Enthalpy (H)
Enthalpy (H) is the total energy amount (Epotential + Ekinetic) of a system during a chemical reaction under constant temperature and pressure conditions. Kinetic energy is the energy of motion (of particles). Potential energy involves stored energy (energy locked up in chemical bonds) The enthalpy of a system cannot be measured - only the change in enthalpy (also known as heat of reaction) can be measured as it will occur in the form of a heat transfer. Enthalpy change represents the difference between the potential energy of the products and of the reactants in a chemical reaction ( Hrxn = Hproducts - Hreactants)
Endothermic vs. Exothermic Breaking bonds requires energy Creating bonds releases energy Endothermic reaction- when a reaction results in a net absorption (gain) of energy. Energy from the environment flows into the system resulting in a decrease in temperature of the surroundings. Exothermic reaction- when a reaction results in a net release (loss) of energy. The temperature of surroundings increases
Representing Enthalpy Change ( H ⁰) The superscript ⁰ (i.e. H⁰) indicates that the reaction occurs at SATP (standard atmospheric temperature and pressure: 25⁰C and 100 kpa
There are three ways to represent the enthalpy change ( H) in a chemical reaction
1. A Thermochemical equation showing the heat of reaction ( H) following the equation H = + (positive) for an endothermic reaction H = - (negative) for an exothermic reaction 3C(s) + 2Fe2O3(s) 4Fe(s) + 3CO2(g) H⁰ = + 495 kj
2. A thermochemical equation showing the heat of reaction in the actual equation as a reactant (on the left) for an endothermic reaction product (on the right) for an exothermic reaction 3C(s) + 2Fe2O3(s) + 495 kj 4Fe(s) + 3CO2(g)
3. Potential Energy (Enthalpy) Diagrams Exothermic Reaction Endothermic Reaction
Sample Problems 1. Write thermochemical equations for the following reactions: a) 2HgO(s) 2Hg(l) + O2(g) if H⁰ = +181.67 kj b) K2O(s) + CO2(g) K2CO3(s) if H⁰ = - 391.1 kj c) 2Al(s) + 3/2O2(g) Al2O3(s) if H⁰ = -1675.7 kj
Solution: a) 2HgO(s) 2Hg(l) + O2(g) H⁰ = +181.67 kj or 2HgO(s) + 181.67 kj 2Hg(l) + O2(g) b) K2O(s) + CO2(g) K2CO3(s) H⁰ = - 391.1 kj or K2O(s) + CO2(g) K2CO3(s) + 391.1 kj c) 2Al(s) + 3/2O2(g) Al2O3(s) H⁰ = -1675.7 kj or 2Al(s) + 3/2O2(g) Al2O3(s) + 1675.7 kj
2. Draw a potential energy diagram for each of the previous situations: a) 2HgO(s) 2Hg(l) + O2(g) if H⁰ = +181.67 kj b) K2O(s) + CO2(g) K2CO3(s) if H⁰ = - 391.1 kj c) 2Al(s) + 3/2O2(g) Al2O3(s) if H⁰ = -1675.7 kj
Standard molar enthalpy of formation ( H⁰f) The quantity of energy that is absorbed or released when one mole of a compound is formed from its elements in their standard states under standard state conditions. (at SATP: 25 C and 100 kpa) (The S.I. Unit is J/mol) H f values have been measured for most compounds and collected into tables (see photocopy). Example: H2(g) + 1/2O2(g) H2O(l) H⁰f = -285.8 kj (This means that 286 kj of energy was released in the formation of one mole of the compound water from the elements hydrogen and oxygen.) C(s) + ½ O 2 (g) CO(g) ΔH f = -110.5KJ
Standard molar enthalpy of combustion ( H⁰comb) the quantity of energy that is released during the combustion of one mole of a given substance A combustion reaction is one in which a substance reacts with oxygen to produce heat. Refer to the photocopied data table 16.3 for values. A typical example is the combustion of methane gas: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 965.1 kj
Calculating Enthalpy Changes From a Balanced Chemical Equation
The energy released or absorbed during a chemical reaction depends on two main factors: 1) the specific reactants (i.e. the chemicals that are reacting) 2) the amount of the reactants (i.e. the # of moles of each - indicated in the balanced chemical equation). Thus, if the amount of the reactants doubles, H will also double.
Sample problem p.644 Methane (CH 4 ) is the main component of natural gas. Natural gas undergoes combustion to provide energy for heating homes and cooking food. a) How much heat is released when 50.00 g of methane forms from its elements? b) How much heat is released when 50.00 g of methane undergoes complete combustion?
Sample Problem #2 Given the reaction: Sample Problems 2NH3(g) + 3Cl2(g) N2(g) + 6HCl(l) H = -461.4 kj How much energy would be released if 35.7 g of NH3(g) were completely reacted?
Solution: From the thermochemical equation, 2NH3(g) + 3Cl2(g) N2(g) + 6HCl(l) H = -461.4 kj when 2 moles of NH3 is completely reacted 461.4 kj of energy is released. We know from the periodic table (molar mass) that one mole of NH3 is 17.04 g. 35.7 g NH3 x 1 mol x - 461.4 kj = - 483 kj 17.04 g 2 mol NH3
Sample Problem #3 Given the reaction: H2(g) + I2(s) 2HI(g) H = +53.0kJ How much energy must be absorbed when 150 g of I2(s) is completely reacted?
Solution: From the thermochemical equation, H2(g) + I2(s) 2HI(g) H = +53.0kJ when one mole of I2(s) is reacted, 53.0 kj is absorbed. The energy absorbed when 150 g of I2(s) is reacted is: 150 g x 1 mol I2(s) x 53.0 kj = 31.3 kj 253.8 g 1 mol I2(s)
Homework Do Practice Problems 19, 21a and 23 on page 645 of the text.
An increase in the enthalpy (internal energy) of a system has three possible outcomes: a) an increase in temperature b) a change in state/phase (e.g. solid liquid) c) a chemical reaction
We will study three events that involve enthalpy (internal energy) changes: Event Symbol T Kinetic Molecular Theory Flow of heat q (EK) Yes Change to movement of molecules Phase change (ie. Solid to liquid) H (Ep) No Change to the intermolecular bonds between particles Chemical reaction H (Ep) No (at SATP) Change to the intramolecular bonds within the particles (bonds are broken and formed) Note that temperature changes cannot always be expected with a change in the amount of heat (enthalpy)