Question

For the reaction of ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{to SO}}}_{\mathbf{\text{3}}}$ at standard conditions,

(a) Calculate${\mathbf{\Delta G}}^{\mathbf{0}}\mathbf{}\mathbf{at}\mathbf{}{\mathbf{25}}^{\mathbf{0}}\mathbf{C}\mathbf{.}$ Is the reaction spontaneous?

(b) Why is the reaction not performed at ${\mathbf{25}}^{\mathbf{0}}\mathbf{C}$?

(c) Is the reaction spontaneous at${\mathbf{500}}^{\mathbf{0}}\mathbf{C}$ ? (Assume${\mathbf{\Delta H}}^{\mathbf{0}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\Delta S}}^{\mathbf{0}}$ are constant with changingT.)

(d) Compare$\mathbf{K}\mathbf{}\mathbf{at}\mathbf{}{\mathbf{500}}^{\mathbf{0}}\mathbf{C}\mathbf{}\mathbf{and}\mathbf{}\mathbf{at}\mathbf{}{\mathbf{25}}^{\mathbf{0}}\mathbf{C}$

(e) What is the highest T at which the reaction is spontaneous?

Short answer

a) $\Delta {G^0}_{Txn}=-141.60\mathrm{kJ}$ since the data-custom-editor="chemistry" $∆{\mathrm{G}}_{\mathrm{rxn}}^{°}$ is negative, the reaction is spontaneous.

b) At this temperature, the rate of the reaction is very low, so the ${\text{SO}}_{\text{3}}$produced very slow and in small, not significant amounts.

c) ${\mathrm{\Delta G}}_{500}=-53.2306\mathrm{kJ}=-53.2\mathrm{kJ}$, the Gibb's energy is negative, so the reaction is also spontaneous.

d) Thus, at higher temperature less product is formed as the K is lower.

e) The highest temperature at which the reaction is spontaneous, is $1055.993\mathrm{K}$.

Step-by-step solution

Concept Introduction

Sulfur dioxide has the formula ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$, while sulphur trioxide has the formula${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}$. Both of them are known as sulphur oxides. The main point of distinction between both ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{and SO}}}_{\mathbf{\text{3}}}$ is that at ambient temperature,${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$is a gas which is colourless, whereas on the other hand${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}$is a colourless to white crystalline solid.

Calculating∆G° at 25°C

Let us calculate $∆\mathrm{G}$.

a) The balanced reaction of ${\text{SO}}_2$conversion to ${\text{SO}}_{\text{3}}$is

${\text{2SO}}_{\text{2}}{\text{(g) +O}}_{\text{2}}\text{(g)}\to {\text{2SO}}_{\text{3}}\text{(g)}$

Since$∆{\mathrm{G}}_{\mathrm{rxn}}^{°}$is a state function, it depends only on the initial and final values, not the path in between the states.

So,

$∆\mathrm{G}°=∆{\mathrm{G}}_{\mathrm{product}}^{°}-∆{\mathrm{G}}_{\mathrm{reactants}}^{°}$

From the reaction equation above,

$∆{\mathrm{G}}_{\mathrm{rxn}}^{°}=\left[2\times ∆{\mathrm{G}}_{{\mathrm{SO}}_3\left(\mathrm{g}\right)}^{°}\right]-\left[2\times ∆{\mathrm{G}}_{{\mathrm{SO}}_2\left(\mathrm{g}\right)}^{°}+∆{\mathrm{G}}_{{\mathrm{O}}_2\left(\mathrm{g}\right)}^{°}\right]$

Using Appendix B,

role="math" localid="1663331214226" $\begin{array}{rcl}∆{\mathrm{G}}_{\mathrm{rxn}}^{°}& =& \left[2\mathrm{mol}\times -371.00\frac{\mathrm{kJ}}{\mathrm{mol}}\right]-\left[2\mathrm{mol}\times -300.20\frac{\mathrm{kJ}}{\mathrm{mol}}+1\mathrm{mol}\times 0.00\frac{\mathrm{kJ}}{\mathrm{mol}}\right]\\ & =& -141.60\mathrm{kJ}\end{array}$

Since the $\begin{array}{rcl}& & ∆{\mathrm{G}}_{\mathrm{rxn}}^{°}\end{array}$is negative, the reaction is spontaneous at standard conditions$\left(298\mathrm{K}\left({25}^0\mathrm{C}\right)\mathrm{and}1\mathrm{atm}]\right)$ .

