Question 2.1: Calculate the equilibrium composition of a mixture of the fo...

The basis is 100 mol of initial gas. To calculate the equilibrium composition, one must know the equilibrium constant for the reaction at 527 K. From the values of \Delta G_{f,i}^{0} and \Delta H_{f,i}^{0} at 298.16 K and equations (2.2.5) and (2.2.6):

\Delta G_{298}^{0}=\left ( 1 \right )\left ( -168.696 \right )+\left ( -1 \right )\left ( 68.156 \right )+\left ( -1 \right )\left ( -228.705 \right )=-8.147 KJ/mol
\Delta H_{298}^{0}=\left ( 1 \right )\left ( -235.421 \right )+\left ( -1 \right )\left ( 52.308 \right )+\left ( -1 \right )\left ( -241.942 \right )=-45.787 KJ/mol

The equilibrium constant at 298.16 K may be determined from equation (2.4.7):

\Delta G^{0}=-RT\ln K_{a}
or
\ln K_{a}=-\frac{-8147}{8.31\left ( 298.16 \right )}=3.29
The equilibrium constant at 527Kmay be determined using equation (2.5.3):
\left [ \frac{\partial \ln K_{a}}{\partial \left ( 1/T \right )} \right ]_{P}=-\frac{\Delta H^{0}}{R}
If one assumes that \Delta H^{0}is independent of temperature,this equation may be integrated to obtain
\ln K_{a,2}-\ln K_{a,1}=\frac{\Delta H^{0}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )
For our case,
\ln K_{a,2}-3.29=\frac{-45.787}{8.31}\left ( \frac{1}{289.16}-\frac{1}{527} \right )=-8.02
or
K_{a,2}=8.83\times 10^{-3} at 527K
Because the standard states are the pure materials at unit fugacity, equation (2.6.5) may be rewritten as
K_{a}=\frac{Y_{C_{2}H_{5}OH}}{Y_{H_{2}O}Y_{c_{2}H_{4}}}\frac{\left ( f/P \right )_{C_{2}H_{5}OH}}{\left ( f/P \right )_{H_{2}O}\left ( f/P \right )_{C_{2}H_{4}}}\frac{1}{P}       (A)
The fugacity coefficients (f ∕P) for the various species may be determined from a corresponding states chart if one knows the reduced temperature and pressure corresponding to the species in question. Therefore:

From the stoichiometry of the reaction it is possible to determine the mole numbers of the various species in terms of the extent of reaction and their initial mole numbers:
n_{i}=n_{i,0}+\nu _{i}\xi

The various mole fractions are readily determined from this table. Note that the upper limit on \xi is 25.0. Substitution of numerical values and expressions for the various mole fractions into equation (A) gives

8.83\times 10^{-3}=\left [ \frac{\frac{\xi }{100.0-\xi }}{\left ( \frac{60.0-\xi }{100.0-\xi } \right )\left ( \frac{25.0-\xi }{100.0-\xi } \right )} \right ]\times \left \{ \left [ \frac{0.280}{0190\left ( 0.885 \right )} \right ] \frac{1}{264.2}\right \}
or
\frac{\xi \left ( 100.0-\xi \right )}{\left ( 60.0-\xi \right )\left ( 25.0-\xi \right )} =8.83\times 10^{-3}\left ( 264.2 \right )\left [ \frac{0.190\left ( 0.885 \right )}{0.=280} \right ]=1.404

This equation is quadratic in \xi. The solution is \xi =10.9. On the basis of 100 mol of starting material, the equilibrium composition is then as follows: