In the previous section, we saw ΔH⊖f and ΔH⊖r. In this section, we will see thermochemical equations and various standard enthalpies.
Thermochemical equations can be explained in 3 steps:
1. We saw a number of reactions in the previous section.
• In those reactions, heat was either absorbed or released.
♦ We denoted the heat as ΔH
2. When we study chemical reactions which involve heat, we write the ΔH together with the equation.
♦ We give the subscripts ‘f’ or ‘r’ which ever is appropriate.
♦ Also we give the superscript ‘⊖’ if it was a standard value.
• That ΔH value and the main equation are separated by a semi colon.
3. An example is given below:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); ΔH⊖r = -1367 kJ mol-1
• This equation describes the combustion of C2H5OH (ethanol) at constant temperature and pressure.
• The negative sign of ΔH⊖r indicates that, it is an exothermic reaction.
• Such equations are called thermochemical equations
Certain rules have to be followed while writing thermochemical equations. There are a total of four rules. Let us see them in detail:
1. Physical state of reactants and products.
• We must specify the physical state of each of the reactants and products.
• In the example in (3) above,
♦ C2H5OH is in the liquid state.
♦ O2 is in the gaseous state.
♦ CO2 is in the gaseous state.
♦ H2O is in the liquid state.
2. Use of balanced equations:
• Only balanced equations should be used in thermochemical equations.
• But coefficients in the balanced equation should be considered as the 'number of moles'. Not as 'number of molecules'.
• For example, the coefficient of C2H5OH in the above equation is ‘one’.
• When one mol C2H5OH undergoes combustion, we get an energy of 1367 kJ
♦ One mol C2H5OH is 46.07 grams.
♦ So when 46.07 grams of C2H5OH burns, we get 1367 kJ
• Obviously, one molecule of C2H5OH will not give so much energy.
3. Magnitude of energy:
• The magnitude of the ΔH⊖r value given after the semi colon should correspond exactly to the number of moles in the balanced equation.
• The importance of this rule can be explained using an example:
(i) Consider the reaction: Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l)
• Let us find the ΔH⊖r of this reaction.
• From the data book, we have:
♦ ΔH⊖r[Fe2O3(s)] = -824.2 kJ mol-1
♦ ΔH⊖r[H2(g)] = 0
♦ ΔH⊖r[Fe(s)] = 0
♦ ΔH⊖r[H2O(l)] = -285.83 kJ mol-1
• So ΔH⊖r = [2(0) + 3(-285.83)] - [1(-824.2) +3(0)] = -33.3 kJ mol-1
(ii) That means, 33.3 kJ is released.
• This much energy is released when one mol Fe2O3(s) reacts with three mol H2(g)
• We will denote it as: ΔH⊖r(1)
(iii) So we can write the thermochemical equation:
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l); ΔH⊖r(1) = -33.3 kJ mol-1
(iv) Let us divide the equation in (i) through out by 2. We get:
1⁄2Fe2O3(s) + 3⁄2H2(g) → Fe(s) + 3⁄2H2O(l)
• The new ΔH⊖r value will be:
[(0) + 3⁄2(-285.83)] - [1⁄2(-824.2) +3⁄2(0)] = -16.6 kJ mol-1
(v) That means, 16.6 kJ is released.
• This much energy is released when 1⁄2 mol Fe2O3(s) reacts with 3⁄2 mol H2(g)
• We will denote it as: ΔH⊖r(2)
(vi) So we can write the new thermochemical equation:
1⁄2Fe2O3(s) + 3⁄2H2(g) → Fe(s) + 3⁄2H2O(l); ΔH⊖f(2) = -16.6 kJ mol-1
• Note that: ΔH⊖r(2) is equal to 1⁄2 of ΔH⊖r(1)
(vii) Thus we can write:
• The energy absorbed or released will depend on the coefficients in the balanced equation.
• As a consequence, the energy absorbed or released will depend on the quantity of reactants taken.
• We cannot say that, the reaction between Fe2O3 and H2 will always give 33.3 kJ energy. It will depend on the quantity of reactants taken.
