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Tang 02 enthalpy and hess' law

This document discusses enthalpy and Hess's Law. It defines enthalpy as the total kinetic and potential energy of a system at constant pressure. The change in enthalpy (ΔH) of a reaction is equal to the enthalpy of the products minus the enthalpy of the reactants. For exothermic reactions where heat is released, ΔH is negative, and for endothermic reactions where heat is absorbed, ΔH is positive. Hess's Law states that the enthalpy change of a reaction is the same whether it occurs in one step or multiple steps. ΔH values can be used to calculate the enthalpy change of unknown reactions using known reaction enthalpies.

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ENTHALPY AND HESS’ LAW
Chemical energy is stored in: – moving electrons – vibration of chemical bonds – rotation and translation of molecules – stored nuclear energy of protons & neutrons – energy stored in chemical bonds ENTHALPY AND HESS’ LAW
Enthalpy enthalpy (H) – total kinetic and potential energy of a system at a constant pressure change in enthalpy (∆H) – change in heat of a system ∆H = Hfinal - Hinitial ∆H = Hproducts - Hreactants ENTHALPY AND HESS’ LAW
Enthalpy For an exothermic reaction: Hproducts < Hreactants ∴∆H is negative Describe what has happened. H reaction progress reactants products ENTHALPY AND HESS’ LAW
Enthalpy For an endothermic reaction: Hproducts > Hreactants ∴∆H is positive Describe what has happened. H reaction progress reactants products ENTHALPY AND HESS’ LAW
Enthalpy How does enthalpy relate to heat with respect to an isolated system? ∆H = q when using a calorimeter. What is the formula for q? ENTHALPY AND HESS’ LAW
Enthalpy What is the formula for q? q = mcΔT Since q = ΔH ΔH = mcΔT ENTHALPY AND HESS’ LAW
Enthalpy How are enthalpy and heat related? ∆Hsystem = -q exothermic reaction ∆Hsystem = +q endothermic reaction This is only true at constant pressure, which is always the case in a calorimeter. Standard conditions for enthalpy changes: temperature at 298K (25ºC) and 100kPa ENTHALPY AND HESS’ LAW
Enthalpy The enthalpy change for a balanced chemical reaction is constant. e.g. 2 H2(g) + O2(g)  2 H2O(g) ∆H° = - 483.6 kJ ∆H° – standard enthalpy of a reaction – must be a balanced chemical equation Read as - 483.6 kJ per 2 mol of H2 being reacted. ENTHALPY AND HESS’ LAW This is called a thermochemical equation
Enthalpy ∆H° values are proportional to coefficient changes of a chemical equation. N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 3 N2O4 (g)  6 NO2 (g) ∆H° = +173.79 kJ ½ N2O4 (g)  NO2 (g) ∆H° = +28.96 kJ ENTHALPY AND HESS’ LAW
Enthalpy N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ Reversing a chemical reaction causes a sign change in front of the ∆H° value. ENTHALPY AND HESS’ LAW
Enthalpy ∆H° values of unknown reactions can be solved when other known reactions are given. ENTHALPY AND HESS’ LAW
Example #1 C(s) + O2(g)  CO2(g) ∆H° = ? C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW What parts of the above reactions are the same as the first reaction?
Rules for solving thermochemical equation questions: For any reaction that can be written in steps, the ∆H° is the same as the sum of the values of the ∆H° for each individual step. Hess’s Law of Summation ENTHALPY AND HESS’ LAW
∆H° Rules 1. If all the coefficients of an eqn are multiplied or divided by a common factor, the ∆H° must be changed likewise. ENTHALPY AND HESS’ LAW N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 3 N2O4 (g)  6 NO2 (g) ∆H° = +57.93kJ x 3 = +173.79 kJ ½ N2O4 (g)  NO2 (g) ∆H° = +57.93kJ ÷ 2 = +28.96 kJ
∆H° Rules 2. When a reaction is reversed, the sign of ∆H° must also be reversed. ENTHALPY AND HESS’ LAW N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ
∆H° Rules 3. When canceling compounds for Hess’s Law, the state of the compounds is important. ENTHALPY AND HESS’ LAW H2(g) + ½ O2(g)  H2O(g) ΔH° = +57.93 kJ K2SO4(s) + 2H2O(l)  H2SO4(aq) + 2KOH(s) ΔH° = +342.4kJ These two CANNOT cancel each other out
C(s) + O2(g)  CO2(g) ∆H° = ? C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #1 C(s) + O2(g)  CO2(g) ∆H° = -393.5 kJ The enthalpies are added together Therefore the enthalpy is -393.5 kJ
2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 This is not as easy to do! STEP 1: Number each given thermochemical equation. 1 2
2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 STEP 2: Arrange the equations so that your desired reactants are on the left side, and your desired products are on the right side 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ 1- 2
2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ 3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ 1- 2 STEP 3: Multiply the equations by factors such that they may match your desired equation. Remember to multiply the enthalpy by the same factor. 3 x
2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ 3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ 1- 2 STEP 4: Add the two equations and enthalpies together. Remember to cancel repeating chemicals. 3 x 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = -822.3 kJ
Enthalpy H2O (s)  H2O (l) What is happening? Is a ∆H° value involved? Why or why not? ENTHALPY AND HESS’ LAW Heat is absorbed by the ice in order to melt. ΔH° is positive, because heat is required
Enthalpy Enthalpy is also involved in physical changes. ENTHALPY AND HESS’ LAW