This reaction is between barium hydroxide and ammonium chloride. Both of these reactants are white crystalline solids. The two white solids are mixed together in a erlenmeyer flask. Before the reaction begins the two solids are at room temperature as is the flask. When the two solids are mixed the first observation is the two solids form a slurry, as a liquid is formed. This is very interesting, but not surprising when you are told water is a product of the reaction. The second observation occurs when the flask is picked up. The temperature has fallen when the two reactants are mixed to a very low temperature. The reaction is;Ba(OH)2.8H2O(s) + 2NH4SCN(s) ---> Ba(SCN)2(aq) + 2NH3(g) + 10H2O(l)As the two white solids are mixed we observed how the temperature of the reaction changed. The temperature is falling-it is getting cooler. Notice also the change which has occurred in the flask. The two solids have changed and we now have a liquid. If we could smell the reaction vessel it would smell of ammonia. Heat is a reactant, it is absorbed by the reactants.(Click on the picture on the right to view the reaction.) |
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For a particulate level view of a solid we will look at a model developed by concord.org. Atoms in the solid state are in fixed positions. In the solid phase atoms, molecules, or particles can vibrate, but they can not rotate or translate. In the solid phase particules are not very dispersed in terms of their position. In terms of energy, or solids atomic energies are limited in range, only in vibrational modes. The limited amount of dispersion in position and energy results in the entropy of solids being very small. |
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For a particulate level view of a liquid we will look at a model developed by concord.org. As we see in this model atoms in the liquid state are not in fixed positions. In the liquid phase atoms, molecules, or particles can translate and vibrate. There is also the possibility of molecules rotating. In the liquid phase particules are more dispersed in terms of their position compared to solids. In terms of energy, for liquids energies can be dispersed over three modes; translation, rotation and vibration. The greater amount of dispersion in position and energy for particles in the liquid phase, results in the entropy of liquids, in general, being larger compared to solid. |
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For a particulate level view of a gas we will look at a model developed by concord.org. As we see in this model atoms in the gas state have the highest amount of dispersion in position and in energy. In the gas phase atoms, molecules, or particles can translate, rotate and vibrate. In the gas phase particules are maximally dispersed in terms of their position. In terms of energy, for gases, energies can be dispersed over three modes; translation, rotation and vibration. The greater amount of dispersion in position and energy for particles in the gas phase, results in the entropy of gas, having the largest entropy values. |
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For a particulate level view of a phase change we will look at a model developed by concord.org. |
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the second law of thermodynamics states that thermodynamically favored natural processes are accompanied by an increase in the entropy of the universe. When considering qualitative prediction in terms of entropy the most important factor is the phase of the substance(s) involved. The next most important factor is the number of particles involved. for example, in a chemical reaction where the phase of the reactants and products are the same we must count the number of reactant and product particles. If there are more particles in the products compared to the reactants, based on the coefficients, the change in entropy will be positive. The third factor to consider, when the phases and numbers of reactants and products are all the same, is the complexity of the reactants or products. Complexity can be determined by the number of different atoms in reactant or product molecules. We will consider a number of examples below to practice.
the third law of thermodynamics says that the entropy of any pure substance at absolute zero is zero.
Substance |
Entropy (joules Kelvins-1) |
C(s) (graphite) |
5.69 |
C(s) (diamond) |
2.44 |
Al(s) |
28.3 |
Ca(s) |
41.6 |
Ba(s) |
62.5 |
Br2(l) |
152 |
H2(g) |
130.6 |
N2(g) |
191 |
O2(g) |
205 |
NO(g) |
210.6 |
F2(g) |
202.7 |
H2O(l) |
69.9 |
H2O(g) |
189 |
CO2(g) |
213.6 |
NH3(g) |
193 |
CH4(g) |
186 |
SO2(g) |
248.5 |
SO3(g) |
256.2 |
C2H6(g) |
229.5 |
C3H8(g) |
269.9 |
HF(g) |
174 |
HCl(g) |
187 |
HBr(g) |
199 |
HI(g) |
206 |
NaCl(s) |
72.3 |
NaCl(aq) |
115.5 |
NaNO3(s) |
116.5 |
NaNO3(aq) |
207 |
NaOH(s) |
64.5 |
NaOH(aq) |
49.8 |
physical states - looking at the table above in general the entropy values for gases are larger than entropy values for liquids and for solids. Particles in the gas phase are more dispersed in energy and in position compared to a particles in the liquid phases and both liquid phase and gas phase are more dispersed than a solid. So we would expect that chemical reactions producing more gases compared to the reactants will have a positive S.
temperature - As discussed earlier about the third law of thermodynamics, the higher the temperature the greater the energy for the substance/system, and the more dispersed that energy can become.
solution process - dissolving solids in water/solvent generally leads to greater dispersal. So comparing NaCl(s) to NaCl(aq) in the aqueous solution the ions are more dispersed compared to the ions in the solid. However, there are some exceptions to this rule, some ions have a very large charge and when added to water, order the water molecules resulting in a more ordered state, or reducing the extent that water molecules are dispersed in the solution. Al3+ is an example of an ion that when dissolved in water produces a more ordered system. When gases dissolve in water the S is always negative. For example, O2(g) has a higher entropy than O2(aq).
molecular complexity - this is a more complicated issue. From the table above we can see some trends, for example, HF, HCl, HBr and HI. The entropy value increases going down the group. The more electrons in the molecule of similar structure the greater the entropy. We can also see a similar trend for CH4(g), C2H6(g), and C3H8(g). In this case the molecule is getting more complex as CH2 groups are added.
In a reaction if the products are more dispersed (in energy and position) compared to the reactants the sign of S is positive (+).
If the products are less dispersed (in energy and position) compared to the reactants the sign of S is negative (-).
1. Phase changes-this is the most important factor when determining the S for a reaction. If the product phase(s) are phase(s) where particles are more dispersed compared to the reactants S will be positive.
Example: What is the sign of S for the reaction;
2HgO(s) ----> 2Hg(l) + O2(g)
S is positive (+) since a solid (reactants) is forming a liquid and a gas (products).
2. n (change in moles)-this is the next most important factor when determining S for a reaction. So if the chemical change can not be determined in terms of phases, doispersal is increased when there are more possibilities of where an element can be. So in a reaction the more moles of substance there are the greater the entropy (disorder).
Example: What is the sign of S for the reaction;
In this reaction the products and the reactants have the same phase, so we can not use phase change to determine the S. There are two moles of products and three moles of reactants. Since there are fewer moles of products the reaction S is negative.
3. Molecular complexity-the last factor to consider, assuming there is no change in phase, and number of moles of product and reactants are equal, is the molecular complexity. The more atoms, the more variety of elements, and/or the more number of electrons all favor greater disorder.
Example: What is the sign of S for the reaction;
H2(g) + Cl2(g) ----> 2HCl(g)
In this reaction the the product and reactants have the same phase and there are equal moles of reactants and products. Molecular complexity must be the criteria used to determine the S. In this case HCl has great molecular complexity compared to H2 or Cl2 because the HCl has different elements. The difference in entropy between the reactants and products is small, but S is still very slightly positive.
Sample Problem
Is the reaction below thermodynamically favored at 298 K?
a) 2H2(g) + O2(g) ----> 2H2O(g) Answer
Sample Problem
Calculate G for the reaction,
Answer
Sample Problem:
Calculate the temperature the reaction,
2H2(g) + O2(g) ----> 2H2O(g)
becomes not thermodynamically favored. Answer