# 4: Second & Third Laws of Thermodynamics (Worksheet)


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Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

Learning Objectives

• Understand the Second and Third Laws of Thermodynamics
• Understand the significance of the thermodynamic functions of entropy and Gibbs free energy
• Understand the relationships between both entropy and Gibbs free energy and the spontaneity of a physical or chemical process
• Understand the relationship between ΔG and K

The three laws of thermodynamics describe restrictions on the behavior of virtually the entire physical world we can experience. Everything that is possible or impossible in a physical, chemical, or biological system is in some way related to these laws. We have previously talked about the First Law of Thermodynamics, which is concerned with the conservation of matter and energy. The Second and Third Laws are concerned with disorder and its relationship to spontaneous and non-spontaneous changes.

## Success Criteria

• Be able to carry out calculations using $$ΔG = ΔH – TΔS$$ and interpret the meaning of the results
• Be able to calculate $$ΔG^o$$ and $$ΔS^o$$ values from tabulated standard free energy data and absolute entropy data
• Be able to calculate approximate values of $$ΔG$$ at non-standard temperatures
• Be able to use $$ΔG = ΔG^o + RT \ln Q$$ to calculate free energy under non-standard conditions
• Be able to calculate $$K$$ from $$ΔG$$

## Second Law of Thermodynamics

The Second Law of Thermodynamics is concerned with a thermodynamic function called entropy, $$S$$. Entropy is a measure of disorder. For example, water vapor has a higher value of S (188.83 J/mol×K) than liquid water (69.91 J/mol×K), because the molecules move more freely in the vapor than in the liquid, making the gaseous state more disordered than the liquid state.

When a system undergoes a chemical reaction or physical change, the overall entropy changes. This overall entropy change, $$ΔS_{total}$$, is the sum of the change in entropy of the system and of its surroundings:

$ΔS_{total} = ΔS_{system} + ΔS_{surroundings}\nonumber$

Some changes occur spontaneously, and others must be forced. A change to a more disordered state is a more probable event, so when a change occurs spontaneously there is an increase in the total entropy. This is the essence of the Second Law of Thermodynamics:

Every spontaneous change results in an increase in total entropy.

The great physical chemist Rudolph Clausius (1822-1888) famously summarized both the First and Second Laws as follows:

The energy of the world is a constant; the entropy strives for a maximum.

Unlike enthalpy, $$H$$, which is governed by the First Law of Thermodynamics, a change in the entropy of the system, which is governed by the Second Law, does not require an equal and opposite change in the surroundings. Some changes only involve an entropy change of the system, but more often we encounter processes that involve entropy changes for both the system and its surroundings. The Second Law only requires that the total entropy must increase for a spontaneous change. Therefore, in a spontaneous process, it is possible for the system to decrease in entropy so long as the surroundings undergo a greater increase in entropy, and vice versa. For example, water freezes spontaneously below its freezing point temperature, even though ice is a more ordered state ($$ΔS_{system} < 0$$). As freezing occurs, the water liberates heat (equal to its heat of fusion) to the surroundings, causing greater thermal motion and more disorder in the molecules of the surroundings ($$ΔS_{surroundings} > 0$$). This represents a greater increase in entropy for the surroundings than the decrease in entropy from the ordering of water molecules in the ice (the system), so the total entropy change is positive ($$ΔS_{total} > 0$$).

### Q1

Does the entropy of the system increase or decrease for the following changes? Indicate whether $$ΔS_{system} > 0$$ or $$ΔS_{system} < 0$$.

