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
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
Standard Conditions
Represented by a “ ° “ beside ΔH or ΔE, etc.
25°C
1 atm
Methods for determining ΔH
1) Calorimetry
2) Application of Hess’ Law
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
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
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
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
Methods for determining ΔH
1) Calorimetry
2) Application of Hess’ Law
3) Enthalpies of Formation
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
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
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 ]
Example 4
Calculate the ΔH° for the combustion of 1 mole of pentaborane (B5H9) reacting with excess oxygen
2B5H9 + 12O2 5B2O3 + 9H2O ΔH° = ?
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.
Homework
pp. 253 #65-67, 72