Topic 5 2 Hesss Law IB SL Chemistry

Topic 5. 2 Hess’s Law IB SL Chemistry Mrs. Page 2015 -2016

APPLICATION & SKILLS • Application of Hess’s law to calculate enthalpy changes. • Calculation of ∆H reactions using ∆Hfө data • Determination of the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes

What is the principle of conservation in relation to Energy and Enthalpy? • The principle of conservation of energy (First Law of Thermodynamics in Physics) states: • Energy can neither be created nor destroyed.

Think About Energy And How This Applies

What does this mean in Chemistry? • The total change in chemical potential energy (enthalpy change) must be equal to the energy lost or gained by the system. • This also means that the total enthalpy change on converting a given set of reactants to a particular set of products is constant, NO MATTER WHICH WAY the change is carried out • This principle holds, whether or not this reaction can actually happen in real life!!

Hess’s Law • The enthalpy change accompanying a chemical reaction is independent of the pathway between the initial and final states. • In other words… ΔHr= ΔH 1 + ΔH 2 UNDERSTANDING: • The enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps.

Hess’s Law • The heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps.

Combustion of methane • ∆H 1 shows the magnitude of the single step reaction. • ∆H 2 and ∆H 3 when combined are equivalent to ∆H 1, demonstrating Hess’s Law.

Nature of Reactions • Most reactions occur in more than one step. • We focus on the OVERALL reaction based on the balanced chemical equation, but there could be many steps involved to reach the final outcome.

REMEMBER: • If you reverse an equation the ∆H must switch signs. § Na (s) + ½ Cl 2(g) Na. Cl (s) ∆H = -411 k. J § Na. Cl (s) Na (s) + ½ Cl 2(g) ∆H = 411 k. J

Standard Conditions • An enthalpy change under standard conditions is called a Standard Enthalpy Change and has the symbol ΔHθ, where θ means under standard conditions. • Standard Enthalpy Change of reaction (ΔHθr) is the enthalpy change (heat given out or taken in) when molar amounts of reactants as shown in the stoichiometric equation react together under standard conditions to give products.

Example of Standard Enthalpy Change of Reaction (ΔHθr) • For example, for the reaction: § N 2(g) + 3 H 2(g) 2 NH 3(g) the enthalpy change of reaction is -92 k. Jmol-1. • This means that 92 k. J of heat energy are given out when 1 mol N 2 reacts with 3 mol N 2 to form 2 mol NH 3. • If the equation is written as: § ½ N 2(g) + 3/2 H 2(g) NH 3(g) § ΔHθr= -46 k. Jmol-1, then the enthalpy change of reaction is for 0. 5 mol N 2 reacting, and the enthalpy change is half as much.

Rules for working out Enthalpy Changes 1 st: Write the equations with given enthalpies. (If you reverse equations remember to change signs) 2 nd: Decide how the given reaction equations should be written to match the original reactions. If you need to multiply or divide any part of the equation, the whole equation must follow the same operation, including ∆H. 3 rd: Solve for the Enthalpy Change by adding the two equations and their enthalpy changes, cancelling common terms to produce the OVERALL equation and its enthalpy change.

Setting up an Enthalpy Change Problem • If you need to multiply or divide any part of the equation, the whole equation must follow the same operation, including ∆H. § Na (s) + ½ Cl 2(g) Na. Cl (s) ∆H = -411 k. J § 2 Na (s) + Cl 2(g) 2 Na. Cl (s) ∆H = -822 k. J

Sample Problem: Given the following enthalpy changes below, calculate the enthalpy change for the reaction. 2 CO(g) + O 2(g) 2 CO 2(g) Given Enthalpies: 2 C(s) + O 2(g) 2 CO(g) ΔHθ=-222 k. Jmol-1 C(s) + O 2(g) CO 2(g) ΔHθ = -394 k. Jmol-1

Sample Problem 1: 2 CO(g) + O 2(g) 2 CO 2(g) Given Enthalpies: 2 C(s) + O 2(g) 2 CO(g) ΔHθ=-222 k. Jmol-1 C(s) + O 2(g) CO 2(g) ΔHθ = -394 k. Jmol-1 • CO on reactant side so flip equation & change sign 2 CO(g) 2 C(s) + O 2(g) ΔHθ = +222 k. Jmol-1 • We need carbons to cancel so multiple second equation by 2 2 C(s) + 2 O 2(g) 2 CO 2(g) ΔHθ = 2(-394 k. Jmol-1 ) • Cancel substances • Rewrite equation to check and add enthalpies 2 CO(g) + O 2(g) 2 CO 2(g) 222 + 2(-394) = -566 k. Jmol-1