4. Reversing the equation:
• When a thermochemical equation is reversed:
♦ sign of the ΔH⊖r is reversed.
♦ magnitude of remains the same.
• For example:
♦ N2(g) + 3H2(g) → 2NH3(g); ΔH⊖r = -91.8 kJ mol-1
♦ is the reverse of
♦ 2NH3(g) → N2(g) + 3H2(g); ΔH⊖r = 91.8 kJ mol-1
• It is clear that:
♦ When 1 mol N2 combines with 3 mol H2 to form 2 mol NH3, 91.8 kJ is released.
♦ To decompose 2 mol NH3 into 1 mol N2 and 3 mol H2, we have to supply 91.8 kJ
Enthalpy changes during phase transformations
• We know that phase transformations involve energy changes.
• So we can specify standard enthalpy changes for such transformations.
• First we will see enthalpy change for fusion. It can be explained in 5 steps:
1. When ice melts to become water, heat is absorbed by the ice.
• This melting takes place at constant pressure and constant temperature (273 K)
2. Using calorimetry, scientists have calculated that:
♦ 1 mole of ice at 273 K requires
♦ 6.00 kJ heat
♦ to transform into water at 273 K
3. Based on this, we can write the thermochemical equation:
H2O(s) → H2O(l); ΔH⊖fus = 6.00 kJ mol-1
• Note that, instead of 'r', the subscript is 'fus'. This is to indicate 'fusion'.
• Fusion is the conversion from solid state to liquid state.
4. ΔH⊖fus is called the standard enthalpy of fusion.
• It is also known as the molar enthalpy of fusion.
◼ It is the amount of energy required for the melting of one mole of a solid substance in standard state.
5. We know that, melting is an endothermic process. So ΔH⊖fus will be always positive.
• Next we will see enthalpy change for vaporization. It can be explained in 5 steps:
1. When water boils to become steam, heat is absorbed by the boiling water.
• This boiling takes place at constant pressure and constant temperature (373 K)
2. Using calorimetry, scientists have calculated that:
♦ 1 mole of water at 373 K requires
♦ 40.79 kJ heat
♦ to transform into steam at 373 K
3. Based on this, we can write the thermochemical equation:
H2O(l) → H2O(g); ΔH⊖vap = 40.79 kJ mol-1
• Note that, instead of 'r', the subscript is 'vap'. This is to indicate 'vaporization'.
• Vaporization is the conversion from liquid state to gaseous state.
4. ΔH⊖vap is called the standard enthalpy of vaporization.
• It is also known as the molar enthalpy of vaporization.
◼ It is the amount of energy required for the vaporization of one mole of a liquid substance at constant temperature and standard pressure (1 bar)
5. We know that, vaporization is an endothermic process. So ΔH⊖vap will be always positive.
• We have seen that, the ΔH⊖vap of water is 40.79 kJ mol-1
• Now is a good time to see the relation between the following two items:
♦ Study of heat in chemistry.
♦ Study of heat in physics.
• The relation can be written in 8 steps:
1. One mole of water at 100 ०C requires 40.79 kJ
2. One mol H2O is 18 grams of H2O
♦ So 18 grams require 40.79 kJ
♦ Then one gram will require (40.79⁄18) kJ
♦ Then one kg will require (40.79⁄18 × 1000) = 2266.11 kJ
3. In physics classes, we see that:
One kg of water at 100 ०C requires 2260 kJ to completely change into water vapour at 100 ०C (In physics, this is called the latent heat of vaporization of water)
• Both values are identical. But the approaches are different.
• The following steps from (4) to (8) explains the difference between the two approaches.
4. In chemistry, we consider the enthalpy change.
♦ Let UA be the internal energy of water at 100 ०C
♦ Let UB be the internal energy of steam at 100 ०C
5. We can write: UB = UA + QP – P(VB – VA)
⇒ UB – UA + P(VB - VA) = QP
⇒ ΔU + W = QP
6. That means, QP (the heat supplied at constant pressure) is utilized for two items:
♦ Increasing the internal energy.
♦ Doing work for expansion against the constant pressure.
