Question

Acrylonitrile is the starting material used in the manufacture of acrylic fibers (U.S. annual production capacityis more than 2 million pounds). Three industrial processes for the production of acrylonitrile are given below. Using data from Appendix 4, calculate $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{,}$$\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$and $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$for each process. For part a, assume that T = 25°C; for part b, T = 70°C; and for part c, T = 700°C. Assume that $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$do not depend on temperature.

1. 
   

b. $\mathit{H}\mathit{C}\mathbf{\equiv }\mathit{C}\mathit{H}\left(g\right)\mathbf{+}\mathit{H}\mathit{C}\mathit{N}\left(g\right)\underset{{\mathbf{70}}^{\mathbf{o}}\mathbf{C}\mathbf{-}{\mathbf{90}}^{\mathbf{o}}\mathbf{C}}{\overset{\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}\mathbf{\times }\mathbf{H}\mathbf{C}\mathbf{l}}{\mathbf{\to }}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{=}\mathit{C}\mathit{H}\mathit{C}\mathit{N}\left(g\right)$

c. $\mathbf{4}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{=}\mathit{C}\mathit{H}\mathit{C}{\mathit{H}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{6}\mathit{N}\mathit{O}\left(g\right)\underset{\mathbf{A}\mathbf{g}}{\overset{{\mathbf{700}}^{\mathbf{o}}\mathbf{C}}{\mathbf{\to }}}\mathbf{4}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{=}\mathit{C}\mathit{H}\mathit{C}\mathit{N}\left(g\right)\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(g\right)\mathbf{+}{\mathit{N}}_{\mathbf{2}}\left(g\right)$

Short answer

a) $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}-183KJ/mol$$\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}-183KJ/mol$

$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}-100J/K·mol$and

localid="1658673363813" $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}-153\text{KJ/mol}$

b) localid="1658673374476" $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}-177KJ/mol$

$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}-129J/K·mol$and

localid="1658673462798" $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}-133KJ/mol$

c) localid="1658673688988" $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}-1336KJ/mol$

$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}88\text{J/k·mol}$and

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}-1422\text{KJ/mol}$

Step-by-step solution

Step 1: Introduction to the Concept

The standard Gibbs free energy change, standard enthalpy change, and standard entropy change have the following relationship:

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathbf{\times }\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}$Gibbs free energy

$\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}$Enthalpy change

$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}$Change in entropy

$\mathit{T}\mathbf{=}$temperature

Step 2: Determination of ΔSo, ΔHoand  role="math" localid="1658675858133" ΔGo

a)

The given reaction is,

$C_2{H}_{4}O\left(g\right)+HCN\left(g\right)\to C{H}_{2}CHCN\left(g\right)+H_{2}O\left(l\right)$

$\Delta H^o=(185KJ/mol+(-286KJ/mol\left)\right)-(-53kJ/mol+135.1KJ/mol)$

$\Delta H^o=-183KJ/mol$

$\Delta S^o=(274J/K·mol+70J/K·mol)-(242J/K·mol+202J/K·mol)$

$\Delta S^o=-100J/K·mol$

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathbf{\times }\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$

$\Delta G^o=-183KJ/mol-(298\text{K}\times \text{0.100KJ/K·mol)}$

$\Delta G^o=-153\text{KJ/mol}$

Therefore$\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$is$\mathbf{-}\mathbf{183}\mathbf{}\mathit{K}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$,$\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$is$\mathbf{-}\mathbf{100}\mathbf{}\mathbf{}\mathit{J}\mathbf{/}\mathbf{}\mathit{k}\mathbf{·}\mathbf{}\mathit{m}\mathit{o}\mathit{l}$

Step 3: Determination of ΔSo, ΔHo and ΔGo

b)

The given reaction is,

$HCCH\left(g\right)+HCN\left(g\right)\underset{{70}^{o}C-{90}^{o}C}{\overset{CaC{l}_2·HCl}{\to }}C{H}_{2}CHCN\left(g\right)$

$\Delta H^o=(185KJ/mol)-(-135.1KJ/mol+227KJ/mol)$

$\Delta H^o=-177KJ/mol$

$\Delta S^o=(274J/K·mol)-(201J/K.mol+2020J/Kmol)$

$\Delta S^o=-129\text{J/K·mol}$

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathbf{\times }\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$

$\Delta G^o=-177\text{KJ/mol-}\left(343\text{K}\times -\text{0.129KJ/K·mol}\right)$

$\Delta G^o=-133KJ/mol$

Therefore $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$is $\mathbf{-}\mathbf{177}\mathbf{}\mathit{K}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathbf{,}\mathbf{}\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$is $\mathbf{-}\mathbf{129}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}\mathbf{·}\mathit{m}\mathit{o}\mathit{l}$and $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$is $\mathbf{-}\mathbf{133}\mathbf{}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$

Step 4: Determination of ΔSo, ΔHoand ΔGo

c)

The given reaction is,

$4C{H}_{2}CHC{H}_3\left(g\right)+6NO\left(g\right)\underset{Ag}{\overset{{700}^{o}C}{\to }}4C{H}_{2}CHCN\left(g\right)+6H_{2}O\left(g\right)+N_2\left(g\right)$

$\Delta H^o=\left(6\times -242KJ/mol+4\times 185KJ/mol\right)-\left(4\times 20.9KJ/mol+6\times 90KJ/mol\right)$

$\Delta H^o=-1336KJ/mol$

$\Delta S^o=\left(192J/K·mol+6\times 189J/K·mol+4\times 274J/K·mol\right)-\left(6\times 211J/K·mol+4\times 266.9J/K·mol\right)$

$\Delta S^o=88J/K·mol$

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathbf{\times }\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$

$\Delta G^o=-1336KJ/mol-\left(973\text{K}\times -\text{0.088}KJ/K·mol\right)$

$\Delta G^o=-1422KJ/mol$

Therefore $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$is $\mathbf{-}\mathbf{1336}\mathbf{}\mathit{K}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{,}\mathbf{}\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$is $\mathbf{88}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}\mathbf{·}\mathit{m}\mathit{o}\mathit{l}$and $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$is $\mathbf{-}\mathbf{1422}\mathbf{}\mathit{K}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}$