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Enthalpy: is the heat absorbed or released by a system at constant pressure .

Enthalpy: is the heat absorbed or released by a system at constant pressure. It is impossible to measure enthalpy directly. Only changes in enthalpy at constant pressure can measured “ΔH” ΔH = H final - H initial. Units of heat energy. We will use joules (J) or kilojoules (kJ).

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Enthalpy: is the heat absorbed or released by a system at constant pressure .

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  1. Enthalpy: is the heat absorbed or released by a system at constant pressure. It is impossible to measure enthalpy directly. Only changes in enthalpy at constant pressure can measured “ΔH” ΔH = Hfinal - Hinitial Units of heat energy. We will use joules (J) or kilojoules (kJ). qp is the heat at constant pressure qp = ΔH( absorbed or realsed heat)

  2. At constant pressure, the change in enthapy ΔH of the system is equall to the energy flow as heat. • This means that for a reaction studied at constant pressure, the flow of heat is a measure of the change in enthalpy for the system. In general for a chemical reaction: ΔH = Hproducts – Hreactants For exothermic reactions, ΔH is negative For endothermic reactions, ΔH is positive

  3. Example : • When 1 mole of methane is burned at constant pressure, 890 KJ of energy is realesed as heat. Calculate ΔH for a process in which 5.8 g sample of methane is burned at constant pressure? • Solution: • ♦Convert 5.8 g of methane to mole • n= mass/ molar mass = 5.8/16=0.36 mol of methane • ♦At constant pressure: • ΔH = qp = - 890 KJ/ mol CH4 • 1 mol of CH4 - 890 KJ • 0.36 mol of CH4 ? • ΔH = qp =((-890 KJ)*(0.36 mol))/ 1 mol = - 320 KJ • Thus, when 5.8 g of CH4 is burned at constant pressure • ΔH = heat flow= - 320 KJ

  4. The following guidelines are helpful in writing and interpreting thermochemical equations: 1- specify the physical states of all reactants and products, because they help determine the actual enthalpy changes 2-If we multiply both sides of a thermochemical equation by a factor n, then DH must also change by the same factor. Returning to the melting of ice

  5. 3-When we reverse an equation, we change the roles of reactants and products. Consequently, the magnitude of DH for the equation remains the same, but its sign changes.

  6. Solution

  7. Standard enthalpy of formation (ΔHfo) The superscript “°” represents standard-state 1 atm and 25°C the subscript “f” stands for formation the standard enthalpy of formation of a compound as the heat change that results when 1 mole of the compound is formed from its elements at a pressure of 1 atm. By convention, the standard enthalpy of formation of any element in its most stable form is zero Molecular oxygen (O 2 ) is more stable than the other allotropic form of oxygen, ozone (O 3 ), at 1 atm and 25°C. Thus, DH°f (O2) = 0, but DH°f (O3) = 142.2 kJ/mol

  8. graphite is a more stable allotropic form of carbon than diamond at 1 atm and 25°C, so we have ΔH°f (C, graphite) = 0 and ΔH°f(C, diamond) = 1.90 kJ/mol n denote the stoichiometric coefficients for the reactants and products,and ∑(sigma) means “the sum of.” Note that in calculations, the stoichiometric coefficients are just numbers without units

  9. ΔH°fvalues can be determined by applying the direct method or the indirect method. The Direct Method graphite and O 2 are stable allotropic forms of the elements, it follows that ΔH ° f (C, graphite) and Δ H ° f (O 2 , g ) are zero

  10. Calculate ΔH0 for the following reaction: 2Al(s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (l) ΔHfo of Fe2O3, Al2O3 and Fe(l) = - 822.2, - 1669.8 and 12.40 kJ/mol Solution ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants) ΔHo = [(ΔHfo (Al2O3))+ (2×ΔHfo (Fe))]- [(2×ΔHfo (Al))+(ΔHfo (Fe2O3))] ΔHo = [ (-1669.8)+ (2×12.40)] – [2×(0)+(-822.2)] = -822.8 kJ