7. Also recall that, the left side of (5) can be written as:
(UB + PVB) – (UA + PVA)
• This is same as HB – HA
• So QP = (HB – HA) = ΔH
• ΔH is enthalpy change.
8. QP for water is the same (40.79 kJ mol-1 or 2260 kJ kg-1) in both physics and chemistry.
♦ In chemistry, we write it in terms of enthalpy change.
♦ In physics, we write it in terms of ‘internal energy change’ and ‘work’.
Solved example 6.14
A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 298K. Given: ΔH⊖vap of water at 298 K = 44.01 kJ mol-1
Solution:
Part (I):
1. Mass of the water film = 18 gram
• Molar mass of water = 18 gram
• So number of moles in the film = mass of film⁄molar mass = 18⁄18 = 1
2. ΔH⊖vap of water at 298 K = 44.01 kJ for one mol
• So heat required by the film = 44.01 kJ
Part (II):
1. We have: ΔU + W = QP (see step 5 in the previous discussion)
• QP is the heat absorbed at constant pressure. In our present case, it is 44.01 kJ
• So, if we can find W, the only unknown will be ΔU
2. So our next task is to find W.
• We have: W = P(VB – VA)
• Recall that, P(VB – VA) is equal to ngRT
Where ng = number of gaseous moles in state B – number of gaseous moles in state A.
(see Eq.6.6 in section 6.2)
3. In our present case, ng = (1-0) = 1
• So we get: W = ngRT
= 1 (mol) × 8.314 × 10-3 (kJ mol-1 K-1) × 298 (K) = 2.477 kJ
• Substituting this in (1), we get:
ΔU + 2.477 = 44.01
⇒ ΔU = 41.53 kJ
Solved example 6.15
Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100°C and 1 bar pressure is converted to ice at 0°C. Given:
♦ Enthalpy of fusion of ice is 6.00 kJ mol-1
♦ Heat capacity of water is 4.2 J g-1 K-1
Solution:
• One mole of water at boiling point (100 ०C) is being converted into ice at 0 ०C
• We will do this problem in two stages:
Stage 1: Water at 100 ०C is converted into water at 0 ०C
1. We have: ΔU + W = QP
• QP is the heat absorbed or released at constant pressure.
♦ In our present case, it is the heat released because, water cools from 100 ०C to 0 ०C
• So, if we can find QP and W, the only unknown will be ΔU
2. First we will find QP
• We know that, heat absorbed or released is given by mCΔT
• One mol water is 18 gram water. So m = 18 gram
• C is given as 4.2 J g-1 K-1
• So QP = 18 (g) × 4.2 (J g-1 K-1) × 100 (K-1) = 7560 J = 7.56 kJ
3. Our next task is to find W
• We have: W = P(VB – VA)
• Recall that, P(VB – VA) is equal to ngRT
Where ng = number of gaseous moles in state B – number of gaseous moles in state A.
(see Eq.6.6 in section 6.2)
4. In our present case, ng = (1-1) = 0
• So we get: W = 0
• Substituting the known values in (1), we get:
ΔU + 0 = 7.56
⇒ ΔU = 7.56 kJ
Stage 2: Water at 0 ०C is converted into ice at 0 ०C
1. We have: ΔU + W = QP
• QP is the heat absorbed or released at constant pressure.
♦ In our present case, it is the heat released because, when ice is formed, heat is released even if it is a process at constant temperature.
• So, if we can find QP and W, the only unknown will be ΔU
2. First we will find QP
• We know that, heat released during formation of ice is same as:
The enthalpy of fusion of ice.
♦ It is given to us: 6.00 kJ mol-1
• In our present case, the quantity is one mole.
• So QP = 6.00 kJ
3. Our next task is to find W.
• We have: W = P(VB – VA)
• Recall that, P(VB – VA) is equal to ngRT
Where ng = number of gaseous moles in state B – number of gaseous moles in state A
(see Eq.6.6 in section 6.2)
4. In our present case, ng = (1-1) = 0
• So we get: W = 0
• Substituting the known values in (1), we get:
ΔU + 0 = 6.00
⇒ ΔU = 6.00 kJ
◼ Combining the two stages, we get:
Total change in internal energy = (7.56 + 6.00) = 13.56 kJ
• Next we will see enthalpy change for sublimation. It can be explained in 5 steps:
1. When dry ice sublimes to become 'gaseous CO2', heat is absorbed by the dry ice.
• This process takes place at constant pressure and constant temperature (195 K).
