free energy pressure
Challenge Assignment?! Need Any Help on Enthalpy and Free Energy Change?! Please Help Me Understand!?
For the reaction
2NaOH (aq) + H2(g) –> 2Na(s) + 2H2O at 298K
(a) Calculate the change in H rxn
(b) Calculate the change in S system
(c) Calculate the change in G rxn
(d) Explain why the reaction is favored by enthalpy, entropy favored, both or neither?
(e) Which direction does the equilibrium shift with the following changes
(i) increased pressure
(ii) increased temperature
(iii) adding hydroxide
CONSTANTS
S J/(mol K) : Na(s)=51.21 H2O(l)=69.96 NaOH(aq)=48.1 H2(g)=130.7
H(formation) kJ/mol: Na(s)=0 H2O(l)=-285.83 NaOH(aq)=-469.15 H2(g)=0
2NaOH (aq) + H2(g) –> 2Na(s) + 2H2O
(a) Calculate the change in H rxn, (products – reactants):
( H2O @ -285.83 ) – (NaOH @ -938.30) = +652.47kJ
(b) Calculate the change in S system. (products – reactants):
(Na102.42 & H2O139.92) – (NaOH@96.2 & H2@130.7) = 15.44Joules
(c) Calculate the change in G rxn
dG = dH -TdS
dG = +652.47kJ – (298K)(15.44Joules)
dG = +652.47kJ – (298K)(0.01544kJ)
dG = +652.47kJ – 4.60kJ
dG = +647.87 kJ
not favored by enthalpy (dH = +), more energetic is less stable & is not helping dG to become negative
favored by entropy (dS = +) having a “+” dS leads to a negative dG , by doing a “-tdS” …
(e) Which direction does the equilibrium shift with the following changes
(i) increased pressure shifts it to the right, (less gases on the right)
(ii) increased temperature shifts it to the right, endothermic reactions use heat as a starting material
(iii) adding hydroxide shifts it to the ritght to use up the added NaOH
check the math, …I have to run to class
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