  11. The Indirect Method ♦ If the reaction proceeds too slowly, or side reactions produce substances other than the desired compound. ΔH °f can be determined by an indirect approach, which is based on Hess’s law Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

  12. the standard enthalpy of formation of carbon monoxide (CO). We might represent the reaction as burning graphite also produces some carbon dioxide (CO2 ), so we cannot measure the enthalpy change for CO directly as shown. So we must apply Hess’s law. The alternative is to regard the reaction as taking place in two steps. ♦ we reverse Equation (b) to get (c)

  13. Notes: ♦ If a reaction is reversed, the sign of ΔH is also reversed. A + BZ ΔH = +231kJ A + BZ ΔH = -231kJ ♦ Enthalpy change ( H) is proportional to the amount of reactants and products. If the coefficients in a balance equation are multiplied by an integer, the value of ΔH is multiplied by the same integer. A + 1/2B 2Z ΔH = + 231kJ 2A + B 4Z ΔH = + 462kJ

  14. Example : Calculate ΔH for the conversion of graphite to diamond: Cgraphite (s) Cdiamond (s) ΔH =? The combustion reaction are: Cgraphite (s) + O2(g)  CO2(g) ΔH1 = - 394 kJ Cdiamond (s) + O2(g)  CO2(g) ΔH2 = - 396 kJ Answer: By reversing reaction (2) and add to reaction (1) gives the required reaction: Cgraphite (s) + O2(g)  CO2(g) ΔH1 = -394 kJ CO2(g)  Cdiamond (s) + O2(g) ΔH3 = +396 kJ _________________________________________________ Cgraphite (s)  Cdiamond (s) ΔH = ΔH1 + ΔH3 = +2 kJ Since ΔH is positive the reaction is endothermic.

  15. Example: Determine ΔH for the following reaction: 3 C graphite (s) + 4H2 (g) C3H8 (g) ΔH = ? The combustion reaction are: C3H8 (g) + 5 O2 (g)  3CO2 (g) + 4 H2O (l) ΔH= - 2220.1 KJ C graphite (s) + O2 (g)  CO2 (g) ΔH= - 393.5 KJ H2 (g) + ½ O2 (g)  H2O (l) ΔH= - 585.8 KJ Solution ♦ Reverse the first equation 3CO2 (g) + 4 H2O (l) C3H8 (g) + 5 O2 (g) ΔH= + 2220.1 KJ ♦ Multiply the second equation by 3 3C graphite (s) + 3O2 (g)3 CO2 (g) ΔH= + 1180.5 KJ Multiply the third equation by 4 4H2 (g) + 2 O2 (g)4H2O (l) ΔH= + 2343.2 KJ 3C graphite (s) + 4H2 (g) C3H8 (g)

  16. Example: Use Hess’s law to determine ΔH for the reaction: C2H4 (g) + Cl2 (g) C2H4Cl2 (l) ΔH =? The combustion reactions are 4 HCl (g) + O2 (g)  2 Cl2 (g) + 2 H2O (l) ΔH = - 202.5 KJ 2 HCl (g) + C2H4 (g) + 1/2 O2 C2H4Cl2 (l) + H2O ΔH = - 319.6 KJ Solution ♦Reverse the first equation and divide by Cl2 (g) + H2O (l) 2 HCl (g) + 1/2O2 (g ) ΔH = + 101.25 KJ ♦Sum the last equation and the second equation Cl2 (g) + H2O (l) 2 HCl (g) + 1/2O2 (g ) ΔH = + 101.25 KJ 2 HCl (g) + C2H4 (g) + 1/2 O2 C2H4Cl2 (l) + H2O ΔH = - 319.6 KJ C2H4 (g) + Cl2 (g) C2H4Cl2 (l) ΔH = - 218.3 KJ

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