Ch. 17: Reaction Energy and Reaction Kinetics Heat of Reaction Heat of Formation Hess’ Law

Heat of Reaction amount of energy released or absorbed during a chemical reaction the difference between the stored energy of reactants and products energy can be a reactant or product same as the change in enthalpy (H) at constant pressure ∆H = Hproducts - Hreactants endo: + ∆H exo: - ∆H

Heat of Reaction / Enthalpy Change

Standard Enthalpy of Formation ∆Hf° change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states ° means that the process happened under standard conditions so we can compare values more easily room temperature (298 K) and 1 atm values given p. 902

Writing Formation Equations always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients) NO2(g): ½N2(g) + O2(g)  NO2(g) ∆Hf°= 34 kJ/mol CH3OH(l): C(s) + 2H2(g) + ½ O2(g) CH3OH(l) ∆Hf°= -239 kJ/mol

Stability Would a large negative value, like ∆Hf°= -239 kJ/mol, indicate a very stable or unstable compound ? If the compound is very stable, lots of energy will be released when it is formed so it will have a large negative value. If the compound is unstable, it will have a positive value and will most likely decompose into its elements

Hess’ Law thermochemical equations can be rearranged and added together to give enthalpy changes for reactions not given in the Appendix Hess’ Law- the overall enthalpy change is equal to the sum of the enthalpy changes for individual steps in a reaction

Rules If a reaction is reversed, the sign of ∆H must be reversed as well. because the sign tells us the direction of heat flow The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way.

Example 1 Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond. Cgraphite (s)  Cdiamond (s) ∆H=?

Example 1 (1) Cgraphite(s) + O2(g)  CO2(g) ∆H=-394kJ/mol (2) Cdiamond(s) + O2(g)  CO2(g) ∆H=-396kJ/mol to get the desired equation, we must reverse 2nd equation: -(2) CO2(g)  Cdiamond(s) + O2(g) ∆H=396kJ/mol Cgraphite (s)  Cdiamond (s) ∆H=-394 + 396 ∆H=2 kJ/mol

2N2(g) + 6H2O(g)  3O2(g) + 4NH3(g) Example 1b Find the enthalpy change for the following reaction. 2N2(g) + 6H2O(g)  3O2(g) + 4NH3(g) Given: NH3(g)  ½N2(g) + 3/2H2(g) ∆H=46 kJ 2H2(g) + O2(g)  2H2O(g) ∆H=-484 kJ

Example 1b NH3: (1) must be reversed and multiplied by 4 H2O: (2) must be reversed and multiplied by 3 2N2(g) + 6H2(g)  4NH3(g) ∆H=-4(46 kJ) 6H2O(g)  6H2(g) + 3O2(g) ∆H=-3(-484 kJ) ∆H= +1268 kJ

Example 2 (1) H2(g) + ½O2(g)  H2O(l) ∆H=-285.8kJ/mol Calculate the change in enthalpy for the following reaction: 2N2(g) + 5O2(g)  2N2O5(g) Given (1) H2(g) + ½O2(g)  H2O(l) ∆H=-285.8kJ/mol (2) N2O5(g) + H2O(l)  2HNO3(l) ∆H=-76.6kJ/mol (3) ½H2(g) + ½N2(g) +3/2O2(g) HNO3(l) ∆H=-174.1kJ/mol

Example 2 -2(∆H1)= (-2)(-285.8kJ/mol)= 571.6 for N2O5: multiply (2) by 2 and reverse it for H2O: multiply (1) by 2 and reverse it for HNO3: multiply (3) by 4 -2(∆H1)= (-2)(-285.8kJ/mol)= 571.6 -2(∆H2)= (-2)(-76.6kJ/mol)= 153.2 4(∆H3)= (4)(-174.1kJ/mol)= -696.4 28.4 kJ/mol

Example 3 Find ∆H for the synthesis of B2H6, diborane: 2B(s) + 3H2(g)  B2H6(g) Given: 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) ∆H2=-2035kJ H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ H2O(l)  H2O(g) ∆H4=44 kJ

Example 3 Start by paying attention to what needs to be on reactants and products side (1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ -(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) -∆H2=2035kJ (3) H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ (4) H2O(l)  3H2O(g) ∆H4=44 kJ

Example 3 Underline what you want to keep- that will help you figure out how to cancel everything else: (1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ -(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) -∆H2=2035kJ (3) H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ (4) H2O(l)  3H2O(g) ∆H4=44 kJ

Example 3 Need 3 H2 (g) so 3 x (3) Need 3 H2O to cancel so 3 x (4) (1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ -(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) -∆H2=-(-2035kJ) 3x(3) 3H2(g) + 3/2O2(g)  3H2O(l) 3∆H3=3(-286kJ) 3x(4) 3H2O(l)  3H2O(g) 3∆H4=3(44 kJ) 2B(s) + 3H2(g)  B2H6(g) ∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ

Entropy - PAP Only Defined as the measure of randomness of particles in a system. Higher energy results in higher entropy of the same particle. i.e. a particle in the gas phase will have a higher entropy than a particle in the liquid phase. At absolute zero, a particle would have zero entropy

Gibbs Free Energy - PAP Only Used to determine the spontaneity of a reaction A positive DG value indicates the reaction is not spontaneous at this condition. A negative DG value indicates the reaction is spontaneous at this condition.

Relationships - PAP Only Using Gibbs Free Energy to find out if a reaction is spontaneous or non-spontaneous. DG DH DS Reaction driven by Favorable temperature Sign Meaning + or - ???????? + Endothermic Increase disorder Entropy High temperatures Non-spontaneous - Decrease disorder Nothing None spontaneous Exothermic Enthalpy and entropy All temperatures Enthalpy Low temperatures