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Heat of Formation

Heat of Formation. If you understood Hess’ Law, you will have NO problem here. QUIZ. Predict the enthalpy change for the reaction: HCl (g) + NaNO 2(s)  HNO 2(g) + NaCl (as) Using the following Equations 2NaCl (s) + H 2 O  2HCl ( aq ) + Na 2 O ( aq ) ∆H˚ 1 = 507kJ

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Heat of Formation

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  1. Heat of Formation If you understood Hess’ Law, you will have NO problem here

  2. QUIZ • Predict the enthalpy change for the reaction: • HCl(g) + NaNO2(s) HNO2(g) + NaCl(as) Using the following Equations • 2NaCl(s) + H2O  2HCl(aq) + Na2O(aq) ∆H˚1 = 507kJ • NO + NO2 + Na2O(aq)  2NaNO2 ∆H˚2 = –427kJ • NO + NO2  N2O + O2 ∆H˚3 = –43kJ • 2HNO2(g)  N2O + O2 + H2O ∆H˚4 = 34kJ

  3. Quiz • Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: • ½H2(g)  +  ½Cl2(g)  →  HCl(g) • 2HCl(g)  +  ½O2(g)  →  H2O(l)  +  Cl2(g)        ΔH = 105 kJ • CH2Cl2(l) +  H2(g)  +  3/2O 2(g)  →  COCl2(g)  +  2H2O(l)    • ΔH = -402.5 kJ • COCl2(g)  +  H2O(l)  →  CH2Cl2(l)  +  O2(g)    ΔH = 47.5 kJ

  4. Ways to Determine ∆H • 1. Calorimetry • 2. Hess’ Law • 3. Standard Enthalpies of Formation

  5. Standard Enthalpies of Formation DEFINITION: The quantity of energy associated with the formation of one mole of a substance from its elements in their standard states

  6. Three Steps to Writing a Formation Equation • Write ONE mole of the product in the state that has been specified. • Write the reactant elements in their standard states. • Balance the equation with coefficients for the reactants yielding ONE mole of product C(s) + O2(g)  CO2(g) ∆H˚f = –393.5kJ/mol

  7. 2. Elements in their standard state • Standard state = most stable form at SATP • SATP = Standard Atmospheric Temperature and Pressure • 25˚C and 100kPa

  8. Elements in the Standard State • Most metals are monoatomic solids: • Ca(s), Au(s), Na(s), Fe(s), C(s) I2(s) • Five common gaseous elements at SATP: • H2(g), O2(g), N2(g), F2(g), Cl2(g) • Two elements are liquid at SATP • Hg(l), Br2(l) • Check your periodic table for states!

  9. 3. Choose Coefficients that balance the left side • Ever since you were taught balancing equations, you learned that you can’t use fractions or decimals…. • Forget that. H2(g) + ½ O2(g) H2O(l) K(s) + 0.5Br2(l)  KBr(s)

  10. Write the formation equation for liquid ethanol • 1. Ethanol = C2H5OH (l)  C2H5OH (l) • 2. C(s) + H2(g) + O2(g) C(s) + H2(g) + O2(g)  C2H5OH (l) 3. Balance 2C(s) + 3H2(g)+ ½O2(g)  C2H5OH (l)

  11. Let’s try a few examples on the board • 1. Benzene • C6H6(l) • 2. Glucose • C6H12O6(s) • 3. Magnesium Hydroxide • Mg(OH)2(s)

  12. What is the enthalpy of formation of an element? H2(g) H2(g) ∆H˚f = ? H2(g) H2(g) ∆H˚f = 0 The standard enthalpy of formation of an element already in its standard state is ZERO.

  13. Logic behind using heats of formation • Hess’s law tells us that we can add together the enthalpies of different reactions to get the enthalpy of a desired reaction provided we can add the chemical equations together to get the desired chemical equation. • Heats of formation basically lets us know the chemical equation and ∆H for any substance

  14. How are we supposed to know all the heats of formation? • You get a chart.

  15. What is the thermochemical equation for the reacton of lime (CaO) with water? CaO(s) + H2O(l) Ca(OH)2(s) ∆H = ? • Ca(s) + ½ O2(g)  CaO(s) ∆H˚f = –634.9kJ/mol • H2(g) + ½ O2(g)  H2O(l) ∆H˚f = –285.8kJ/mol • Ca(s) + H2(g) + O2(g)  Ca(OH)2(s) ∆H˚f = –986.1kJ/mol

  16. ∆H The enthalpy of a reaction is the enthalpy of the products minus the enthalpy of the reactants.

  17. The main component in natural gas used in home heating or laboratory burners is methane. What is the molar enthalpy of combustion of methane fuel? • ASSUME combustion yields CO2(g) and H2O(l) Check your chart: • ∆H˚f methane = – 74.4kJ/mol • ∆H˚foxygen = 0kJ/mol • ∆H˚fcarbon dioxide = – 393.5kJ/mol • ∆H˚fwater (liquid) = – 285.8kJ/mol

  18. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) • ∆H = Σ∆Hproducts – Σ∆Hreactants • ∆H = Σn∆Hf products– Σn∆Hf reactants • ∆H = [n∆H˚fCO2 + n∆H˚fH20] – [n∆H˚fCH4+ n∆H˚f02] • ∆H = [1mol(– 393.5kJ/mol) + 2mol(– 285.8kJ/mol)] – [1mol(– 74.4kJ/mol) + 2mol(0kJ/mol) • ∆H = –890kJ

  19. 3. Working Backwards • The standard enthalpy of combustion of benzene (C6H6(l)) to carbon dioxide and liquid water is –3273kJ/mol. What is the standard enthalpy of formation of benzene, given the tabulated values for carbon dioxide and liquid water? • (Note that here you know the ∆H of the reaction but need to find a ∆H˚f )

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