State Function revisited….State Function revisited…. Dependent ONLY on a system’s state at a...

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Hess’ Law and Enthalpy of Formation

description

Standard ConditionsStandard Conditions  Represented by a “ ° “ beside Δ H or Δ E, etc.  25°C  1 atm

Transcript of State Function revisited….State Function revisited…. Dependent ONLY on a system’s state at a...

Page 1: State Function revisited….State Function revisited….  Dependent ONLY on a system’s state at a given moment in time.  Only initial and final states

Hess’ Law and Enthalpy of Formation

Page 2: State Function revisited….State Function revisited….  Dependent ONLY on a system’s state at a given moment in time.  Only initial and final states

State Function revisited….

Dependent ONLY on a system’s state at a given moment in time. Only initial and final states Not based on the path to get to a given

condition

Ex. Energy, Enthalpy

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Standard Conditions

Represented by a “ ° “ beside ΔH or ΔE, etc.

25°C

1 atm

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Methods for determining ΔH

1) Calorimetry

2) Application of Hess’ Law

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Hess’ Law

Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step ΔHrxn = ΣΔHA+B+C (sum of ΔH for each step)

Allows us to break a chemical reaction down into multiple steps to calculate ΔH Add the enthalpies of the steps for the enthalpy for

the overall chemical reaction

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Guidelines for using Hess’ Law

Must use data and combine each step in a way that gives the chemical reaction with the unknown ΔH

Set up steps so chemical compounds not in the final reaction are cancelled

Reverse a reaction if necessary and change the sign on ΔH

Check for correct mole ratios

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Example 1:

H2O(l) H2O (g) ΔH° = ?

Based on the following: H2 + ½ O2 H2O(l) ΔH° = -285.83 kJ/mol

H2 + ½ O2 H2O(g) ΔH° = -241.82 kJ/mol

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Example 2:

C(s) + 4H2 C3H8 (g) ΔH° = ?

Based on the following: 2H2 + O2 2H2O ΔH° = -571.7 kJ/mol

C3H8 + 5 O2 3CO2 + 4H2O(g) ΔH°= -2220.1kJ/molC(s) + O2 CO2 (g) ΔH° = -393.5 kJ/mol

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Methods for determining ΔH

1) Calorimetry

2) Application of Hess’ Law

3) Enthalpies of Formation

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Enthalpy of Formation (ΔHf°)

Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state.

Elements present in “most thermodynamically stable state” 25°C°, 1atm

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Example 3

MgO (s) + CO2 (g) MgCO3 (s)

MgO + CO2 Mg (s) + C (s) + 3/2 O2

Mg (s) + C (s) + 3/2 O2 MgCO3

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Apply Hess’ Law----

Really ΔHf (products) - ΔHf (reactants)

Calculate ΔHrxn based on enthalpy of formation (ΔHf)

aA + bB cC + dDΔH° =[c (ΔHf°)C + d(ΔHf°)D] - [a (ΔHf°)A + b (ΔHf°)B ]

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Example 4

Calculate the ΔH° for the combustion of 1 mole of pentaborane (B5H9) reacting with excess oxygen

2B5H9 + 12O2 5B2O3 + 9H2O ΔH° = ?

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Example 5

Isopropyl alcohol (rubbing alcohol) undergoes a combustion reaction

2(CH3)2CHOH + 9O2 6CO2 + 8H2O

ΔH° = -4011 kJ/mol

Calculate the standard enthalpy of formation for isopropyl alcohol.

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Homework

pp. 253 #65-67, 72