A unique value for a specified state of a system. Path-Dependent functions. Enthalpy (∆H ) and Internal Energy ( ∆U) changes in a chemical reaction

1. Function of State
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. Functions of state
A system is describe : T, P, amount of substance [specified the state of the system.]
Any property that has a unique value for a specified state of a system is said to be a
function of state, or a state function.
For example, Density of pure water, at (293.15 K) and at100 kPa, is in a specified state.
Obtain H2O from three different samples of water:
i) purified by extensive distillation of groundwater;
ii) synthesized by burning pure H2 (g) in pure O2 (g) and
iii) prepared by driving off the water of hydration from CuSO4 . 5H2O and condensing the
gaseous water to a liquid.
The densities of the three different samples will all be the same 0.99820 g/ml:
the value of a state function depends on the state of the system, and
not on the path how that state was established

3. The internal energy of a system is a function of state
Consider, for example, heating 10.0 g of ice at 0 oC to a final temperature of 50 oC.
The internal energy of the ice at 0 oC has one unique value,
the liquid water at 50 oC has another,
Difference in internal energy between the two states also has a unique value, ∆U = U2 –U1,
and this difference is something that can precisely measure.
The value of a function of state depends on the state of the system, and
not on how that state was established
Thus, the overall change in internal energy =
= ∆𝑈 + ∆𝑈 = 𝑈2 − 𝑈1 + 𝑈1 − 𝑈2 = 0
It is the quantity of energy (as heat) that must be transferred from the surroundings to the
system during the change from state 1 to state 2, as
𝑆𝑡𝑎𝑡𝑒 1 𝑈1
∆𝑈
𝑆𝑡𝑎𝑡𝑒 2 𝑈2
−∆𝑈
𝑆𝑡𝑎𝑡𝑒 1 (𝑈1)

4. Path-Dependent Functions
Consider a process is occurring at 0.100 mol of He at 298 K and under a pressure of 2.40
atm as state 1, and under a pressure of 1.20 atm as state 2. The change from state 1 to
state 2 occurred in a single step.
Suppose that in another instance, we allowed the expansion to occur through an
intermediate stage. That is, suppose the external pressure on the gas was first reduced
from 2.40 atm to 1.80 atm (at which point, the gas volume would be 1.36 L). Then, in a
second stage, reduced from 1.80 atm to 1.20 atm, thereby arriving at state 2.
The amount of work done by the gas in a single-stage expansion was -1.24 x 102 J
The amount of work done in the two stage process is the sum of two pressure volume
work for each stage of the expansion.

5. 𝑤 = −1.80 𝑎𝑡𝑚 𝑥 1.36 𝐿 − 1.02𝐿 − 1.20 𝑎𝑡𝑚 ( 2.04 𝐿 − 1.36 𝐿)
𝑤 = −0.61 𝐿 𝑎𝑡𝑚 − 0.82 𝐿 𝑎𝑡𝑚 =
= −1.43 𝐿 𝑎𝑡𝑚 𝑥
101 𝐽
1𝐿 𝑎𝑡𝑚
= −1.44 𝑥 102 𝐽
slightly more work is done in the two-stage expansion.
Work is not a function of state; it is path dependent
A reversible process is one that can be made to reverse its direction when an
infinitesimal change is made in a system variable.

6. Heats of Reaction: ∆U and ∆H
According to the first law of thermodynamics, we can also say that
∆𝑈 = 𝑞 + 𝑤 (a)
We have previously identified a heat of reaction as qrxn and so
∆𝑈 = 𝑞 𝑟𝑥𝑛 + 𝑤 (𝑏)
Consider the combustion reaction carried out in a bomb calorimeter
The original reactants and products are confined within the bomb, and we say that
the reaction occurs at constant volume.
Because the volume is constant, ∆𝑉 = 0 and no work is done. That is, w = -P ∆𝑉 = 0
Denoting the heat of reaction for a constant-volume reaction as qV
∆𝑈 = 𝑞 𝑟𝑥𝑛 + 𝑤 = 𝑞 𝑟𝑥𝑛 + 0 = 𝑞 𝑟𝑥𝑛 = 𝑞 𝑣
The heat of reaction measured in a bomb calorimeter (∆𝑉=0) is equal to ∆U

