CHEM 210 1nd Edition Lecture 22 Outline of Last Lecture I. ThermodynamicsA. EnergyB. EnthalpyOutline of Current Lecture II. Heat EquationsIII. Hess’s LawCurrent LectureWe can use the given heat of a reaction to find amounts or grams of the elements. For example if we have the equation,Al2O3 (s) 2Al (s) + 3/2 O2 (g) ΔH= 1676 kJWe first look to see at the equation and see that 1676 kJ gives us 2 moles of Al. We then convert to Joules from kJ.1.000x 10^3 kJ x (2 mol Al) / 1676 kJ) x (26.98 g Al) / (1 mol Al) = 32.20 g AlWe also used stoichiometry to cancel out the units of Joules and convert from moles to grams using the molar mass of Al.Hess’s LawThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Hess’s Law is a law that states that multiple equations can be used and manipulated in order to get to the final answer as well as just achieving the final answer with one equation. Ex: CO (g) + NO (g) CO2 (g) + ½ N2 (g) ΔH = ?For the problem, we need to find the change in enthalpy or ΔH. We are given these two equations:1. CO (g) + ½ O2 (g) CO2 (g) ΔH = 2832. N2 (g) + O2 (g) 2NO (g) ΔH = 180.6Using these two equations, we can manipulate one to cancel out with some components of another. In this case, we would multiply the second equation by ½ to cancel out the ½ O2 from both equations.1. CO(g) + ½ O2 (g) CO2 (g) ΔH = 2832. NO(g) ½ N2 (g) + ½ O2 (g) ΔH = -90.3 CO(g) + NO(g) CO2 (g) + ½ N2 (g) ΔH = -373.3
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