Chapter 6.2 - Enthalpy

In the previous section, we saw the first law of thermodynamics. We also saw the work done in various thermodynamic processes. In this section, we will see enthalpy.

Details about enthalpy can be written in 20 steps:
1. Consider an isochoric (constant volume) process
• We have seen that, no work is done during an isochoric process
• So we can write: U_B = U_A + Q + 0
⇒ ΔU = U_B – U_A = Q
2. That means, the 'heat' is equal to 'change in internal energy'. We can write:
• If during an isochoric process, heat is supplied to the system, all that heat will be utilized to increase the internal energy of the system.
    ♦ That is., ‘increase in internal energy’ will be equal to the ‘heat supplied’.
• If during an isochoric process, heat is released from the system, all that heat will be at the expense of the internal energy of the system.
    ♦ That is., 'decrease in internal energy' will be equal to the 'heat released'.
3. In chemistry, most of the reactions are carried out in open flasks or test tubes.
    ♦ So the system is open to the atmosphere.
• So a constant pressure (the atmospheric pressure) is acting on the system.
• That means, the systems that we encounter in chemistry are not isochoric but isobaric.
4. We know that, work done in an isobaric process is given by PΔV
• So applying the first law of thermodynamics, we get:
U_B = U_A + Q_P – P ΔV
    ♦ Where
          ✰ P is the atmospheric pressure.
          ✰ Q_P is the heat supplied at constant atmospheric pressure.
• Thus we get:
ΔU = U_B – U_A = Q_P – P ΔV
5. We know that, ΔV is the change in volume experienced by the system.
    ♦ In the initial state A, the volume of the system will be V_A
    ♦ In the final state B, the volume of the system will be V_B
• So change in volume ΔV will be equal to (V_B – V_A)
• So the result in (4) will becomes: U_B - U_A = Q_P – P(V_B – V_A)
⇒ U_B – U_A = Q_P – P V_B + P V_A
⇒ U_B + P V_B – U_A – P V_A = Q_P
• Thus we get Eq.6.2: Q_P = (U_B + P V_B) – (U_A + P V_A)
6. Eq.6.2 gives an easy method to calculate Q_P
• The quantities inside the brackets can be considered as a new property.
• Consider the second item on the right side. It is 'U_A + P V_A'
• Let us analyze U_A + PV_A
Clearly, it is the sum of two items:
(i) U_A, which is the internal energy of the system at the initial state.
(ii) PV_A, which is the product of the pressure and volume of the system at the initial state.
7. Note that, dimensionally, the product ‘PV_A’ ie equivalent to ‘energy’.
• U_A is also energy.
• So the addition of U_A and PV_A is dimensionally correct.
• The sum gives an energy.
■ This energy is called enthalpy
• It is derived from the Greek word ‘enthalpein’ which means: heat
8. Enthalpy is represented by the letter H
• So we can write:
U_A + P V_A = H_A
• H_A is the enthalpy possessed by the system at the initial state.
9. Similarly we can write:
U_B + P V_B = H_B
• H_B is the enthalpy possessed by the system at the final state.
10. In general we can write:
Eq.6.3: H = U + PV
11. We know that, U is a state variable.
• Now let us check whether enthalpy is a state variable or not.
    ♦ We have: H = U + PV
    ♦ We know that, U, P and V are state variables
    ♦ So H will be a state variable
12. Now, Eq.6.2 can be modified as: Q_P = H_B – H_A
From this we get Eq.6.4: Q_P = ΔH
• That means:
Heat in an isobaric process
= The change in enthalpy
= The final enthalpy minus initial enthalpy
13. Let us check whether Q_P is a state variable or not
    ♦ We have Q_P = H_B – H_A
    ♦ We just saw that, H is a state variable.
    ♦ So Q_P will be a state variable.
14. It is interesting to note that, heat (Q) is a path variable. But heat at constant pressure (Q_P) is a state variable.
15. In the previous sections, we have seen that:
    ♦ We are not interested in the actual values U_A and U_B
    ♦ What we want is the difference (ΔU) between U_A and U_B
• Now, we see that, H_A and H_B depends on U_A and U_B as follows:
      H_A = U_A + PV_A and H_B = U_B + PV_B
      [See steps (8) and (9)]
• So it seems that, we cannot ignore the actual values U_A and U_B, because, they are necessary to calculate H_A and H_B
• But fortunately, we do not need the actual values H_A and H_B either.
    ♦ What we want is the difference (ΔH) between H_A and H_B
• When we calculate ΔH, we get:
(U_A + PV_A) - (U_B + PV_B)
= (U_A – U_B) + P(V_A – V_B)
• So with regard to ΔH also, the difference ΔU is what we want.
16. An important point to note:
• When the reaction takes place at constant pressure, the heat absorbed/liberated is equal to the difference in enthalpy ΔH.
    ♦ This we proved in step (12)
17. Based on this information, we can analyze exothermic and endothermic reactions.
■ First we will see the condition in which ΔH is positive. This can be written in 3 steps:
(i) We have: Q_P = H_B – H_A
    ♦ If Q_P is positive, it implies that: H_B > H_A
    ♦ It also implies that ΔH is positive.
(ii) H_B > H_A implies that final enthalpy is greater than initial enthalpy.
    ♦ That means, heat is absorbed during the reaction.
    ♦ But a reaction in which heat is absorbed is an endothermic reaction.
(iii) So we can write:
If ΔH is positive, it implies that, the reaction is endothermic.
■ Next we will see the condition in which ΔH is negative. This can be written in 3 steps:
(i) We have: Q_P = H_B – H_A
    ♦ If Q_P is negative, it implies that: H_B < H_A
    ♦ It also implies that ΔH is negative.
(ii) H_B < H_A implies that final enthalpy is lesser than initial enthalpy.
    ♦ That means, heat is released during the reaction.
    ♦ But a reaction in which heat is released is an exothermic reaction.
(iii) So we can write:
If ΔH is negative, it implies that, the reaction is exothermic.
18. The relation between ΔH and ΔU when the process is isochoric:
• This can be explained in 3 steps:
(i) Consider the result in (15): ΔH = (U_A – U_B) + P(V_A – V_B)
    ♦ Thus we get Eq.6.5: ΔH = ΔU + PΔV
(ii) For an isochoric process, ΔV = 0
    ♦ So for an isochoric process, ΔH = ΔU
(iii) In reactions where there are solids and/or liquids, there will not be appreciable changes in volumes.
    ♦ In such cases, ΔV can be taken as zero.
    ♦ So for reactions involving only solids and/or liquids, ΔH will be equal to ΔU.
19. The relation between ΔH and ΔU when gases are involved:
• This can be explained in 2 steps:
(i) When gases are involved, there will be volume changes.
    ♦ Then ΔH will not be equal to ΔU.
(ii) That means, there will be some difference between ΔH and ΔU.
    ♦ This difference between ΔH and ΔU will be equal to PΔV.
    ♦ This is evident from Eq.6.5
20. So, when gases are involved in a reaction, we will have to calculate PΔV.
• Let us see how it is calculated. It can be written in 6 steps:
(i) In the initial state A, only reactants will be present.
• That is., at state A, all the molecules present in the system will be those of the reactants.
• There will not be any molecules of the products.
(ii) Let V_A be the total volume of all the reactant molecules.
• Using ideal gas equation, we can write: PV_A = n_ART
    ♦ Where n_A is the total number of moles of the reactants.
(iii) Let V_B be the total volume of all the product molecules.
• Using ideal gas equation, we can write: PV_B = n_BRT
    ♦ Where n_B is the total number of moles of the products.
(iv) Subtracting PV_A from both sides of (iii), we get:
PV_B – PV_A = n_B RT – PV_A
(v) But from (ii), PV_A = n_ART
• So the result in (iv) becomes:
PV_B – PV_A = n_BRT – n_ART
⇒ P (V_B – V_A) = (n_B – n_A) RT
(vi) Thus we get Eq.6.6: PΔV = Δn_gRT
    ♦ Where Δn_g is the difference between the two items:
          ✰ Total number of gaseous moles in the final state B.
          ✰ Total number of gaseous moles in the initial state A.