1. Water is boiled
2. $$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$$
3. $$N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)$$

## Gibbs Free Energy

For a chemical reaction run at constant temperature and pressure, the reaction’s effect on the entropy of the surroundings can be calculated by the equation

$ΔS_{surroundings} = –\dfrac{ΔH_{system} }{T}\label{eq1A}$

The $$ΔH _{system}$$ is the enthalpy change of the system, which transfers heat to or from the surroundings. The negative sign is inserted in the equation above to show the effect on the surroundings. Note that if $$ΔH$$ is negative, the entropy change for the surroundings will be positive, favoring a spontaneous process. This is consistent with the observation that many exothermic reactions occur spontaneously. However, not all exothermic reactions occur spontaneously, and not all endothermic reactions occur non-spontaneously. Recall that it is the total entropy change, $$ΔS_{total}$$, that determines whether or not a reaction will be spontaneous. As we have seen,

$ΔS_{total} = ΔS_{system} + ΔS_{surroundings}.\label{eq2A}$

Substituting the expression $$–ΔH _{system}/T$$ from Equation \ref{eq1A} for $$ΔS_{surroundings}$$ in Equation \ref{eq2A}, we may write

$ΔS_{total} = ΔS_{system} – \dfrac{ ΔH _{system} }{T}\nonumber$

Multiplying through by $$–T$$, this may be rewritten

\begin{align*} –TΔS_{total} &= –TΔS_{system} + ΔH_{system} \\[4pt] &= ΔH_{system} – TΔS_{system} \end{align*}

We can replace $$–TΔS_{total}$$ with a function first proposed by the American mathematician J. Willard Gibbs (1839-1903), called the Gibbs free energy, $$G$$:

$G = H – TS \nonumber$

For a chemical reaction at constant pressure and temperature, we define the change in Gibbs free energy as $$ΔG = –TΔS_{total}$$. Thus, our previous equation, $$–TΔS_{total} = ΔH _{system} – TΔS _{system}$$, becomes

$ΔG_{system} = ΔH_{system} – TΔS_{system} \label{Eq2}$

where all terms refer to the system.

The sign convention for $$ΔG$$ is consistent with what we have seen for other energy terms (e.g., $$ΔU$$, $$ΔH$$), by which heat is liberated when the sign is negative. Because $$ΔG$$ is defined on the basis of $$–TΔS_{total}$$, and because $$ΔS_{total}$$ indicates whether or not a reaction is spontaneous, we can make the following generalizations regarding the sign of $$ΔG$$ :

• If $$ΔG_{system} < 0$$ (i.e., negative), the reaction is spontaneous as written.
• If $$ΔG_{system} > 0$$ (i.e., positive), the reaction is non-spontaneous as written, but is spontaneous in the reverse direction.
• If $$ΔG_{system} = 0$$, the reaction is at equilibrium.

From Equation $$\ref{Eq2}$$, we can see the factors that favor a spontaneous reaction:

• Reactions with $$ΔH < 0$$ (exothermic) favor spontaneity.
• Reactions that increase randomness ($$ΔS > 0$$) favor spontaneity.

These two factors may work in opposition in certain cases, and the spontaneity determined from Equation \ref{Eq2} depends upon the relative magnitudes of $$ΔH$$ and $$TΔS$$.

### Q2

Fill in the missing values for the following reactions occurring at 25 oC, and determine if the reaction is spontaneous or non-spontaneous.

Reaction ΔH (kJ/mol) ΔS (J/mol×K) ΔG (kJ/mol) Spontaneous?
$$H_2(g) + Br_2(g) \rightarrow 2 HBr(g)$$ –72.46 +114.09
$$2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l)$$ –571.66 –326.34
$$2 N_2(g) + O_2(g) \rightarrow 2 N_2O(g)$$ +163.2 –147.99
$$N_2O_4(g) \rightarrow 2 NO_2(g)$$ +58.02 +176.61

## Standard Gibbs Free Energy of Formation

Like enthalpy, we can define a standard Gibbs Free Energy of formation, $$ΔG^o_f$$, which is the value of $$ΔG^o$$ when one mole of the substance in its standard state is formed from the stoichiometric amounts of its component elements, each in their standard states. As with $$ΔH^o_f$$, the standard state is 25 oC and 1 atm. These values of $$ΔG^o_f$$ can be used in the same way we used $$ΔH^o_f$$ values to calculate the change for an overall reaction; i.e.,

$ΔG^o = \sum nΔG^o_f (products) – \sum mΔG^o_f (reactants) \nonumber$

where $$n$$ and $$m$$ are the stoichiometric coefficients for each product and reactant, respectively. As with $$ΔH^o_f$$, $$ΔG^o_f$$ = 0 for all elements in their standard states.