7. For a reaction at constant volume, ∆U = qV
The first law of thermodynamics,
for the same reaction at constant pressure
∆𝑈 = 𝑞 𝑃 + 𝑤
∆𝑈 = 𝑞 𝑉 = 𝑞 𝑃 + 𝑤
𝑞 𝑉 = 𝑞 𝑃 + 𝑤
𝑞 𝑉 = 𝑞 𝑃 − 𝑃∆𝑉
𝑞 𝑃 = ∆𝑈 + 𝑃∆𝑉

8. Another state function, enthalpy, H, is the sum of the internal energy and the pressure
volume product of a system: The enthalpy change, for a process between initial and final
states is
∆𝐻 = 𝐻𝑓 − 𝐻𝑖 = 𝑈𝑓 + 𝑃𝑓 𝑉𝑓 − 𝑈𝑖 + 𝑃𝑖 𝑉𝑖
∆𝐻 = 𝑈𝑓 − 𝑈𝑖) + 𝑃𝑓 𝑉𝑓 − 𝑃𝑖 𝑉𝑖
∆𝐻 = ∆ 𝑈 + ∆𝑃𝑉
If the process is carried out at a constant temperature and pressure and with work limited
to pressure volume work, the enthalpy change is
and the heat flow for the process under these conditions is
∆H = qP
[Change in Enthalpy = the heat of reaction at constant pressure]
∆𝐻 = ∆ 𝑈 + 𝑃∆𝑉

9. Enthalpy (∆H ) and Internal Energy ( ∆U) Changes in a
Chemical Reaction
the heat of reaction at constant pressure is ∆H, and
the heat of reaction at constant volume is ∆U,
are related by the expression:
∆𝑈 = ∆𝐻 − 𝑃∆𝑉
The last term in this expression is the energy associated with the change in volume of
the system under a constant external pressure.
To assess just how significant pressure volume work is, consider the following reaction
2𝐶𝑂 𝑔 + 𝑂2 𝑔 → 2𝐶𝑂2 𝑔

10. If the heat of this reaction is measured under constant-pressure conditions at a
constant temperature of 298 K, we get that 566.0 kJ of energy has left the
system as heat: ∆H = -566.0 kJ.
the ideal gas equation = P∆V = RT (nf – ni )
Here, is the number of moles of gas in the products (2 mol CO2) and is the
number of moles of gas in the reactants (2 mol CO + 1 mol O2).Thus
P∆V = 0.0083145 kJ mol-1 K-1 * 298 K * [2 – (2 + 1)] mol = -2.5 kJ
The change in internal energy is
∆𝑈 = ∆𝐻 − 𝑃∆𝑉
∆𝑈 = −566.0 𝑘𝐽 − (−2.5𝑘𝐽)
∆𝑈 = −563.5𝑘𝐽
This calculation shows that the
term P∆ V is quite small compared
to ∆H and
∆U and ∆H are almost the same.

11. Example 7.7
How much heat is associated with the complete combustion of 1.00 kg of sucrose, C12 H22O11 ?
The amount of heat generated is given as ∆H = -5.65 x 103 kJ/mol
Solution
Express the quantities in moles
𝑚𝑜𝑙 ? = 1.00 𝑘𝑔 𝐶12 𝐻22 𝑂11 𝑥
1000 𝑔 𝐶12 𝐻22 𝑂11
1 𝑘𝑔 𝐶12 𝐻22 𝑂11
𝑥
1 𝑚𝑜𝑙 𝐶12 𝐻22 𝑂11
342.3 𝑔𝐶12 𝐻22 𝑂11
= 2.92 𝑚𝑜𝑙 𝐶12 𝐻22 𝑂11
The conversion factor is -5.65 x 103 kJ, of heat is associated with the combustion of
1 mol C12 H22O11
𝑘𝐽 ? = 2.92 𝑚𝑜𝑙 C12 H22O11 𝑥
−5.65 𝑥 103 𝑘𝐽
1 𝑚𝑜𝑙 C12 H22O11
= −1.65 𝑥 104 𝑘𝐽 x
The negative sign denotes that heat is given off in the combustion

