Changes in Enthalpy During Chemical Reactions A change in enthalpy during a reaction depends on many variables, temperature being one of the most important

Standard Thermodynamic Temperature To standardize the enthalpies of reactions, data are often presented for reactions in which both reactants and products have the standard thermodynamic temperature of 25.00 degrees Celsius or 298.15 K. A thermodynamic value for a reaction is usually presented by using the chemical equation: 1/2H2(g) + 1/2Br2(l)  HBr(g) ΔH = -36.4kJ

Calorimetry The experimental measurement of an enthalpy change for a reaction is called calorimetry. A Calorimeter is an instrument used to measure the enthalpy change of a reaction Enthalpy changes of combustion reactions, which are always exothermic, are determined using a bomb calorimeter.

Bomb Calorimeter Consists of a vessel in which high pressure oxygen and sample are ignited electrically. The heat energy is transferred to a water bath that surrounds the vessel. The temperature change of the water along with the known specific heat of water are used to calculate the enthalpy change. Used by nutritionists to find the energy content of foods. Nutritionists assume that all the energy released through combustion is available to the body as we digest it. Calorimetric measurements can be made with very high precision.

Adiabatic Calorimetry An adiabatic calorimeter uses an insulated vessel. Adiabatic means “not allowing energy to pass through.” Energy cannot enter or escape the vessel. Adiabatic calorimetry is used for reactions that are not ignited, such as for reactions in aqueous solution.

Hess’s Law The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process. OR The amount of heat release or absorbed in a chemical reaction does not depend on the number of steps in the reaction.

Hess’s Law Applied For the following equation: P4(s) + 10Cl2(g)  4PCl5(g) ΔH = -1596 kJ Phosphorous pentachloride may also be prepared in two steps as: Step 1: P4(s) + 6Cl2(g)  4PCl3(g) ΔH = -1224 kJ Step 2: PCl3(g) + Cl2(g)  PCl5(g) ΔH = -93 kJ The second reaction must take place 4 times so for each occurrence 93 kJ are released, therefore the total enthalpy change in the two steps would be calculated -1224 kJ + (4 X -93 kJ) = -1596 kJ the same as the single chemical reaction.

Hess’s Law and Algebra Chemical equations can be manipulated using rules of algebra to get a desired equation. When equations are added or subtracted, enthalpy changes must be added or subtracted. When equations are multiplied by a constant, the enthalpy must be multiplied by that constant.

When CO is formed from solid carbon C, and carbon dioxide CO2 using the equations below: 2C(s) + O2(g)  2CO(g) ΔH = -221 kJ C(s) + O2(g)  CO2(g) ΔH = -393 kJ But carbon dioxide is a reactant not a product as shown in the second reaction so we cannot simply add the enthalpies. The second reaction is actually the reverse with CO2 the reactant -C(s) + -O2(g)  -CO2(g) ΔH = -(-393 kJ) or CO2(g)  C(s) + O2(g) ΔH = 393 kJ

2C(s) + O2(g)  2CO(g) ΔH = -221 kJ CO2(g)  C(s) + O2(g) ΔH = 393 kJ 2C(s)+O2(g)+CO2(g) 2CO(g)+C(s)+O2(g) ΔH = 172 kJ Final equation : C(s)+CO2(g) 2CO(g) ΔH = 172 kJ

Standard Enthalpies of Formation The enthalpy change in forming 1 mole of a substance from elements in their standard states. The standard enthalpy of formation for all pure elements is 0 Using the standard enthalpies of formation, the enthalpy change for any reaction can be calculated.

For the equation: SO2(g) + NO2(g)  SO3(g) + NO(g) For the reactants take the standard enthalpy change for that compound times a negative since you are breaking it into the elements rather than forming it from the elements. Therefore SO2 and NO2 would have enthalpy changes of 296.8 kJ/mol and -33.1kJ/mol respectively. For the products the standard enthalpies of formation are taken from the table.

To calculate the enthalpy change for the reaction: SO2(g) + NO2(g)  SO3(g) + NO(g) Use the formula: ΔHreaction = ΔHproducts - ΔHreactants ΔHreaction = (-395.8 kJ/mol + 90.3 kJ/mol) – (296.8kJ/mol + -33.1kJ/mol) ΔHreaction = -41.8 kJ/mol