Enthalpy diagram

Phase is a form of matter that is uniform with respect to chemical composition and the state of aggregation on both microscopic and macroscopic length scales.
Determination of Hess’s Law (Direct or Indirect)

1. Enthalpy Diagram 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. Enthalpy Diagrams

3. What Determines the Relative Stability of the Solid, Liquid, and Gas Phases? Phase A form of matter that is uniform with respect to chemical composition and the state of aggregation on both microscopic and macroscopic length scales. For example, liquid water ice and liquid water Water solidifies T is lowered from 300 to 250 K at 1 atm. It vaporizes to form a gas. When heated to 400 K at atmospheric pressure, Dry ice (CO 2 ), sublimes at 1 bar an normal temperature The solid phase is the most stable state of a substance at sufficiently low temperatures, and that the gas phase is the most stable state of a substance at sufficiently high temperatures.

4. Enthalpy Diagrams Enthalpy diagrams are a representation of the relative internal energy of a system before and after a reaction. Horizontal lines are drawn to show the energy of the system at a particular time . The formulae of all the elements or compounds present in the system including their state: solid, liquid or gas. A line at a higher level merely shows that the enthalpy of the system has increased due to the addition of heat from the surroundings , which has caused a physical change (e.g. melting) or chemical change (i.e. a reaction) in the system.

5. In an exothermic reaction : Heat is evolved, The enthalpy is decreased, ∆H is negative

6. Indirect Determination of Hess ’ s Law ∆H is an Extensive Property Consider the standard enthalpy change in the formation of NO (g) from its elements at 25 o C   To express the enthalpy change in terms of one mole of we divide all coefficients and the value by two  

7. ∆H Changes Sign When a Process is Reversed if a process is reversed, the change in a function of state reverses sign . Thus, ∆H for the decomposition of 1 mole of NO (g) = - ∆H for the formation of 1 mole of NO (g)   Imagine that this reaction as proceeding in two steps as given below  

8. Imagine that the reaction as proceeding in two steps as:    

9.     the overall  

10. Hess’ s law If a process occurs in stages or steps (even if only hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for all the individual steps . Hess s law is simply a consequence of the state function property of enthalpy. Regardless of the path taken in going from the initial state to the final state, ∆H has the same value.

11. Applying Hess’s Law Example 7.9 Use the heat of combustion data (given) to determine ∆H° for reaction = ?   Solution = -2219.9 kJ      

12. Because our objective in reaction is to produce C 3 H 8 (g), the next step is to find a reaction in which C 3 H 8 (g) is formed ---- the reverse of reaction (a) = -2219.9 kJ   ∆H° = -(-2219.9) kJ = 2219.9 kJ   Now, we turn our attention to the reactants, C(graphite) and H 2 (g). To get the proper number of moles of each, we must multiply equation (b) by 3 and equation (c) by 4.   =-1143kJ‘  

13. Now, combine all the three modified equations, Add, Cancel out the similar available on the opposite sides of arrows, Get, over all reaction equation   In this example, 3 unrelated combustion reactions were used to determine the ‘enthalpy of reaction’ of another reaction.

14. Standard Enthalpies of Reaction Use Hess s law to calculate the standard enthalpy of reaction for the decomposition of sodium bicarbonate, a minor reaction that occurs when baking soda is used in baking.   From Hess s law, we see that the following four equations yield the above given equation when added together. (g) -2 x [NaHCO 3 (s)]    

15.     for reaction (a) is the negative of twice . Equations (b), (c) and (d) represent the formation of one mole each of Na 2 CO 3 (s), CO 2 (g) and H 2 O (l) and Thus, we can express the value of for the decomposition reaction as   ∆H° = ∆ H f °[Na 2 CO 3 ] (s) + ∆ H f °[H 2 O] (l) +∆ H f °[CO 2 ] (g) - 2 * ∆ H f °[NaHCO 3 ] (s) The enthalpy change for the overall reaction is the sum of the standard enthalpy changes of the individual steps

17. Calculating from Tabulated Values of ∆H o Example 7.11 Let us apply equation (7.21) to calculate the standard enthalpy of combustion of ethane, a component of natural gas Solution       = -1559.7 kJ  

18. Calculating an Unknown Value Use the data here and in Table 7.2 to calculate of benzene, C 6 H 6 (l)   Solution   ? - 393.5 - 2858     = ? [Solve]  

19. Ionic Reactions in Solutions ions of a single type cannot be created in a chemical reaction because cations and anions are produced simultaneously. Therefore, a particular ion need to be chosen to which an enthalpy of formation of zero can be assigned, in its aqueous solutions. The ion we arbitrarily choose for our zero is H + . Now let us see how we can use expression (7.21) and data to determine the enthalpy of formation of OH - ( aq ).    

21. Problem Statement Ammonia is used in the industrial preparation of nitric acid according to the equation: 4 NH 3 (g) + 5O 2 (g) → 4 NO(g) + 6 H 2 O(g) What is the standard enthalpy change ∆H for this reaction?   ∆H o = [4 × (90.3) + 6 × (-241.8)] – [4 × (-45.9) + 5 × (0)] = -906 kJ Solution