# Including the Surroundings

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:23

In order to determine whether a reaction is spontaneousCapable of proceeding without an outside source of energy; refers to a reaction in which the products are thermodynamically favored (product-favored reaction). or not, it is not sufficient just to determine ΔSm, the entropyA thermodynamic state function, symbol S, that equals the reversible heat energy transfer divided by temperature; higher entropy corresponds to greater dispersal of energy on the molecular scale. See also standard entropy. difference between products and reactants. As an example, let us take the reaction

2Mg(s) + O2(g) → 2MgO(s)    1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg., 298 K      (1)

Since this reaction occurs at the standard pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. and at 298 K, we can find ΔS from the standard molar entropies:

ΔS = ΔSm°(298 K) = 2Sm°(MgO) – 2Sm°(Mg) – Sm°(O2)

= ( 2 × 26.8 – 2 × 32.6 – 205.0) J K–1 mol–1 = –216.6 J K–1 mol–1

This result would suggest that the reaction is not spontaneous, but in fact it is. Once ignited, a ribbon of magnesium metalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. burns freely in air to form solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. magnesium oxide in the form of a white powder. The reaction is plainly spontaneous even though ΔS is negative. Why is this not a contradiction of the second law?

The answer is that we have failed to realize that the entropy change which the magnesium and oxygen atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. undergo as a result of the reaction is not the only entropy change which occurs. The oxidationThat part of a chemical reaction in which a reactant loses electrons; simultaneous reduction of a reactant must occur. of magnesium is a highly exothermicDescribes a process in which energy is transferred to the surroundings as a result of a temperature difference. reaction, and the heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. which is evolved flows into the surroundings, increasing their entropy as well. There are thus two entropy changes which we must take into account in deciding whether a reaction will be spontaneous or not: (1) the change in entropy of the system actually undergoing the chemical changeA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also referred to as chemical reaction., which we will indicate with the symbol ΔSsys; and (2) the change in entropy of the surroundings, ΔSsurr, which occurs as the surroundings absorb the heat energyA system's capacity to do work. liberated by an exothermic reaction or supply the heat energy absorbed by an endothermicIn chemical thermodynamics, describes a process in which energy is transferred from the surroundings to the system as a result of a temperature difference. reaction.

Of these two changes, the first is readily obtained from tables of entropy values. Thus, for the oxidation of magnesium, according to Eq. (1), ΔSsys has the value already found, namely, –216.6 J K–1 mol–1.

The second entropy change, ΔSsurr, can also be derived from tables, as we shall now show.

When a chemical reactionA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also called chemical change. occurs at atmospheric pressure and its surroundings are maintained at a constant temperatureA physical property that indicates whether one object can transfer thermal energy to another object. T, then the surroundings will absorb a quantity of heat, qsurr equal to the heat energy given off by the reaction.

qsurr = –ΔH      (2)

(The negative sign before ΔH is needed because qsurr is positive if the surroundings absorb heat energy, but ΔH is negative if the system gives off heat energy for them to absorb.) If we now feed Eq. (2) into Eq. (1) in the section on measuring entropy, we obtain an expression for the entropy change of the surroundings in terms of ΔH:

$\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}=\frac{-\Delta H}{T}$      (3)

Using this equation it is now possible to find the value of ΔSsurr from tables of standard enthalpies of formation. In the case of the oxidation of magnesium, for example, we easily find that the enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure. change for

2Mg(s) + O2(g) → 2MgO(s)      1 atm, 298 K

is given by

ΔH = ΔHm°(298 K) = 2 × ΔHf° (MgO) = 2 × –601.8 J K–1 mol–1= –1204 J K–1 mol–1

Substituting this result into Eq. (3), we then find

$\Delta S_{\text{surr}}=\frac{-\Delta H}{T}=\frac{\text{1204 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}}=\text{4040 J K}^{-\text{1}}\text{ mol}^{-\text{1}}$

If a reaction is spontaneous, then it is the total entropy change ΔStot, given by the sum of ΔSsurr and ΔSsys, which must be positive in order to conform to the second law. In the oxidation of magnesium, for example, we find that the total entropy change is given by

ΔStot = ΔSsurr + ΔSsys = (4040 – 216.6) J K–1 mol–1 = 3823 J K–1 mol–1

This is a positive quantity because the entropy increase in the surroundings is more than enough to offset the decrease in the system itself, and the second law is satisfied. In the general case the total entropy change is given by

ΔStot = ΔSsurr + ΔSsys = $\frac{-\Delta H}{T}$ + ΔSsys

The second law requires that this sum must be positive; i.e.,

$\frac{-\Delta H}{T}$ + ΔSsys > 0      (4)

This simple inequality gives us what we have been looking for: a simple criterion for determining whether a reaction is spontaneous or not. Since both ΔH and ΔS can be obtained from tables, and T is presumably known, we are now able to predict in advance whether a reaction will be uphill or downhill.

