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State Function revisited….State Function revisited….  Dependent ONLY on a system’s state at a given moment in time.  Only initial and final states 

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Presentation on theme: "State Function revisited….State Function revisited….  Dependent ONLY on a system’s state at a given moment in time.  Only initial and final states "— Presentation transcript:

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2 State Function revisited….State Function revisited….  Dependent ONLY on a system’s state at a given moment in time.  Only initial and final states  Not based on the path to get to a given condition Ex. Energy, Enthalpy

3 Standard ConditionsStandard Conditions  Represented by a “ ° “ beside Δ H or Δ E, etc.  25°C  1 atm

4 Methods for determining Δ H 1)Calorimetry 2)Application of Hess’ Law

5 Hess’ LawHess’ Law  Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step  Δ H rxn = ΣΔ H A+B+C (sum of Δ H for each step)  Allows us to break a chemical reaction down into multiple steps to calculate Δ H  Add the enthalpies of the steps for the enthalpy for the overall chemical reaction

6 Guidelines for using Hess’ Law  Must use data and combine each step in a way that gives the chemical reaction with the unknown Δ H  Set up steps so chemical compounds not in the final reaction are cancelled  Reverse a reaction if necessary and change the sign on Δ H  Check for correct mole ratios

7 Example 1:Example 1: H 2 O (l)  H 2 O (g) Δ H° = ? Based on the following: H 2 + ½ O 2  H 2 O (l) Δ H° = -285.83 kJ/mol H 2 + ½ O 2  H 2 O (g) Δ H° = -241.82 kJ/mol

8 Example 2:Example 2: C (s) + 4H 2  C 3 H 8 (g) Δ H° = ? Based on the following: 2H 2 + O 2  2H 2 O Δ H° = -571.7 kJ/mol C 3 H 8 + 5 O 2  3CO 2 + 4H 2 O (g) Δ H°= -2220.1kJ/mol C (s) + O 2  CO 2 (g) Δ H° = -393.5 kJ/mol

9 Methods for determining Δ H 1)Calorimetry 2)Application of Hess’ Law 3)Enthalpies of Formation

10 Enthalpy of Formation ( Δ H f °)  Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state.  Elements present in “most thermodynamically stable state”  25°C°, 1atm

11 Example 3Example 3  MgO (s) + CO 2 (g)  MgCO 3 (s) MgO + CO 2  Mg (s) + C (s) + 3/2 O 2 Mg (s) + C (s) + 3/2 O 2  MgCO 3

12 Apply Hess’ Law----Apply Hess’ Law----  Really Δ H f (products) - Δ H f (reactants)  Calculate Δ H rxn based on enthalpy of formation ( Δ H f )  aA + bB  cC + dD Δ H° =[c ( Δ H f °) C + d( Δ H f °) D ] - [a ( Δ H f °) A + b ( Δ H f °) B ]

13 Example 4Example 4  Calculate the Δ H° for the combustion of 1 mole of pentaborane (B 5 H 9) reacting with excess oxygen  2B 5 H 9 + 12O 2  5B 2 O 3 + 9H 2 O Δ H° = ?

14 Example 5Example 5  Isopropyl alcohol (rubbing alcohol) undergoes a combustion reaction 2(CH 3 ) 2 CHOH + 9O 2  6CO 2 + 8H 2 O Δ H° = -4011 kJ/mol Calculate the standard enthalpy of formation for isopropyl alcohol.

15 Homework  pp. 253 #65-67, 72


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