The molar free energy of liquid Na-K (at 384 K) alloy is plotted below: a. What

Question

The molar free energy of liquid Na-K (at 384 K) alloy is plotted below:

a. What is the free energy of mixing at yK = 0.6? Show your work both graphically

and numerically.

b. What are the approximate activities of Na and K?

c. Assume the entropy of mixing to be ideal—what does this mean regarding the

enthalpy of mixing?

d. Using the same assumption from (c), what is the enthalpy of mixing at yK = 0.7?

Remember the enthalpy of mixing is composition-dependent. Explain why this is

an ideal solution, regular solution, or neither.

e. If 4 kmol of a K0.2Na0.8 and 2 kmol of a K0.8Na0.2 solution are mixed at constant

temperature and pressure, will they mix spontaneously or will work be required?

Explanation / Answer

a) If Yk = 0.6 , YNa = 0.4
extreme R.H.S of the graph YNa=1 and extreme L.H.S of the graph YK = 1
from the graph  mix G = -1800 J/mole
mix G = RT (YNa lnYNa + YK ln Yk) = 8.314 * 384 K (0.4 ln (0.4) + 0.6 * ln (0.6) ) = -2148.64 J/mol

b) For a binary solution at constant temperature and pressure
   YK d ln(K) + YNa d ln(Na) = 0

c) If we assume the entropy of mixing to be ideal, then molecules of ideal gas are spread out enough
that they do not interact with one another when mixed, which implies that no heat is absorbed or
  produced and results in a mixH of zero.

d) YK = 0.7 , YNa = 0.3
   From the graph = -1700 J/mol = mix G
  mix S = -4.427083 J/mol K = -R (YNa lnYNa + YK ln Yk)
   mix G =  mix H - Tmix S
   -1700 J/mol = mix H - (384) ( -4.427083)
     mix H = -3399.9 J/mol
calculating mix H from the formula,
mix G = RT (YNa lnYNa + YK ln Yk) = 8.314 *384 (0.7 ln(0.7) + 0.3ln(0.3)) = -1950.23 J/mol
     mix S = -R (YNa lnYNa + YK ln Yk) = -5.07872 J/mol K
   -1950.23  =  mix H - 384 * (-5.07872)
   mix H = -3900.45 J/mol
The previous assumption is only for mixing of two ideal gases where enthalpy of mixing will be 0
This is just a regular solution as you can observe the value for enthalpy of mixing is not 0.

e) From the graph, for 4kmol mixture of K0.2Na0.8   mix G= -1200 J/mol
   for 2 kmol mixture of   mixG = -1650 J/mol
   Since mixG is negative the mixing process is spontaneous.
  

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