91Ó°ÊÓ

Methanol is created by hydrogenation of CO and CO₂ as well as a reversed water—gas shift reaction, from petroleum product (synthesis gas).

a) The reaction given is:

${\text{CO(g)+2H}}_{\text{2}}\text{(g)}→{\text{CH}}_{\text{3}}\text{OH(l)}$

We can compute the ${\text{Δ±á}}_{\text{rxn}}^{\text{°}}{\text{andΔ³§}}_{\text{rxn}}^{\text{°}}$to estimate the reaction spontaneity to show that the reaction is thermodynamically possible.

Both enthalpy and entropy are state functions, which mean they solely depend on the system's starting and ultimate states:

$\begin{array}{l}{\text{Δ±á}}_{\text{ren}}^{\text{°}}{\text{=Δ±á}}_{\text{products}}^{\text{°}}{\text{-Δ±á}}_{\text{reactants}}^{\text{°}}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}{\text{=Δ³§}}_{\text{products}}^{\text{°}}{\text{-Δ³§}}_{\text{reactants}}^{\text{°}}\end{array}$

The enthalpy can be calculated as following:

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{Δ±á}}^{\text{°}}\left({\text{CH}}_{\text{3}}\text{OH(l)}\right)\right]\text{-}\left[{\text{Δ±á}}^{\text{°}}{\text{(CO(g))+2×Δ±á}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{1³¾´Ç±ô×}\left(\text{-238.60}\frac{\text{kJ}}{\text{mol}}\right)\right]\text{-}\left[\text{1³¾´Ç±ô×}\left(\text{-110.50}\frac{\text{kJ}}{\text{mol}}\right)\text{+2³¾´Ç±ô×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=-128.10kJ}\\ \end{array}$

The entropy can be calculated similarly:

$\begin{array}{c}{\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{Δ³§}}^{\text{°}}\left({\text{CH}}_{\text{3}}\text{OH(l)}\right)\right]\text{-}\left[{\text{Δ³§}}^{\text{°}}{\text{(CO(g))+2×Δ³§}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{1³¾´Ç±ô×}\left(\text{197.50}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\right]\text{-}\left[\text{1³¾´Ç±ô×}\left(\text{127.00}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\text{+2³¾´Ç±ô×}\left(\text{130.60}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\right]\\ \end{array}$

${\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=-190.70}\frac{\text{J}}{\text{K}}\text{<0}$

Since, the entropy is negative$\mathrm{Δ}{S}_{rxn}^{°}<0$, the reaction is spontaneous and exothermic (heat is released,$\mathrm{Δ}{H}_{rxn}^{°}<0$).

b) To test the spontaneity dependency on temperature, we must determine the temperature at which the reaction is spontaneous, assuming that the enthalpy and entropy do not change considerably with temperature. In the case of a spontaneous reaction,

${\text{Δ³Ò}}_{\text{Txn}}\text{<0}$

Thus,

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}{\text{=Δ±á}}_{\text{rxn}}{\text{-T×Δ³§}}_{\text{rxn}}\text{<0}\\ {\text{Δ±á}}_{\text{rxn}}{\text{-T×Δ³§}}_{\text{Txn}}\text{<0}\\ {\text{T×Δ³§}}_{\text{rxn}}{\text{<Δ±á}}_{\text{rxn}}\\ \text{T<}\frac{{\text{Δ±á}}_{\text{rxn}}}{{\text{Δ³§}}_{\text{rxn}}}\end{array}$

By using the values obtained in a), we can estimate the temperature:

$\begin{array}{l}\text{T<}\frac{{\text{-128.10×10}}^{\text{3}}\frac{\text{J}}{\text{mol}}}{\text{-190.70}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}}\\ \text{T<671.735K}\end{array}$

At temperature below 671.735 K the reaction is spontaneous.

c) At the specified temperature, all components are in a gaseous (vapour) state in the methanol oxidation reaction:

${\text{2CH}}_{\text{3}}{\text{OH(g)+O}}_{\text{2}}\text{(g)}→{\text{2CH}}_{\text{2}}{\text{O(g)+2H}}_{\text{2}}\text{O(g)}$

The ${\text{Δ³Ò}}_{\text{Txn}}^{\text{°}}$cannot be calculated directly, since the values given in Appendix B are defined for temperature. Thus, the enthalpy and entropy should be separately calculated first.

The enthalpy can be calculated as following:

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{2Δ±á}}^{\text{°}}\left({\text{C}}_{\text{2}}\text{O(g)}\right){\text{+2Δ±á}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right)\right]\text{-}\left[{\text{2Δ±á}}^{\text{°}}\left({\text{CH}}_{\text{3}}\text{OH(g)}\right){\text{+2×Δ±á}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{2³¾´Ç±ô×}\left(\text{-116.00}\frac{\text{kJ}}{\text{mol}}\right)\text{+2³¾´Ç±ô×}\left(\text{-241.826}\frac{\text{kJ}}{\text{mol}}\right)\right]\text{-}\left[\text{2³¾´Ç±ô×}\left(\text{-201.20}\frac{\text{kJ}}{\text{mol}}\right)\text{+1³¾´Ç±ô×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=-313.252kJ}\end{array}$

The entropy can be calculated similarly:

$\begin{array}{c}{\text{Δ³§}}_{\text{Txn}}^{\text{°}}\text{=}\left[{\text{2Δ³§}}^{\text{°}}\left({\text{CH}}_{\text{2}}\text{O(g)}\right){\text{+2Δ³§}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right)\right]\text{-}\left[{\text{2Δ³§}}^{\text{°}}\left({\text{CH}}_{\text{3}}\text{OH(g)}\right){\text{+2×Δ³§}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{2³¾´Ç±ô×}\left(\text{219.00}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\text{+2³¾´Ç±ô×}\left(\text{188.72}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\right]\text{-}\left[\text{2³¾´Ç±ô×}\left(\text{238.00}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\text{+1³¾´Ç±ô×}\left(\text{205.00}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right)\right]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=134.44}\frac{\text{J}}{\text{K}}\end{array}$

Finally, Gibb's free energy change can be calculated at ${\text{100}}^{\text{°}}\text{C(373K)}$:

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}{\text{=Δ±á}}_{\text{rxn}}^{\text{°}}{\text{-T×Δ³§}}_{\text{rxn}}^{\text{°}}\\ {\text{Δ³Ò}}_{\text{rxn}}\text{=-313.252}\frac{\text{kJ}}{\text{mol}}\text{-373°­Ã—}\left(\text{134.44}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\right){\text{×10}}^{\text{-3}}\frac{\text{kJ}}{\text{J}}\\ {\text{Δ³Ò}}_{\text{rxn}}\text{=-363.398kJ}\end{array}$

Therefore, at${\text{100}}^{\text{°}}{\text{C(373K)Δ³Ò}}_{\text{rxn}}\text{=-363.398kJ/mol}$