EXAMPLE 1 Using the Table of Some Standard Enthalpies of Formation at 25°C and the Table of Standard Molar Entropies find ΔHm°(298 K) and ΔSm°(298 K) for the reaction

N2(g) + 3H2(g) → 2NH3(g)      1 atm

Predict whether the reaction will be spontaneous or not at a temperature of (a) 298 K, and (b) 1000 K.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We find from the Table of Some Standard Enthalpies of Formation that

ΔHm°(298 K) = ΣΔHf°(products) – ΣΔHf°(reactants)

= 2ΔHf°(NH3) – ΔHf°(N2) – 3ΔHf°(H2)

= (–2 × 46.1 – 0.0 – 0.0) kJ mol–1 = –92.2 kJ mol–1

and from the Table of Standard Molar Entropies:

ΔSm°(298 K) = ΣΔSm°(products) – ΣΔSm°(reactants)

= 2ΔSm°(NH3) – ΔSm°(N2) – 3ΔSm°(H2)

= (2 × 192.2 – 191.5 – 3 × 130.6) J K–1 mol–1 = –198.9 J K–1 mol–1

a) At 298 K the total entropy change per mol N2 is given by

ΔStot = $\frac{-\Delta H_{m}^{ o}}{T}$ + ΔSsys° = $\frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}}$ – 198.9 J K–1 mol–1

= (309.4 – 198.9) J K–1 mol–1= 110.5 J K–1 mol–1

Since the total entropy change is positive, the reaction is spontaneous:

b) Since tables are available only for 298 K, we must make the approximate assumption that neither ΔH nor ΔS varies greatly with temperature. Accordingly we assume

ΔHm°(1000 K) = ΔHm°(298 K) = –92.2 kJ mol–1

and      ΔSm°(1000 K) = ΔSm°(298 K) = – 198.9 J K–1 mol–1

Thus for 1 mol N2 reacted,

ΔStot = $\frac{-\Delta H_{m}^{ o}}{T}$ + ΔSsys° = $\frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{1000 K}}$ – 198.9 J K–1 mol–1

= (92.2 – 198.9) J K–1 mol–1= –106.7 J K–1 mol–1

Since the total entropy change is negative at this high temperature, we conclude that N2 and H2 will not react to form NH3, but that rather NH3 will decompose into its elements.

Apart from enabling us to predict the direction of a chemical reaction from tables of thermodynamic data, the inequality [Eq. (4)] shows that three factors determine whether a reaction is spontaneous or not: the enthalpy change ΔH, the entropy change ΔS, and the temperature T. Let us examine each of these to see what effect they have and why.

The enthalpy change, ΔH      As we well know, if ΔH is negative, heat will be released by the reaction and the entropy of the surroundings will be increased, while if ΔH is positive, the surroundings will decrease in entropy. At room temperature we usually find that this change in the entropy of the surroundings as measured by –ΔH/T is the major factor in determining the direction of a reaction since ΔS is almost always small by comparison. This explains why at room temperature most spontaneous reactions are exothermic. On the molecular level, as we saw in the sections on enthalpy, heat and energy, an exothermic reaction corresponds to a movement from a situation of weaker bonding to a situation of stronger bonding. The formation of more and/or stronger bonds is thus a big factor in tending to make a reaction spontaneous.

The entropy change, ΔSsys      If the system itself increases in entropy as a result of the reaction (i.e., if ΔSsys is positive), this will obviously contribute toward making the total entropy change positive and the reaction spontaneous. As we saw in the previous section of this chapter, reactions for which ΔS is positive correspond to the relaxation of some of the constraints on the motion of the atoms and molecules in the system. In particular, dissociationThe breaking apart of one species into two or more smaller species; often applied to ions in a crystal lattice, which dissociate when the ionic solid dissolves in water. Dissociation refers to separation of particles that already exist; ionization refers to the formation of ions from neutral species, as in the ionization of a weak acid in aqueous solutoin. reactions and reactions in which the amount of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. in the gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. phase increases correspond to reactions for which ΔS is positive.

The temperature, T      Because it alters the magnitude of –ΔH/T relative to ΔS, the temperature regulates the relative importance of the enthalpy change and the entropy change in determining whether a reaction will be spontaneous or not. As we lower the temperature, the effect of a reaction on the entropy of its surroundings becomes more and more pronounced because of the operation of the “boy in the living room” effect. As we approach absolute zeroThe minimum possible temperature: 0 K, -273.15 &deg;C, -459.67 &deg;F., the value of –ΔH/T begins to be an infinitely large positive or negative quantity, and ΔS becomes insignificant by comparison.

At a very low temperature, therefore, whether the reaction is spontaneous or not will depend on the sign of ΔH, i.e., on whether the reaction is exothermic or endothermic. By contrast, as we raise the temperature to very high values, the “boy in the bedroom” effect takes over and the reaction affects the entropy of its surroundings to an increasingly smaller extent until finally it is only the value of ΔSsys which determines the behavior of the reaction. In short, whether a reaction is spontaneous or not is controlled by the sign of ΔH at very low temperatures and by the sign of ΔS at very high temperatures.

