l13 Chapt7-3 web

Chemistry 5

Chapter-7 Thermochemistry Part-3 23 October 2002

Enthalpies of Reaction ƒ A thermochemical equation combines • balanced chemical equation • reaction enthalpy, ∆H ∆H of reaction– heat change at constant P– ignores work

done by/on the system in moving from initial to final state.

ƒ Reaction enthalpy The reaction enthalpy is the change in enthalpy for stoichiometric number of moles of reactants in the chemical equation: CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(l) ∆H = -890 kJ ƒ Things to pay attention to: • ∆H is extensive Enthalpy change is directly proportional to amounts of substances in system. • ∆H depends on direction ∆H changes sign when a process is reversed; that is, ∆Η is a state function, and reverse intial/final states. • State of reactants/products Enthalpy change depends on the state– gas, liquid or solid– of substances in chemical equation. ∆H is state function and changing state of reactant/product changes intial/final state.

Enthalpies of Reaction-- Example ƒ Thermochemical Equation: CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(l) ∆H = -890 kJ ƒ Change Number of Moles 2CH4(g) + 4O2(g) Æ 2CO2(g) + 4H2O(l) ∆H = -1780 kJ (∆H is extensive property) ƒ Change Direction of Reaction CO2(g) + 2H2O(l) Æ CH4(g) + 2O2(g) ∆H = +890 kJ (∆H is state function) ƒ Change State of Reactants and Products CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(g) ∆H = -802 kJ It takes heat (∆Hvap) to change water from liquid to gas state, and thus heat of reaction is reduced. How do we generalize this idea?

Combining Thermochemical Eqns ƒ Thermochemical Equations: CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(l) ∆H = -890 kJ 2H2O(l) Æ 2H2O(g)

∆H = +88 kJ

CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(g) ∆H = -802 kJ ƒ Hess’s Law The overall reaction enthalpy is the sum of reaction enthalpies of the steps into which the reaction is divided. If a process occurs in steps– even if these are hypothetical– then the enthalpy change for the overall process is the sum of the enthalpy changes of the individual steps! initial state

ENTHALPY

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

final state

Hess’s Law– Example-2 S(s) + 3/2O2(g) Æ SO3(g)

S solid

∆H = ?

+O2

+ 3/2 O2 SO2 gas SO3 gas

+ 1/2O2

ƒ Hess’s Law 1. S(s) + O2(g) Æ SO2(g)

∆H1 = -320.5 kJ

2. SO2(g) + 1/2O2(g) Æ SO3(g) S(s) + 3/2O2(g) Æ SO3(g)

∆H2 = -75.2 kJ

∆Hrxn = ∆H1 + ∆H2 = -320.5 + -75.2 = -395.7 kJ

Standard Enthalpies ƒ Standard Enthalpy of Formation, ∆Hfo of a substance is the enthalpy change that occurs in the formation of one mole of the substance in the standard state from reference form of the elements in their standard states. ƒ Pure Elements in Reference Form? Generally, most stable form of elements in standard state; for example: H is H2(g), O is O2(g), Na is Na(s) • pure elements, reference form: The standard enthalpy of formation of a pure element in its reference form is zero, O.

∆Hfo: Examples ƒ

ƒ Formation Reactions: C(s, graphite) + O2(g) Æ CO2(g) C(s, graphite) + 2H2(g) Æ CH4(g) 1/2N2(g) + 3/2H2(g) Æ NH3(g) 1/2N2(g) + 1/2O2(g) Æ NO(g) 1/2N2(g) + O2(g) Æ NO (g) 2 Na(s) + 1/2Cl2(g) Æ NaCl(s)

Standard Enthalpies of RXN ƒ

∆Hfo & ∆Horxn • Standard enthalpies of formation tabulated since they can be used to calculate enthalpy of reaction. • How? Use state function property of H– does not matter what pathway we take– and design path through elements in their standard forms.

∆H = Σvp∆Hfo(products) – Σvr∆Hfo(reactants) ƒ Example: NO(g) + (1/2) O2(g) Æ NO2(g) • ∆Horxn = ∆Hf(NO2) – ∆Hf(NO) – 1/2∆Hf(O2) • determine enthalpies of formation

ENTHALPY

NO(g) + (1/2) O2(g)

NO2(g) (1/2) N2(g) + O2(g)

RXN of Al & Iron Oxide ƒ Overall Chemical Reaction 2 Al + Fe2O3 Æ Al2O3 + 2Fe Is reaction exothermic or endothermic? ƒ Enthalpy of Reaction: ∆H = Σvp∆Hfo(products) – Σvr∆Hfo(reactants) = ∆Hf(Al2O3) + 2∆Hf(Fe) – 2∆Hf(Al) – ∆Hf(Fe2O3) = -1676 + 2(0) – 2(0) - -824 = -852 kJ/mol ƒ Demonstration– Observations? •

After starting, the reaction proceeds vigorous

•

A lot of heat is released!!

Al(s) Al2O3(s) Fe(s) Fe2O3(s)

∆Hfo (kJ/mol) 0 -1676 0 -824

Ionic Reactions in Water? ƒ Standard State • conditions: Many chemical reactions proceed in aqueous solution (e.g., biochemical reactions) and involve ionic species, such as protons and metal cations. Standard State: 1M concentration • reference state: To determine enthalpies of formation for ions there is new issue– cannot create ion of single type in chemical reaction. Therefore, define a reference state to which compare enthalpies of formation of other ions:

∆Hfo (H+) = 0 ƒ

Enthalpies of Ionic Reactions ƒ When 50.0ml of 0.100M Na2SO4(aq) is mixed with 25.0ml of 0.200M AgNO3(aq), a white ppt forms. • Net ionic reaction? 2Ag+(aq) + SO42-(aq) Æ Ag2SO4(s) • Given ∆Hfo(Ag2SO4(s)) = -715 kJ/mol, calculate ∆Hrxn; is reaction endo or exothermic? ∆Hrxn = ∆Hf(Ag2SO4) – 2∆Hf(Ag+) – ∆Hf(SO42-) = -715 – 2(106) – (-909) = -18 kJ/mol Slightly Exothermic total enthalpy for experiment = 2.5x10-3mol x -18kJ/mol = -4.5x10-2 kJ • If the solution starts at 25.0oC, will the final temperature be high or lower, and by how much? Slightly Higher Assume: (i) specific heat = that of water, 4.18 J/g-K (ii) density = that of water 1g/ml (iii) solve using q = m x specific heat x ∆T

Food & Enthalpies of Reactions ƒ

ƒ Overall Thermochemical Reactions: • Carbohydrate– sucrose: C12H22O11 + 12O2 Æ 12CO2(g) + 11H2O(l) • Fat C12H24 + 36O2 Æ 12CO2(g) + 12H2O(l)

ƒ +/- of different types of fuels? ∆H(carbohydrate) ~ 17 kJ/g ∆H(fat) ~ 38 kJ/g There is approximate 2x more energy store per gram in fat than carbohydrates…….hence, if you are a bear sleeping for the winter it is much better to store needed fuel as fat!