of our changing entropy of our surroundings through the utilization of the following formula.

ΔS = - ΔH T

Entropy Calculations Example

Here it says determine the change in the entry of the universe for the following reaction at 32°C. So here they're giving us this chemical reaction and we're going to say that our enthalpy of the reaction change in enthalpy of our reaction is -140 kilojoules and the entropy of our reaction is 3.6 joules per Kelvin.

So remember here that our total change in entropy, which is the same thing as the entropy of our universe, equals the change in the entropy of our surroundings plus the change in the entropy of our reaction. Here we already know that changing entropy of our reaction. That's 3.6 Joules per Kelvin. What we need to do is determine the change in the entropy of our surroundings, which remember is equal to negative delta H of R reaction divided by temperature in Kelvin.

So we're going to take that number and plug it in and we're going to say here, since entropy change of our reaction is in joules, I'm going to convert these kilojoules into joules by multiplying them by 1000 temperature. We add 273.15 to this 32°C and that gives me 305.15 Kelvin. So we have a negative of a negative. Remember the negative was already there, and then we have an additional negative.

Here the change in the entropy of my surroundings will be a positive value. It comes out to be +458.79 Joules per Kelvin. We take that and we plug it here into our formula and now we're going to have the change in the entropy of our universe, or total. That comes out to a +462.39 Joules per Kelvin. So this would be our final answer.

Δ S = Δ S + Δ S Δ S = - Δ H T

Entropy of the System

In this video, we'll take a look at how to calculate the entropy of our system, which represents our chemical reaction. Here we're going to say that each substance has a standard molar entropy, represented by X0 associated with it. Now, these values will always be provided in some way. That's because there's so many compounds, so many substances, each with their own unique molar entropy. There's no way you can memorize all of them.

Now this is important. Unlike standard molar enthalpies, which number RΔH0 for substances in their natural state, we're going to say that your entropy, standard molar entropy, does not equal 0. It's always going to be a value greater than 0. Now here we have the entropy of reaction formula. Here it says that the change in the standard entropy of our reaction, which is our system, equals. So when we talk about standard entropy of our reaction, that's typically in units of joules per Kelvin.

Here we have Σ which is summation and this would be N times standard entropy of our products minus Σ mole standard entropy of our reactants. Now remember, Σ just stands for sum of. So we're taking all the products, lumping them together, taking all the reactants, lumping them together. And here represents the moles of substance. And then we're going to say here entropy with not equals your standard mode entropy of a substance, and this will be in joules over moles times K.

When we talk about standard conditions, remember that we're talking about a temperature of 25°C and a pressure of one atmosphere. So basically we're saying here that the entropy, standard entropy of our reaction is equal to products minus reactants. We're going to utilize this formula to help us solve for the entropy of our system in the following questions. So let's pay attention and see how we utilize this formula.

Entropy Calculations Example

Here it says to calculate the change in the standard entropy of our reaction for the following reaction in 25°C. Here we have two moles of nitrogen monoxide reacting with one mole of oxygen gas to produce 2 moles of nitrogen dioxide gas. Here we're told that the enthalpy of reaction is equal to -114.14 kilojoules.

All right, so here they want us to calculate this. And remember, change in the standard entropy of our reaction is just equal to products, minus reactants. Here we're given the standard molar entropies of each of the compounds found within this reaction. We look at our products. We're going to say we have two moles of nitrogen dioxide, and according to our chart, each one has a standard MOA entropy of 240.1 joules over moles times K.

Plug that in, moles cancel out. Minus my reactants. We have two moles of nitrogen monoxide each one is 2108 joules over moles times K. Again, moles cancel out. Plus we have one mole of oxygen gas with a value of 205.2 Joules over moles times K. So again, moles cancel out. Here we can see that the moles cancel out, so our units at the end will be in joules per Kelvin.

When we plug this in, we get 146.6 Joules over Kelvin. This will be our final answer. Now, you might also notice that I gave us the enthalpy of the reaction. Here we don't even need to use it because the formula to calculate the entropy of our system doesn't need this entropy value. So again, sometimes professors give you additional information and that doesn't mean you have to use it. In this case, I gave you enthalpy of the reaction and it's just important. Is it just telling us at which the reaction is releasing energy? But it's not important to answer the question here.

