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5 kjmol 3944 kjmol gibbs free energy g the standard

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Unformatted text preview: reemole of the compound is formed from its elements in its their their standard states. Standard states Standard free energy of formation of its zero any element in its stable form is zero C ( graphite) + O2 ( g ) → CO2 ( g ) ∆H ° f ∆G ° f 0 0 0 0 -393.5 (KJ/mol) -394.4 (KJ/mol) Gibbs Free Energy (G) The standard free-energy of reaction ( ∆G o ) : standard freerxn 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 )] rxn f f f f ∆G o = ∑ n∆G o (products) − ∑ m∆G o (reactants) rxn f 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 ) ∆G ° f 0 0 2 NH 3 ( g ) -16.66 (KJ/mol) ∆G ° =2mol × (-16.66 KJ/mol)-[0KJ/mol+3mol × 0KJ/mol] RXN = - 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) R = + 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 ° less (b) Calculate the standard free energy change for the reaction, Is the prediction from part (a) correct? C3 H 8 ( g ) + 5O2 ( g ) ∆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 a) at 25°C: NH4NO3(s) → NH4NO3(aq), given: NH NH4NO3(s): ∆Hf° = –365.6 kJ/mole ; S° = 151.1 J/mo...
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