### Q3

Given the following $$ΔG^o_f$$ values, calculate the standard free energy for the combustion of one mole of $$\ce{C2H6(g)}$$, and determine if the reaction is spontaneous or non-spontaneous:

• $$\ce{C2H6(g)}: –32.9 kJ$$
• $$\ce{H2O(l)}: –237.1 kJ$$
• $$\ce{CO2(g)}: –394.4 kJ$$

### Absolute Entropies and the Third Law

If we lower the temperature of a substance, molecular motion will be diminished and greater ordering will occur. At a temperature of absolute zero we might suppose that a perfect crystal, representing the ultimate order, would have an absolute entropy of zero ($$S = 0$$). This reasoning lead Walther Nernst in 1906 to formulate what is known as the Third Law of Thermodynamics:

At the absolute zero of temperature, a perfect crystalline substance would have an absolute entropy of zero.

However, absolute zero is an unattainable temperature and no substance forms a perfect crystal, so all substances have non-zero absolute entropies at all real temperatures.

Unlike enthalpy and free energy, absolute entropies, $$S$$, can be defined and calculated. Values are obtained from the temperature variation of heat capacities. However, when using these, it is important to realize that absolute entropies are not changes in entropy, $$ΔS$$.

The standard absolute entropy of a substance, $$S_o$$, is the entropy of the substance in its standard state at 25 °C and 1 atm. By the Third Law of Thermodynamics, these values are always positive numbers; i.e., $$S_o > 0$$. The change in entropy under standard condition for a reaction, $$ΔS^o$$, can be calculated from absolute standard entropy data as

$ΔS^o = \sum nS^o (products) – \sum mS^o (reactants) \nonumber$

where $$n$$ and $$m$$ are the stoichiometric coefficients for each product and reactant, respectively. Note, that the absolute entropy of an element is not zero, and the absolute entropy of a compound cannot be calculated from the absolute entropies of its elements. Standard absolute entropies of substances are routinely tabulated along with $$ΔH^o_f$$ and $$ΔG^o_f$$ data.

### Q4

Calculate $$ΔH^o$$, $$ΔS^o$$, and $$ΔG^o$$ for the following reaction at 25 °C.

$\ce{2 H2O2(l) -> 2 H2O(l) + O2(g)}\nonumber$

Given the following data:

Substance $$ΔH^o_f\, (kJ/mol)$$ $$S^o\, (J/mol\, K)$$
$$\ce{H2O(l) }$$ –285.83 69.91
$$\ce{O2(g)}$$ 0 205.0
$$\ce{H2O2(l)}$$ –187.8 109.6

## Gibbs Free Energy and Temperature

From the relationship $$ΔG = ΔH – TΔS$$, we can see that the value of the Gibbs free energy of a reaction depends upon the absolute temperature, $$T$$. But the values for ΔH and $$S$$ generally show only small changes with temperature. This allows us to use data for $$ΔH^o$$ and $$S^o$$ to estimate $$ΔG$$ values at non-standard temperatures (i.e., $$T \neq 298\, K$$). When we use $$ΔH^o$$ and $$S^o$$ values at nonstandard temperatures, however, we should realize that the $$ΔG^o$$ values obtained are only estimates. Nonetheless, these values usually lead to correct deductions about a reaction’s spontaneity at or near the chosen temperature.

### Q5

Consider the reaction

$\ce{CaCO3(s) -> CaO(s) + CO2(g)}\nonumber$

for which $$ΔH^o = +178.1 \,kJ/mol$$, $$ΔG^o = +130.2 \,kJ/mol$$, and $$ΔS^o = +160.5\, J/K \, mol$$. a. Is this reaction spontaneous at 25 °C? b. Assuming that $$ΔH$$ and $$ΔS$$ do not change significantly with changing temperature, is this reaction spontaneous at 1200 K?