12. Problem statement
Hydrogen peroxide decomposes according to the following thermochemical reaction:
H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ
Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.
Solution
the molar mass of H2O2 = 2 x 1 for hydrogen + 2 x 16 for oxygen= 34.0
∆𝐻 = 1.00 𝑔 𝐻2 𝑂2 𝑥
1 𝑚𝑜𝑙 H2O2
34 𝑔 𝑜𝑓 H2O2
𝑥
−98.2 𝑘𝐽
1 𝑚𝑜𝑙 H2O2
= −2.89 𝑘𝐽

13. Enthalpy Change Accompanying a Change in State of Matter
Molar heat of vaporization
The heat required to vaporize a fixed quantity of liquid [mole] is called the enthalpy (or
heat) of vaporization
𝐻2 𝑂 𝑙 → 𝐻2 𝑂 𝑔 44.0 𝑘𝐽 𝑎𝑡 298 𝐾
The energy requirement to melt one mole of a solid is called the enthalpy (or heat) of
fusion.
𝐻2 𝑂 𝑠 → 𝐻2 𝑂 𝑙 6.01 𝑘𝐽 𝑎𝑡 273.15 𝐾

14. Example 7.8
Calculate for the process in which 50.0 g of water is converted from liquid at 10.0 oC to
vapor at 25.0 oC
Solution
Step 1: Heating water from 10.0 to 25.0 oC.
This heat requirement can be determined by the method
𝑘𝐽 ? = 50.0 𝑔 𝐻2 𝑂 𝑥
4.18 𝐽
𝑔 𝐻2 𝑂 ℃
𝑥 25.0 − 10.0 ℃ 𝑥
1𝑘𝐽
1000𝐽
= 3.14𝑘𝐽
Step 2: Vaporizing water at 25 OC
For this part of the calculation, the quantity of water must be expressed in moles so that
we can then use the molar enthalpy of vaporization at 25 OC; 44.0 kJ/mol.

15. 𝑘𝐽 ? = 50.0 𝑔 𝐻2 𝑂 𝑥
1 𝑚𝑜𝑙 𝐻2 𝑂
18.02 𝑔 𝐻2 𝑂
𝑥
44.0𝑘𝐽
1 𝑚𝑜𝑙𝐻2 𝑂
= 122 𝑘𝐽
Total enthalpy change
∆𝐻 = 3.14 𝑘𝐽 + 122 𝑘𝐽 = 125 𝑘𝐽
Note that the enthalpy change is positive, which reflects that the system (i.e., the water)
gains energy.
The reverse would be true for condensation of water at 25.0 °C and cooling it to 10.0 °C

16. the law of conservation of energy
In interactions between a system and its surroundings,
the total energy remains constant energy is neither created nor destroyed.
Thus, heat gained by a system is lost by its surroundings, and vice versa
𝑞 𝑠𝑦𝑠𝑡𝑒𝑚 = − 𝑞 𝑠𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
the temperature change is expressed as ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 . where Tf is the final temperature
and Ti is the initial temperature.
When the temperature of a system increases (Tf > Ti), ∆𝑇 is positive.
A positive q signifies that heat is absorbed or gained by the system.
When the temperature of a system decreases (Tf < Ti), ∆𝑇 is negative.
A negative q signifies that heat is evolved or lost by the system.
𝑞 𝑠𝑦𝑠𝑡𝑒𝑚 + 𝑞 𝑠𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = 0
Applied to the exchange of heat, this means that