Question

Ethanol can be synthesized by direct hydration of ethylene: C2H4(g) + H2O(g) ----> C2H5OH(g). at t = 300degC and P = 60 atm. Calculate the yield of the rxn for conditions as above and the equimolar (1:1) input mixture. Use following data: DfH298 (KJ mol-1): C2H4 = 52.4; H20 = -241.8; C2H5OH = -234.8. DfG298 (Kj mol-1): C2H4 = 68.4; H20 = -228.6; C2H5OH = -167.9

Answer 1

I am assuming that when you are saying yield of reaction, you are really talking about the reaction quotient Q, since all thermodynamic parameters are given. If we need to calculate the actual percentage yield of the reaction, we may require the actual masses of the reactants and the products that we started with. I will show how to calculate the reaction quotient Q which is a measure of the extent of reaction.

The reaction is

C₂H₄ (g) + H₂O (g) -----à C₂H₅OH

The enthalpy of formation at 298K is

ΔH⁰ = [ΣΔH_f]_products – [ΣΔH_f]_reactants = (-234.8 kJ/mol) – {(52.4 kJ/mol) + (-241.8 kJ/mol)} = {(-234.8) – (-189.4)} kJ/mol = -45.4 kJ/mol

The standard Gibb’s free energy is

ΔG⁰ = [ΣΔG_f]_products – [ΣΔG_f]_reactants = (-167.9 kJ/mol) –{(68.4 kJ/mol) + (-228.6 kJ/mol)} = {(-167.9) – (-160.2)} kJ/mol = -7.7 kJ/mol

With these two values, we need to calculate the standard entropy change. We know that

ΔG⁰ = ΔH⁰ – TΔS⁰ where T = 298 K is the standard temperature. Therefore,

ΔS⁰ = (ΔG⁰ – ΔH⁰)/T = {(-7.7) – (-45.4)} kJ/mol/298 K = 126.5 J/mol.K

Now, we know that only ΔH⁰ and ΔS⁰ are independent of temperature whereas ΔG is temperature dependent. Hence, at 300°C, i.e, T = (273 + 300) = 573 K, we have

ΔG = ΔH – TΔS (we omit the 0 superscript, since this is not the standard state; however the values for ΔH and ΔS remain the same as before)

or, ΔG = (-45.4 kJ.mol) – (573 K)(126.5 J/mol. K) = {(-45.4) – (72.48)} kJ/mol = -117.88 kJ/mol

We also know

ΔG = ΔG⁰ + RTlnQ where Q is the reaction quotient.

Therefore,

(-117.88 kJ/mol) = (-7.7 kJ/mol) + (8.314 J/mol.K)(573 K) lnQ

or, -110.18 kJ/mol = 4.76 kJ/mol.lnQ

or lnQ = -110.18/4.76 = -23.15

or, Q = e^-23.15 = 8.86*10^-11 (ans)