2. Using calorimetry, scientists have calculated that:
♦ 1 mole of dry ice at 195 K requires
♦ 25.2 kJ heat
♦ to transform into gaseous CO2 at 195 K
3. Based on this, we can write the thermochemical equation:
CO2(s) → CO2(g); ΔH⊖sub = 25.2 kJ mol-1
• Another example is naphthalene. It sublimes slowly
♦ The ΔH⊖sub value is 73.0 kJ mol-1
• Note that, instead of 'r', the subscript is 'sub'. This is to indicate 'sublimation'.
• Sublimation is the direct conversion from solid state to gaseous state.
4. ΔH⊖sub is called the standard enthalpy of sublimation.
• It is also known as the molar enthalpy of sublimation.
◼
It is the amount of energy required for the sublimation of one mole of
a solid substance at constant temperature and standard pressure (1
bar).
5. We know that, sublimation is an endothermic process. So ΔH⊖sub will be always positive.
• Based on the sublimation temperature of dry ice, we can write some interesting points about heat. It can be written in 6 steps:
1. We saw that, dry ice sublimes at 195 K.
• In Celsius scale, this temperature is (273.15 – 195) = - 78.15 ०C
2. Imagine a room with some blocks of dry ice.
• Let the temperature inside the room be -80 ०C
• This is lower than -78.15 ०C. So the dry ice can easily remain in the solid state.
3. Let the temperature be gradually increased to -75 ०C
• As soon as the temperature reaches -78.15 ०C, the dry ice will begin to sublime.
4. Remember that, sublimation of dry ice requires 25.2 kJ mol-1
• So our present sublimation indicates that, heat is flowing into the dry ice.
5. -78.15 ०C is too cold for us humans
• We see that, heat can flow even at such low temperatures.
• Due to the heat flow, the dry ice sublimes at -78.15 C
• -78.15 ०C is too hot for dry ice.
6. We can write:
• We use the words heat, hotness and heat flow commonly in our day to day life.
• The hotness experienced by one person may not be so hot for another person.
• For example, a tourist visiting a 'place with hot climate' may find the heat unbearable. But the same heat may not affect a native.
• However, in science, those words have precise definitions.
• Next we will see enthalpy change for combustion. It can be explained in 5 steps:
1. When cooking gas (butane) undergoes combustion, heat is released.
2. Using calorimetry, scientists have calculated that:
♦ 1 mole of butane releases
♦ 2658.0 kJ heat
3. Based on this, we can write the thermochemical equation:
C4H10(g) + 13/2 O2(g) → 4CO2 (g) + 5H2O(l); ΔH⊖c = -2658.0 kJ mol-1
• Note that, instead of 'r', the subscript is 'c'. This is to indicate 'combustion'.
• Combustion is the reaction with oxygen.
• Another example is the combustion of glucose. The thermochemical equation is:
C6H12O6(s) + 6O2(g) → 6CO2 (g) + 6H2O(l); ΔH⊖c = -2802.0 kJ mol-1
4. ΔH⊖c is called the standard enthalpy of combustion.
• It is also known as the molar enthalpy of combustion.
◼
It is the amount of energy released when one mole of
a substance undergoes combustion. The reactants and products should be in their standard states.
5. We know that, combustion is an exothermic process. So ΔH⊖c will be always negative.
• We have seen:
Enthalpy of formation, reaction, fusion, vaporization, sublimation and combustion.
• We have also seen calorimetry and thermochemical equations.
• Based on those topics, we will see some solved examples. The link is given below:
Solved examples 6.16 to 6.21
• Next we have to see:
Enthalpy of atomization, bond, lattice and solution.
• But before that, we have to see Hess’s law. We will see it in the next section.
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