Methanol Powered Cars 2 CH 3 OH(l) + 3 O 2(g) 2 CO 2(g) + 4 H 2 O(g) ΔHrxn = ? 2 CH 4(g) + O 2(g) 2 CH 3 OH(l) ΔHrxn = -328 k. J CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) ΔHrxn = -802. 5 k. J 2 CH 3 OH(l) 2 CH 4(g) + O 2(g) ΔHrxn = +328 k. J 2(CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g)) ΔHrxn = -1605 k. J 2 CH 3 OH(l) + 2 CH 4(g) + 4 O 2(g) 2 CH 4(g) + O 2(g) + 2 CO 2(g) + 4 H 2 O(g) 3 2 CH 3 OH + 3 O 2 2 CO 2 + 4 H 2 O ΔHrxn = -1277 k. J

You Try: What’s the enthalpy change for: H 2(g) + O 2(g) H 2 O 2(aq) Given enthalpies: 2 H 2 O 2(aq) 2 H 2 O(l) + O 2(g) ΔH= -200 k. Jmol-1 2 H 2(g) + O 2(g) 2 H 2 O(l) ΔH= -600 k. Jmol-1

Sample Problem 3 What’s the enthalpy change for: Fe(s) + Cl 2(g) Fe. Cl 2(s) Given the enthalpy changes for: 2 Fe(s) + 3 Cl 2(g) 2 Fe. Cl 3(s) ΔH= -800 k. Jmol-1 2 Fe. Cl 2(s) + Cl 2(g) 2 Fe. Cl 3(s) ΔH= -120 k. Jmol-1

Harder Yet Given the following data: N 2(g) + O 2(g) 2 NO(g) ∆H 1 = + 180. 7 k. Jmol-1 2 NO(g) + O 2(g) 2 NO 2(g) ∆H 2 = - 113. 1 k. Jmol-1 2 N 2 O(g) 2 N 2(g) + O 2(g) ∆H 3 = - 162. 3 k. Jmol-1 Use Hess’s law to calculate ∆H 4 for the following reaction. N 2 O (g) + NO 2 (g) 3 NO (g)

Harder Yet Given the following data: N 2(g) + O 2(g) 2 NO(g) ∆H 1 = + 180. 7 k. Jmol-1 2 NO(g) + O 2(g) 2 NO 2(g) ∆H 2 = - 113. 1 k. Jmol-1 2 N 2 O(g) 2 N 2(g) + O 2(g) ∆H 3 = - 162. 3 k. Jmol-1 Use Hess’s law to calculate ∆H 4 for the following reaction. N 2 O (g) + NO 2 (g) 3 NO (g) -1 2 N 2 O(g) 2 N 2(g) + O 2(g) ∆H 3 = - 162. 3 k. Jmol Flip: 2 NO 2(g) 2 NO(g) + O 2(g) N 2(g) + O 2(g) 2 NO(g) N 2 O (g) + NO 2 (g) 3 NO (g) ∆H 2 = +113. 1 k. Jmol-1 ∆H 1 = + 180. 7 k. Jmol-1

Using Enthalpies of Combustion Data Book Table 13 Find ∆Hr for C 2 H 4(g) + H 2(g) C 2 H 6(g) ∆Hc = -1411 k. Jmol-1 3 O 2(g) -286 ½ O 2(g) -1561 +1561 CO 2(g) + H 2 O(g) ∆Hr = -1411 + -286 + 1561 = -86 k. J mol-1 Enthalpy Cycle Diagram

Your Turn • Use the enthalpy of combustion data in the data book to calculate the enthalpy of formation of sucrose C 12 H 22 O 11 (Show the enthalpy cycle diagram to support your work) 12 C(s) + 11 H 2(g) + 5½ O 2(g) C 12 H 22 O 11(s) -2234 k. Jmol-1

HOMEWORK • Hess’s Law Worksheet