Enthalpy & Thermochemical Equations Thornburg 2014

Enthalpy The amount of heat gained or lost by a system at a constant pressure is called Enthalpy Chemists use the symbol ∆H for the change in enthalpy Because most of the reactions we study occur with no change in pressure, the heat lost or gained by a system is equal to the enthalpy change q = ∆H Thornburg 2014

The enthalpy of a system can be calculated using this equation: ∆H = m x C x ∆T m = mass C = specific heat ∆T = the change in temperature Thornburg 2014

Calculating Enthalpy Practice Problem: Clyde the Bear performs an experiment by reacting 20 grams of water which contains NaOH with 50 grams of water containing HCl. The initial temperature for both samples of water is 23 ˚ C. Clyde observes the reaction in his calorimeter and records the highest temperature at 44 ˚ C. Clyde knows the specific heat for water is 4.18J/g x ◦ C. Help Clyde calculate the change in enthalpy for his reaction. Thornburg 2014

∆H = m x C x ∆T ∆H = ? m = 20 g + 50 g = 70 g C = 4.18J/g x ˚ C ∆T = 44 ˚ C - 23 ˚ C = 21 ˚ C ∆H = 70 g x 4.18J/g x ˚ C x 21 ˚ C ∆H = J Thornburg 2014

Thermochemical Equations Like all chemists, Clyde the Bear wants to determine exactly how much heat is released or absorbed when his chemical reaction occurs A chemical equation that includes the exact amount of energy released or absorbed during the reaction is called a thermochemical equation Example: 1HCl + 1NaOH → 1NaCl + 1H KJ Thornburg 2014

Practice Problem Clyde the Bear reacts 50 grams of HCl with an excess amount of NaOH. How much heat should this reaction release? Example: 1HCl + 1NaOH → 1NaCl + 1H KJ 50 g HCl x 1 mole HCl = 1.37 moles HCl 36.5 g HCl 1.37 moles HCl x 60 KJ = 82.2 KJ 1 mole HCl Thornburg 2014