For the following reaction at 27 °C, calculate ∆S°_rxn, ∆S_surr, and ∆S_tot. Determine if reaction is favorable.

Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g) ∆H_rxn = 98.8 kJ

• A. ∆S°_rxn = 0.1415 J/K, ∆S_surr = 0.3292 J/K, ∆S_tot = −0.1877 J/K, not favorable
• B. ∆S°_rxn = 141.5 J/K, ∆S_surr = −329.17 J/K, ∆S_tot = −187.67 J/K, not favorable
• C. ∆S°_rxn = 70.75 J/K, ∆S_surr = 164.58 J/K, ∆S_tot = −93.84 J/K, favorable
• D. ∆S°_rxn = 283.0 J/K, ∆S_surr = 658.34 J/K, ∆S_tot = −375.34 J/K, favorable

Here’s what students ask on this topic:

Which of the following has the largest standard molar entropy, S° (298.15 K)?

To determine which substance has the largest standard molar entropy (S°) at 298.15 K, you would typically compare the S° values of the substances in question. Entropy is a measure of the disorder or randomness in a system, and in general, the standard molar entropy increases with the complexity of the molecule, the number of atoms in the molecule, and the number of possible microstates (ways the energy can be distributed within the molecule).

Here are some general trends that can help you determine which substance might have the largest S°:

1. Gases have higher entropy than liquids, which in turn have higher entropy than solids.
2. Among substances in the same state of matter, those with larger, more complex molecules tend to have higher entropy.
3. Substances with more atoms or a greater variety of atoms generally have higher entropy.
4. For isomers, the more branched isomers tend to have higher entropy than linear ones.

Without specific substances to compare, these guidelines can help predict which might have the highest standard molar entropy. If you have a list of specific substances, you would look up their standard molar entropy values in a reference source such as a textbook or a chemical data database to make a direct comparison.

Which of the following have a standard molar entropy, $s^{°}$, of zero?

The standard molar entropy ($S^{°}$) of zero is a unique value assigned to a perfectly ordered crystalline substance at absolute zero temperature (0 Kelvin) according to the third law of thermodynamics. At this temperature, the entropy, which is a measure of disorder or randomness in a system, is considered to be at its minimum because the particles in the substance have the least amount of motion and are in their most ordered state.

No real substances actually have a standard molar entropy of zero at temperatures above absolute zero because there is always some degree of thermal motion contributing to entropy. However, for the sake of calculations and conventions in thermodynamics, the standard molar entropy of an element in its most stable form at 1 bar of pressure and a specified standard temperature (usually 298.15 K) is often compared to this theoretical zero point.

It's important to note that while the third law provides a reference point, actual standard molar entropy values for substances at standard temperature and pressure are always positive.

Which process is necessarily driven by an increase in the entropy of the surroundings?

The process that is necessarily driven by an increase in the entropy of the surroundings is a spontaneous process. In thermodynamics, a process is considered spontaneous if it occurs without needing to be driven by an external force. The second law of thermodynamics states that for a process to be spontaneous, the total entropy change of the system and its surroundings must increase. This means that while the entropy of a system (for example, the substances involved in a chemical reaction) might decrease, the entropy of the surroundings must increase by a greater amount, leading to an overall increase in the total entropy of the universe. This net increase in entropy is what drives spontaneous processes. Examples of such processes include the melting of ice at room temperature and the diffusion of a perfume from one part of a room to another.

Which of the following would lead to the highest gain in entropy in the surroundings?

The gain in entropy in the surroundings is highest when a process involves the transfer of a large amount of heat to the surroundings at a low temperature. According to the second law of thermodynamics, entropy, which is a measure of disorder or randomness, increases with the transfer of heat.

When a process occurs at a lower temperature, the same amount of heat transfer results in a larger change in entropy than it would at a higher temperature. This is because entropy change (ΔS) is calculated by the formula:

$\Delta S=\frac{q}{T}$

where q is the heat transferred and T is the temperature in Kelvin. As T is in the denominator, a lower T results in a higher ΔS for the same amount of heat q.

Therefore, a process that transfers a large amount of heat to the surroundings at the lowest possible temperature would lead to the highest gain in entropy in the surroundings.