### Q6

For the vaporization of cyclohexane,

$\ce{C6H12(l) <=> C6H12(g)}\nonumber$

$$ΔH^o = +33.1\, kJ/mol$$ and $$ΔS^o = 93.84\, J/K\,mol$$. Assuming that these values do not change significantly with increasing temperature, estimate the boiling point temperature of cyclohexane. [Hint: Recall that boiling means that the liquid and vapor are in equilibrium, and therefore $$ΔG = 0$$.]

## Calculating ΔG Under Non-Standard Conditions

Tabulated data for Gibbs free energies are values under standard conditions. Very often we are interested in a chemical system under non-standard conditions. The value of $$ΔG$$ under nonstandard conditions can be calculated from $$ΔG^o$$ by the equation

$ΔG = ΔG^o + RT \ln Q \label{GibbsQ}$

in which $$R = 8.314 J/K\,mol$$, $$T$$ is in units of kelvin, and $$Q$$ is the reaction quotient, which we defined in our discussions of the equilibrium constant. Recall that $$Q$$ has the same form as K, except the concentrations or pressures are not assumed to be equilibrium values.

Under standard conditions, all substances have unit activities; i.e., their effective concentrations are 1. This means that under standard conditions, all concentrations and pressures in the $$Q$$ expression are 1, so $$\ln Q = \ln (1) = 0$$. Thus, under standard conditions Equation \ref{GibbsQ} reduces to

$ΔG = ΔG^o \nonumber$

as it should. But most often we deal with chemical systems in which the concentrations and pressures of reactant and product species are not 1, even at 25 °C. To calculate $$ΔG$$ under such non-standard conditions of concentration and pressure, we need to evaluate $$Q$$ and use Equation \ref{GibbsQ}.

### Q7

Under standard conditions, $$ΔG^o = –32.74 \,kJ/mol$$ for the reaction

$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g) }\nonumber$

What is the value of $$ΔG$$ for the reaction at 298 K when the partial pressures of a mixture are 0.0100 atm for $$\ce{N2(g)}$$, 0.0100 atm for $$\ce{H2(g)}$$, and 5.00 atm for $$\ce{NH3(g)}$$? Is the reaction spontaneous or non-spontaneous under these conditions?

## Free Energy and the Thermodynamic Equilibrium Constant

At equilibrium, $$ΔG = 0$$ and $$Q = K$$. We can use these restrictions to derive the relationship between $$ΔG$$ and $$K$$ from Equation \ref{GibbsQ}

$ΔG = ΔG^o + RT \ln K = 0. \label{master}\nonumber$

Equation \ref{master} is a powerful equation that coupled equilibrium conditions to standard conditions (almost never the same). Rearranging, we have

$ΔG^o = –RT \ln K\nonumber$

from which we obtain We see the following generalizations from these equations:

 $$ΔG^o > 0$$ $$K < 1$$ $$ΔG^o = 0$$ $$K = 1$$ $$ΔG^o < 0$$ $$K > 1$$

$$K$$ in this equation is the thermodynamic equilibrium constant, defined in terms of the activities of participants in their standard states. This rigorous form of $$K$$ inherently has no units. For gas phase reactions, $$K$$ is approximately $$K_p$$, because the standard state of gas species at 25 °C is defined in terms of one atmosphere. Otherwise, the type of $$K$$ calculated from $$ΔG^o$$ depends upon the definition of standard states used in the determination of the Gibbs free energy. If all reactants and products are ions in solution, $$K$$ approximately corresponds to $$K_c$$, because the standard state of ions in solution is defined at 25 °C in terms of mol/L.

### Q8

Under standard conditions, $$ΔG^o = -32.74\, kJ$$ for the reaction

$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g).}\nonumber$

What is the value of $$K$$, the thermodynamic equilibrium constant, at 25 °C?

4: Second & Third Laws of Thermodynamics (Worksheet) is shared under a not declared license and was authored, remixed, and/or curated by Robert Carter.