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We will see some solved examples:
Solved example 6.6
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100°C is 41kJ mol_^-1 . Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C.
Solution:
• We can solve this problem using two methods
    ♦ Method I is by using the basics.
    ♦ Method II is by using the equations that we derived.
Method I:
1. Let A be the initial state.
• In this state, the system is:
    ♦ 1 mol of water at 100 °C and 1 bar pressure
2. Let B be the final state.
• In this state, the system is:
    ♦ 1 mol of water vapour at 100 °C and 1 bar pressure.
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. But enthalpy change is given as 41 KJ
• So we can write:
[(U_B - U_A) + P(V_B - V_A)] = 41
5. In the above equation, there are 3 terms: (U_B - UA), P(V_B - V_A) and 41
• So, if we can find P(V_B - V_A), we can calculate (U_B - U_A)
6. P(V_B - V_A) can be calculated in steps:
(i) From ideal gas equation, we have:
    ♦ P_AV_A = n_ART_A
    ♦ P_BV_B = n_BRT_B
• In our present case:
    ♦ P_A = P_B = P = 1 bar
    ♦ T_A = T_B = T = 100 °C = 373 K
(ii) So P(V_B - V_A) = (n_B - n_A)RT
    ♦ n_A = number of gaseous moles in the initial state = 0
    ♦ n_B = number of gaseous moles in the final state = 1
    ♦ So (n_B - n_A) = 1
(iii) Substituting the known values, the right side of (ii) becomes:
(1 mol) × (8.3 J mol_^-1 K_^-1) × (373 K) = 3095.9 J = 3.096 kJ
7. We can use the value obtained above, in the place of P(V_B - V_A) in (4)
• We get: [(U_B - U_A) + 3.096] = 41
• So (U_B - U_A) = ΔU = (41 - 3.096) = 37.904 kJ

