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THERMOCHEMISTRY

THERMOCHEMISTRY. ENTHALPY: The Heat of Reaction and Hess’s Law. Enthalpy. The term used for the total energy of a system when it is at constant pressure is called the enthalpy of a system. It is symbolized as  H .

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THERMOCHEMISTRY

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  1. THERMOCHEMISTRY ENTHALPY: The Heat of Reaction and Hess’s Law

  2. Enthalpy • The term used for the total energy of a system when it is at constant pressure is called the enthalpy of a system. It is symbolized as H. • The word enthalpy comes from Greek endon - in, within and thalpein - to warm • Enthalpy is a state function so when a system reacts at constant pressure and absorbs or evolves energy we say that it experiences an enthalpy change.

  3. Enthalpy: a state function • As a state function, its value depends solely on the difference in enthalpy between the initial and final states and not on the mechanism by which the system undergoes this change.  H = Hfinal - Hinitial • For a chemical reaction, the initial state refers to the reactants and the final state to the products, so for a chemical reaction we can write:  H = Hproducts - Hreactants

  4. The Importance of  H • Our formal definition of H,in an absolute sense, serves only one purpose. This purpose is to define the meaning of positive and negative values of H. • It also gives us an idea of the magnitude of the energy absorbed or released.

  5. a) If  H is negativethis reaction is (-) Exothermic b) If  H is positivethis reaction is (+) Endothermic system system surroundings surroundings Sign Conventions for Enthalpy Energy is leaving the system Energy is being absorbed by the system

  6. + - Exothermic Endothermic Endothermic or Exothermic? • Determine whether a reaction would be endothermic or exothermic under the following conditions: a) ifHproducts > Hreactantsthen H is__ (+/-) and the reaction is ________________. b) if Hproducts < Hreactantsthen H is__ (+/-) and the reaction is ________________.

  7. Thermochemical Equations • Thermochemical reactions are generally written at standard thermochemical temperature and pressure. This is adapted to the normal laboratory conditions of 25C and 1 atmosphere. • To show that  H is at this standard condition it is designated as: H • The units of  H  are usually kilojoules (kJ)

  8. Notice that all physical states of the reactants and products are cited in this equation. This sign indicates an Exothermic Reaction Example: A Thermochemical Equation N2(g) + 3 H2(g)  2 NH3(g)H= -92.38 kJ How can we calculate the enthalpy of this reaction given only 1.5 mol of H2? Divide the entire equation by 2 including H H= - 92.38 kJ/2 = - 46.19 kJ

  9. RULES FOR MANIPULATINGTHERMOCHEMICAL EQUATIONS: • When an equation is reversed (written in the opposite direction) the sign of H  must also be reversed. • Formulas canceled from both sides of an equation must be for the substance in identical physical states. • If all the coefficients of an equation are multiplied or divided by a common factor, the value of H  must be likewise changed.

  10. Example of a Hess’s Law Calculation • Given two chemical reactions (#1 and #2) and the H  for each, find the H  for a third (target) reaction. #1: Fe2O3 + 3CO(g) 2 Fe(s) + 3CO2(g) H=-26.7 kJ #2: CO(g) + 1/2 O2(g)  CO2(g)H=-283.0 kJ Calculate the H for the following reaction: 2 Fe(s) + 3/2 O2(g) Fe2O3(s)H= ? This is the target equation

  11. STEP 1: Begin by reviewing the position of the reactants and the products in the desired reaction and compare them with the given equations. TARGETEQUATION: 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) H= ? EQ #1: Fe2O3 + 3CO(g)  2 Fe(s) + 3CO2(g) H=-26.7 kJ To match the target equation correctly, equation #1 must have 2 Fe on theleft. However, it has 2 Fe on the product sideof the reaction arrow. SOLUTION: Reverse the entire equationas well as the +/- sign onH 

  12. Solution: Given equation: Fe2O3(s) + 3CO(g) 2 Fe(s) + 3CO2(g) H =- 26.7 kJ 2 Fe(s) + 3CO2(g)  Fe2O3(s) + 3CO(g) Reversing the equation: Reverses the sign of enthalpy: H = + 26.7 kJ

  13. STEP 2: • There must be 3/2 O2 on the left, and we must be able to cancel 3 CO and 3 CO2 when the equations are added. If we multiply the second equation by 3 we will obtain the necessary coefficients. #2: 3 [CO(g) + 1/2 O2(g)  CO2(g) ] H= -283.0 kJ 3CO(g) + 3/2 O2(g)3 CO2(g) Remember to multiply the H  by 3 also.  H = 3 (-283.0 kJ) = - 849.0 kJ

  14. STEP 3: • Now put the equations together and cancel out anything that appears on both sides of the equations (very similar to adding redox equations together). 2 Fe(s) + 3CO2(g)  Fe2O3(s) + 3CO(g) H = + 26.7 kJ 3CO(g) + 3/2 O2(g) 3 CO2(g)H = - 849.0 kJ 2Fe(s) +3CO2(g) +3CO(g) + 3/2O2(g)  Fe2O3(s) + 3CO(g)+ 3 CO2(g) NET: 2 Fe(s) + 3/2 O2(g) Fe2O3(s) H = - 822.3 kJ

  15. Hess’s Law • Stated in terms of standard enthalpies of formation: H = (sum of Hfof all the products) - (sum of Hfof all the reactants)

  16. HEATS OF FORMATION AND HESS’S LAW: • The standard enthalpy of formation, Hf, is the amount of heat absorbed or evolved when one mole of the substance is formed at 25C and 1 atm from its elements (Standard Thermochemical Conditions) and in its most stable form (solid, liquid, gas). Tables are provided.

  17. Using HfTables to Calculate H • Sample HfTable – Note that Hf is in kJ/mol Note also that Hf for elements in their standard states = 0

  18. Example of Hess’ Law • Some chefs keep baking soda, NaHCO3 handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire and the heat decomposes it to give CO2, which further smothers the flame. The equation for the decomposition of NaHCO3 is: 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)

  19. Calculation Process Note the minus sign • Recalling that: H = (sum of Hfof all the products) - (sum of Hfof all the reactants) • And utilizing the Table Values of all products and reactants:-1131 kJ/mol Na2CO3 -286 kJ/mol H2O -394 kJ/mol CO2-947.7 kJ/mol NaHCO3

  20. Complete the Final Calculation 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g) -947.7kJ/mol-1131kJ/mol-286kJ/mol-394kJ/mol H =(- 1131 kJ/mol + - 286 kJ/mol + - 394 kJ/mol)- (2 x - 947.7 kJ/mol)= -1131 + -286 + -394 + 1895.4 = +84 kJ Note that you must multiply the enthalpy by the coefficient Note also that you must correct for a sign change: -(-) = +

  21. Summary • Enthalpy, H, is a state function and is expressed in kiloJoules. • There are two types of calculations based on Hess’s Law that deal with enthalpy illustrated in this tutorial: • Those in which one manipulates several different chemical reactions to combine and form the desired reaction. H’s for each reaction has been provided to calculate the desired H. • Enthalpy calculations utilizing a Table of Standard Enthalpies of Formation which can be used with the following equation: H = (sum of Hfof all the products) - (sum of Hfof all the reactants)

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