The hydrogenation of styrene to produce ethylbenzene reaches equilibrium in the
ID: 523657 • Letter: T
Question
The hydrogenation of styrene to produce ethylbenzene reaches equilibrium in the reaction below.
C6H5CH=CH2 + H2 C6H5C2H5
The system initially contains 3 mol of H2 for each mole of styrene. Assume ideal gases. For styrene, Gf,298.15 = 213.18kJ/mol and Hf,298.15 = 147.36kJ/mol.
a) Calculate the standard Gibb's energy of the reaction at 298.15 K. (kJ/mol)
b) Calculate the standard enthalpy of the reaction at 298.15 K. (kJ/mol)
c) Calculate KaR at the reference temperature.
d) Calculate Ka at 600 °C using the shortcut van't Hoff equation.
e) Assuming the system initially contains 3 mol of H2 for each mole of styrene, derive an algebraic expression for the total moles at reaction equilibrium using the reaction coordinate. Use E to represent the reaction coordinate.
f) Develop an algebraic expression for the mole fraction of styrene at equilibrium. Use E for the reaction coordinate.
g) Develop an algebraic expression for the mole fraction of hydrogen at equilibrium. Use E for the reaction coordinate.
h) Develop an algebraic expression for the mole fraction of ethylbenzene at equilibrium. Use E for the reaction coordinate.
j) Solve for the numerical value of the reaction coordinate using MATLAB or Excel if the pressure is 1 bar.
k) Using the reaction coordinate calculated above, calculate the mole fractions of each species at reaction equilibrium. (Ystyrene, Yhydrogen, Yethylbenzene)
l) If the pressure was raised to 11 bar, calculate the reaction coordinate at 600°C.
m) Using the reaction coordinate calculated at 11 bar, calculate the mole fractions of each species at reaction equilibrium. (Ystyrene, Yhydrogen, Yethylbenzene)
Explanation / Answer
for the reaction, C6H5CH=CH2 + H2 C6H5C2H5
gibss free energy change= 1*gibbs free energy change of ethyl benzene- { 1* gibbs free energy change of styrene +1* gibbs free energy change of H2}
where 1, 1, and 1 are coefficients of ethyl benzenem styrene, and H2 respectively.
since gibss free energy change of H2= 0 and that of ethyl benzne= 19.7 Kj/mole
gibbs free energy change of reaction = 19.7-213.18=-193.48 Kj/mole
2. enthapy change similarly= -13.1-147.36= -160.46 Kj/mole
3. deltaG0= -RTlnK, lnK= -deltaG/RT= 193.48*1000/(298*8.314) , K= 8.225*1033
4. ln (K2/K1)= (deltaH/R)*(1/T1-1/T2)
T1= 298 K, T2= 600+273= 873K
ln(K2/K1)= (-160.46*1000/8.314)*(1/298-1/873)
K2/K1= 3*10-19, K2= 8.225*1033* 3*10-19 = 2.45*1015
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