Method II:
1. The pressure is 1 bar and temperature is 373 K
    ♦ Under these conditions, 1 mol of water changes into steam.
    ♦ During this process, the pressure and temperature remains the same.
    ♦ 1 mol of water at 100 °C is initial state A.
    ♦ 1 mol of water vapour at 100 °C is final state B.
2. Given that, the enthalpy change for this process is 41 kJ
    ♦ So ΔH = 41 kJ
3. We are asked to find the internal energy change ΔU
    ♦ We have Eq.6.5 which connects ΔH and ΔU: ΔH = ΔU + PΔV
4. So our next aim is to find PΔV
    ♦ We have Eq.6.6: PΔV = Δn_gRT
5. Δn_g = number of gaseous moles in the final state - number of gaseous moles in the initial state = (1 - 0) = 1
6. Substituting the known values in (4), we get:
PΔV = (1 mol) × (8.3 J mol^_-1 K^_-1) × (373 K) = 3095.9 J = 3.096 kJ
7. So from (3), we get:
ΔU = (ΔH - PΔV) = (41 - 3.096) = 37.904 kJ

Solved example 6.7
The molar enthalpy change for fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔU for the melting of ice at 0.0°C
Solution:
• We can solve this problem using two methods
    ♦ Method I is by using the basics.
    ♦ Method II is by using the equations that we derived.

Method I:
1. Let A be the initial state
• In this state, the system is:
    ♦ 1 mol of ice at 0.0 °C and 1.00 atm pressure
2. Let B be the final state.
• In this state, the system is:
    ♦ 1 mol of water at 0.0 °C and 1.00 atm pressure
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. But enthalpy change is given as 6.01 KJ
• So we can write:
[(U_B - U_A) + P(V_B - V_A)] = 6.01
5. In the above equation, there are 3 terms: (U_B - UA), P(V_B - V_A) and 6.01
• So, if we can find P(V_B - V_A), we can calculate (U_B - U_A)
6. P(V_B - V_A) can be easily calculated because, the values of P, VA and VB are given:
    ♦ P = 1 atm = 1.013 × 10^₅ N m^_-2.
    ♦ Initial volume V_A = Volume of 1 mol ice = 0.0197 L = 0.0197 × 10^_-3 m^₃
    ♦ Final volume V_B = Volume of 1 mol water = 0.0180 L = 0.0180 × 10^_-3 m^₃.
We get: P(V_B - V_A) = 1.013 × 10^₅ × (0.0180 - 0.0197) × 10^_-3 = -0.17221 J = -0.00017221 kJ
7. We can use the value obtained above, in the place of P(V_B - V_A) in (4)
• We get: [(U_B - U_A) -0.00017221] = 6.01
• So (U_B - U_A) = ΔU = (6.01 + 0.00017221) = 6.01 kJ

Method II:
1. The pressure is 1 atm (1.013 × 10_⁵ N m_^-2) and temperature is 0.0 °C (273 K)
    ♦ Under these conditions, 1 mol of ice changes into water.
    ♦ During this process, the pressure and temperature remains the same.
2. Given that, the enthalpy change for this process is 6.01 kJ
    ♦ So ΔH = 6.01 kJ
3. We are asked to find the internal energy change ΔU
    ♦ We have Eq.6.5 which connects ΔH and ΔU: ΔH = ΔU + PΔV
4. So our next aim is to find PΔV
• In this problem, the initial and final volumes are given:
    ♦ Initial volume V_A = Volume of 1 mol ice = 0.0197 L = 0.0197 × 10_^-3 m_³
    ♦ Final volume V_B = Volume of 1 mol water = 0.0180 L = 0.0180 × 10_^-3 m_³
• Thus PΔV = 1.013 × 10_⁵ × (0.0180 - 0.0197) × 10_^-3 = -0.17221 J = -0.00017221 kJ
5. So from (3), we get:
ΔU = (ΔH - PΔV) = [6.01 - (-0.00017221)] = 6.01 kJ
6. In this problem, gases are not involved. Water is in the liquid state and ice is in the solid state. We have seen that, when gases are not involved, ΔH  will be equal to ΔU (see step 18).

Solved example 6.8
At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔU for the conversion of C (graphite) to C (diamond) under these conditions
Solution:
• We can solve this problem using two methods
    ♦ Method I is by using the basics.
    ♦ Method II is by using the equations that we derived.

Method I:
1. Let A be the initial state.
• In this state, the system is:
    ♦ 1 mol of graphite at 298 K and 1 atm pressure
2. Let B be the final state.
• In this state, the system is:
    ♦ 1 mol of diamond at 298 K and 1 atm pressure
3. We have seen that:
    ♦ Enthalpy in the initial state, H_A = (U_A + PV_A)
    ♦ Enthalpy in the final state, H_B = (U_B + PV_B)
    ♦ So enthalpy change = (H_B - H_A) = [(U_B - U_A) + P(V_B - V_A)]
4. When a process takes place at constant pressure, the heat supplied is equal to the enthalpy change (Eq.6.4)
• But heat supplied is given as 1.850 KJ
• So we can write:
[(U_B - U_A) + P(V_B - V_A)] = 1.850
5. In the above equation, there are 3 terms: (U_B - UA), P(V_B - V_A) and 1.850
• So, if we can find P(V_B - V_A), we can calculate (U_B - U_A)
6. P(V_B - V_A) can be easily calculated because, the values of P, V_A and V_B are given:
    ♦ P = 1 atm = 1.013 × 10^₅ N m^_-2.
    ♦ Initial volume V_A = Volume of 1 mol graphite = 0.00534 L = 0.00534 × 10^_-3 m^₃
    ♦ Final volume V_B = Volume of 1 mol diamond = 0.00342 L = 0.00342 × 10^_-3 m^₃
We get: P(V_B - V_A) = 1.013 × 10^₅ × (0.00342 - 0.00534) × 10^_-3 = -0.1944 J = -0.0001944 kJ
7. We can use the value obtained above, in the place of P(V_B - V_A) in (4)
• We get: [(U_B - U_A) -0.0001944] = 1.850
• So (U_B - U_A) = ΔU = (1.850 + 0.0001944) = 1.850 kJ

Method II:
1. The pressure is 1 atm (1.013 × 10_⁵ N m_^-2) and temperature is 298 K
    ♦ Under these conditions, 1 mol of graphite changes into diamond.
    ♦ During this process, the pressure and temperature remains the same.
2. When pressure remains the same, the heat is equal to ΔH (see Eq.6.4)
• Given that, 1.850 kJ of heat is required.
    ♦ So Q_P = ΔH = 1.850 kJ
3. We are asked to find the internal energy change ΔU
    ♦ We have Eq.6.5 which connects ΔH and ΔU: ΔH = ΔU + PΔV
4. So our next aim is to find PΔV
• In this problem, the initial and final volumes are given:
    ♦ Initial volume V_A = Volume of 1 mol graphite = 0.00534 L = 0.00534 × 10_^-3 m_³
    ♦ Final volume V_B = Volume of 1 mol diamond = 0.00342 L = 0.00342 × 10_^-3 m_³
• Thus PΔV = 1.013 × 10_⁵ × (0.00342 - 0.00534) × 10_^-3 = -0.1944 J = -0.0001944 kJ
5. So from (3), we get:
ΔU = (ΔH - PΔV) = [1.850 - (-0.0001944)] = 1.850 kJ
6. In this problem, gases are not involved. Graphite is in the solid state and diamond is also in the solid state. We have seen that, when gases are not involved, ΔH  will be equal to ΔU (see step 18)

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In the next section we will see two more solved examples. We will also see extensive and intensive properties

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