Since ΔH can be positive or negative and so can ΔS, there are four possible combinations of these two factors, each of which exhibits a different behavior at high and low temperatures. All four cases are listed and described in the next table, and they are also illustrated by simple examples in Fig. 1. In this figure the surroundings are indicated by a shaded border around the reaction system. If the reaction is exothermic, the border changes from gray to pink, indicating that the surroundings have absorbed heat energy and thus increased in entropy. When the border changes from pink to gray, this indicates an endothermic reaction and a decrease in the entropy of the surroundings. In each case the change in entropy of the system should be obvious from an increase or decrease in the freedom of movement of the molecules and/or atoms.

Effect of Temperature on the Spontaneity of a Reaction.

 Case Sign of ΔH Sign of ΔS Behavior 1 – + Spontaneous at all temperatures 2 – – Spontaneous only at low temperatures 3 + + Spontaneous only at high temperatures 4 + – Never spontaneous
Figure 1 Shows two of four different thermodynamic types of chemical reactions. In both cases, ΔH is positive.
Figure 2 Shows the other two thermodynamic types of chemical reactions. In both cases, ΔH is negative.

Case l: Reaction is exothermic, and ΔSsys is positive    The example illustrated in Fig. 16.7(I) is the decomposition of ozone to oxygen.

2O3(g) → 3O2(g)      1 atm

for which ΔHm°(298 K) = –285 kJ mol–1 and ΔSm°(298 K) = +137 J K–1 mol–1. A reaction of this type is always spontaneous because the entropy of both the surroundings and the system are increased by its occurrence.

Case 2: Reaction is exothermic, and is ΔSsys negative    The example illustrated in Fig. 16.7(II) is the reaction of magnesium metal with hydrogen gas to form magnesium hydride:

Mg(s) + H2(g) → MgH2(g)      1 atm

for which ΔHm°(298 K) = –76.1 kJ mol–1 and ΔSm°(298 K) = –132.1 J K–1 mol–1. Reactions of this type can be either spontaneous or nonspontaneous depending on the temperature. At low temperatures when the effect on the surroundings is most important, the exothermic nature of the reaction makes it spontaneous. At high temperatures the effect of ΔSsys predominates. Since ΔSsys is negative (free H2 molecules becoming fixed H ions), the reaction must become nonspontaneous at high temperatures. Experimentally, solid MgH2 will not form from its elements above 560 K, and any formed at a lower temperature will decompose.

Case 3: Reaction is endothermic, and ΔSsys is positive    The example illustrated in Fig. 16.7(III) is the vaporizationThe formation of a vapor from a liquid; evaporation or boiling. of liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container bromine:

Br2(l)→ Br2(g)1 atm

for which ΔHm°(298 K) = +31.0 kJ mol–1 and ΔSm°(298 K) = 93.1 J K–1 mol–1. This example is usually classified as a physical rather than as a chemical change, but such distinctions are not important in thermodynamics. As in the previous case this reaction can be spontaneous or nonspontaneous at different temperatures. At low temperatures, bromine will not boil because the entropy increase occurring in the bromine as it turns to vaporThe gaseous state of a substance that typically exists as a liquid or solid; a gas at a temperature near or below the boiling point of the corresponding liquid. is not enough to offset the decrease in entropy, which the surroundings experience in supplying the heat energy which is needed for the change in state. At higher temperatures the entropy effect on the surroundings becomes less pronounced, and the positive value of ΔSsys makes the reaction spontaneous. At 101.3 kPa (1atm) pressure, bromine will not boil below 331 K (58°C), but above this temperature it will.

Case 4: Reaction is endothermic, and ΔSsys is negative    The example illustrated in Fig. (IV) is the reaction between silver and nitrogen to form silver azide, AgN3:

2Ag(s) + 3N2(g) → 2AgN3(s)

for which ΔHm°(298 K) = +620.6 kJ mol–1 and ΔSm°(298 K) = –461.5 J K–1 mol–1. Reactions of this type can never be spontaneous. If this reaction were to occur, it would reduce the entropy of both the system and the surroundings in contradiction of the second law. Since the forward reaction is nonspontaneous, we expect the reverse reaction to be spontaneous. This prediction is borne out experimentally. When silver azide is struck by a hammer, it decomposes explosively into its elements!

EXAMPLE 2 Classify the following reactions as one of the four possible types (cases) just described. Hence suggest whether the reaction will be spontaneous at (i) a very low temperature, and (ii) at a very high temperature.

 ΔHm°(298 K)/ kJ mol–1 ΔSm°(298 K)/ J K–1 mol–1 a) N2(g) + F2(g) → 2NF3(g) –249 –277.8 b) N2(g) + 3Cl2(g) → 2NCl3(g) +460 –275 c) N2F4(g) → 2NF2(g) +93.3 +198.3 d) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g) –2044.7 +101.3

Solution

a) This reaction is exothermic, and ΔS is negative. It belongs to type 2 and will be spontaneous at low temperatures but nonspontaneous at high temperatures.

b) Since this reaction is endothermic and ΔS is negative, it belongs to type 4. It cannot be spontaneous at any temperature.

c) Since this reaction is endothermic but ΔS is positive, it belongs to type 3. It will be spontaneous at high temperatures and nonspontaneous at ow temperatures. (All dissociation reactions belong to this class.)

d) This reaction belongs to type 1 and is spontaneous at all temperatures.