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of the compound is formed from its elements in
their standard states.
Standard free energy of formation of
any element in its stable form is zero
C ( graphite) + O2 ( g ) → CO2 ( g )
0 -393.5 (KJ/mol)
-394.4 (KJ/mol) Gibbs Free Energy (G)
The standard free-energy of reaction ( ∆G o ) :
the free-energy change for a reaction when
freeit occurs under standard states.
standard aA + bB cC + dD
∆G o = [c ⋅ ∆G o (C ) + d ⋅ ∆G o ( D)] − [a ⋅ ∆G o ( A) + b ⋅ ∆G o ( B )]
f ∆G o = ∑ n∆G o (products) − ∑ m∆G o (reactants)
f ∆rG = ∆r H − T ∆r S At room temperature Sample
Sample exercise: (a) By using the data from Appendix C
to calculate the standard free energy change for the
following reaction at room temperature N 2 ( g ) + 3H 2 ( g )
f 0 0 2 NH 3 ( g )
-16.66 (KJ/mol) ∆G ° =2mol × (-16.66 KJ/mol)-[0KJ/mol+3mol × 0KJ/mol]
= - 33.32 KJ (b) What is the ∆G for the reverse of the above reaction 2 NH 3 ( g ) N 2 ( g ) + 3H 2 ( g ) ∆G ° XN =[0KJ/mol+3mol × 0KJ/mol]-2mol × (-16.66 KJ/mol)
= + 33.32 KJ (a)
(a) Without using the data fron Appendix C, Predict
whether the ∆G° for the following reaction is more
negative or less negative than ∆H °, when given:
C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2O (l ) ∆H ° = −2220 KJ ∆rG = ∆r H − T ∆r S ∆ r S = S products − Sreac tan ts
∆r S < 0 the ∆G° for the reaction is less negative than ∆H °
(b) Calculate the standard free energy change for the
reaction, Is the prediction from part (a) correct? C3 H 8 ( g ) + 5O2 ( g )
f -23.47 0 3CO2 ( g ) + 4 H 2O(l )
-394.4 -237.13 ∆G ° = [3 × (−394.4)+4 × (−237.13)]-[(-23.47)+5 × (0)]
= − 2108 KJ the G° is less negative than H °
less Determine the effect of T on spontaneity of reaction!
a) Calculate the standard free energy change for reaction
at 25°C: NH4NO3(s) → NH4NO3(aq), given:
NH4NO3(s): ∆Hf° = –365.6 kJ/mole ; S° = 151.1 J/mo...
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This document was uploaded on 02/26/2014.
- Fall '14