prelim_2_2006_solutions

prelim_2_2006_solutions - MS&E 303 – Fall 2006...

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Unformatted text preview: MS&E 303 – Fall 2006 Prelim #2 Solutions 1. (10 pts) (i) Sketch the activity versus composition for a single phase system (solid) that exhibits negative deviation from ideality. (ii) Explain, with reference to the drawing, the meaning of Raoult’s and Henry’s laws. (iii) Explain how the activity coefficient is related to Henry’s constant. See hand sketches for i and part of ii. (ii) Raoult’s Law requires that the activity of the majority component in an alloy approach the ideal solution case (equals its molar fraction) as you approach 100%. Henry gives the other limit, stating only that the activity is proportional to the concentration in the dilute limit. (iii) As the activity coefficient is just the ratio of the activity to the mole fraction, in the dilute limit the activity coefficient and Henry’s constant are equal. lim X A → γ A = γ A (0) = γ o A = k A 2. (20 pts) Explain why, for a chemical reaction, that K A = exp •- Δ G o RT ‚ . (Basically derive this from some reasonable low level without all the math details. It should start at roughly the partial molar free energy level.) In your answer, precisely define and explain the meaning of Δ G o . For a reaction to be in equilibrium, the net change in G for a shift of reactants from the left to the right must be precisely zero. As the shift involves addition and removal of atoms (moles) from a system, this then states Δ G = 0 = X products n i G i- X reactants n j G j The partial molar free energy can be written as a sum of a reference state G o i and a correction term RT ln a i . Putting the two together, we get Δ G = 0 = X products n i G o i- X reactants n j G o j + RT X products n i ln a i- X reactants n j ln a j = Δ G o ( T ) + RT ln K A The standard state free energy of the reaction Δ G o ( T ) is the conceptual free energy change had the reaction occured with all components in their standard states. The standard state may be defined as we choose, though the default is a pure material in the equilibrium phase at the temperature of interest and one atmosphere of pressure. 3. (10 pts) At what partial pressure of oxygen would Fe 3 O 4 spontaneously decompose at 1000 K. The standard state free energy of formation for Fe 3 O 4 is Δ G o ( T ) =- 1 , 102 , 200 + 307 . 4 T . Direct application of equilbrium reaction. At 1000 K, the standard state free energy change is -794,800 and K A = 3 . 278 × 10 41 . The reaction is 3Fe( s ) + 2O 2 ( g ) ⇐⇒ Fe 3 O 4 ( s ) K A = a ( s ) Fe 3 O 4 ( a ( s ) Fe ) 3 ( a ( g ) O 2 ) 2 = 1 p 2 O 2 p O 2 = 1 √ K A = 1 . 747 × 10- 21 atm If the oxygen partial pressure is below this value, then Fe 3 O 4 will decompose to try to push it up. Above this level, Fe will oxidize to pull down the p O 2 . We can replace the a (...
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prelim_2_2006_solutions - MS&E 303 – Fall 2006...

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