Drill 350 J are released as ice ( Specific Heat = 2.1 J / (g oC) ) cools from - 5.0 oC to -32 oC. What is the mass of ice?

Thermochemical Equations/ Hess’ Law

Objective SWBAT: Correctly manipulate thermochemical equations to predict the enthalpy of reaction (Hess’s Law)

Thermochemical Equations Include enthalpy (heat) accompanying the reaction Coefficients represent moles – it is possible to have fractions – Example: ½ means half a mole of the substance State of matter is specified (because they influence the overall amount of energy as heat gained or lost)

Examples of Thermochemical equations Δ H is directly proportional to the amount of substance produced or reacting in a reaction H2(g) + ½ O2(g)  H2O(l) Δ H = -285.8 kJ 2 H2(g) + O2(g)  2 H2O(l) Δ H = -571.6 kJ

Laws of Thermochemistry Δ H for a reaction is equal in magnitude but opposite in sign from the reverse reaction HgO(s)  Hg(l) + ½ O2(g) Δ H = 90.7 kJ Hg(l) + ½ O2(g)  HgO(s) Δ H = -90.7 kJ

Thermochemical Equations Practice Complete the practice at your desk

Hess’s Law Notes

Hess’s Law The overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.

Hess’s Law Example Determine ∆H of the reaction Sn(s) + 2 Cl2(g)  SnCl4(l) Sn(s) + Cl2(g)  SnCl2(s) ∆H= -83.6 kJ SnCl2(s) + Cl2(g)  SnCl4(l) ∆H = -46.7 kJ

Answer ∆H = -130. kJ

Hess’s Law Example 2

Solution Determine what we must do to the three given equations to get our target equation: first eq: multiply it by two to get 2C second eq: do nothing third eq: flip it so as to put CH4 on the product side Rewrite all three equations with changes applied: + 15.3 kJ

Drill Explain Hess’s Law in your own words.

Hess’s Law Practice Problems Complete the practice problems at your desk. Problems not completed in class will become your homework