Function of state

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 H 2 O from three different samples of water: purified by extensive distillation of groundwater; synthesized by burning pure H 2 (g) in pure O 2 (g) and prepared by driving off the water of hydration from CuSO 4 . 5H 2 O 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 o C to a final temperature of 50 o C. The internal energy of the ice at 0 o C has one unique value, the liquid water at 50 o C has another, Difference in internal energy between the two states also has a unique value, ∆U = U 2 –U 1 , 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 =   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  

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 10 2 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.   =   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 (a) We have previously identified a heat of reaction as q rxn and so     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 , and no work is done. That is, w = -P = 0   Denoting the heat of reaction for a constant-volume reaction as 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 = q V The first law of thermodynamics, for the same reaction at constant pressure          

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       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 = q P [Change in Enthalpy = the heat of reaction at constant pressure]  

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:   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  

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 ( n f – n i ) Here, is the number of moles of gas in the products (2 mol CO 2 ) and is the number of moles of gas in the reactants (2 mol CO + 1 mol O 2 ).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       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, C 12 H 22 O 11 ? The amount of heat generated is given as ∆H = -5.65 x 10 3 kJ/ mol Solution Express the quantities in moles   The conversion factor is -5.65 x 10 3 kJ, of heat is associated with the combustion of 1 mol C 12 H 22 O 11 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 : H 2 O 2 (l ) → H 2 O(l) + 1/2 O 2 (g); Δ H = -98.2 kJ Calculate the change in enthalpy, Δ H, when 1.00 g of hydrogen peroxide decomposes. Solution the   molar mass of H 2 O 2   = 2 x 1 for hydrogen + 2 x 16 for oxygen= 34.0  

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   The energy requirement to melt one mole of a solid is called the enthalpy (or heat) of fusion .  

14. Example 7.8 Calculate for the process in which 50.0 g of water is converted from liquid at 10.0 o C to vapor at 25.0 o C Solution Step 1: Heating water from 10.0 to 25.0 o C. This heat requirement can be determined by the method   Step 2: Vaporizing water at 25 O C 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 O C; 44.0 kJ/mol.

15.   Total enthalpy change   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   the temperature change is expressed as where T f is the final temperature and T i is the initial temperature. When the temperature of a system increases (T f > T i ), is positive. A positive q signifies that heat is absorbed or gained by the system . When the temperature of a system decreases (T f < T i ), is negative. A negative q signifies that heat is evolved or lost by the system.     Applied to the exchange of heat, this means that