Therefore, we get role="math" localid="1663331383883" $∆{\mathrm{G}}_{\mathrm{rxn}}^{°}=-141.60\mathrm{kJ}$.

Explanation

b) The reaction of ${\text{SO}}_{\text{2}}{\text{to SO}}_{\text{3}}$is not performed at data-custom-editor="chemistry" $25°\mathrm{C}$. Because the reaction rate is very slow at this temperature and thus the ${\text{SO}}_{\text{3}}$ is created slowly and in little, non-significant amounts.

Finding if the Reaction is Spontaneous

c) To compute the spontaneity of the reaction at $500°\mathrm{C}\left(773\mathrm{K}\right)$the enthalpy and entropy of the reaction at this temperature must first be recalculated, because the Gibb's energy values given in Appendix B are only valid for $25°\mathrm{C}\left(298\mathrm{K}\right)$.

Since ${\mathrm{\Delta H}}_{\mathrm{rxn}}^{°}\mathrm{and}{\mathrm{\Delta S}}_{\mathrm{rxn}}^{°}$are also state functions,similarly to a), it can be represented as,

Using Appendix B,

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{rxn}}^{°}& =& \left[2\mathrm{mol}·-396.00\frac{\mathrm{kJ}}{\mathrm{mol}}\right]-\left[2\mathrm{mol}·-296.80\frac{\mathrm{kJ}}{\mathrm{mol}}+1\mathrm{mol}·0.00\frac{\mathrm{kJ}}{\mathrm{mol}}\right]\\ ∆{\mathrm{S}}_{\mathrm{rxn}}^{°}& =& =\left[2\mathrm{mol}·256.66\frac{\mathrm{J}}{\mathrm{mol}·\mathrm{K}}\right]-\left[2\mathrm{mol}·248.10\frac{\mathrm{J}}{\mathrm{mol}·\mathrm{K}}+1\mathrm{mol}·205\frac{\mathrm{J}}{\mathrm{mol}·\mathrm{K}}\right]\\ ∆{\mathrm{H}}_{\mathrm{rxn}}^{°}& =& -198.40\mathrm{kJ}∆{\mathrm{S}}_{\mathrm{rxn}}^{°}\\ & =& -187.88\mathrm{J}/\mathrm{K}\end{array}$

Then, $∆{\mathrm{G}}_{\mathrm{Txn}}^{°}$can be recalculated at $773\mathrm{K}\left(500°\mathrm{C}\right)$:

localid="1663332258977" $\begin{array}{rcl}∆{\mathrm{G}}_{500°}& =& ∆{\mathrm{H}}_{\mathrm{rxn}}^{°}-\mathrm{T}∆{\mathrm{S}}_{\mathrm{rxn}}^{°}\\ & =& -198.40\mathrm{kJ}-773\mathrm{K}·-187.88\frac{\mathrm{J}}{\mathrm{K}}·{10}^{-3}\frac{\mathrm{kJ}}{\mathrm{J}}\\ & =& -53.2306\mathrm{kJ}\\ & =& -53.2\mathrm{kJ}\end{array}$

At$500°\mathrm{C}$, Gibb's energy is negative, so the reaction is also spontaneous.

Comparing K at 500°C and at 25°C

d) To compare the equilibrium constant K at different temperatures, express it from

$\begin{array}{rcl}∆{\mathrm{G}}_{\mathrm{rxn}}& =& -\mathrm{R}\times \mathrm{T}\times \mathrm{lnK}\\ \mathrm{lnK}& =& \frac{∆{\mathrm{G}}_{\mathrm{rxn}}}{{\displaystyle -\mathrm{R}\times \mathrm{T}}}\\ \mathrm{K}& =& {\mathrm{e}}^{\mathrm{lnK}}\end{array}$

At $25°\mathrm{C}$, from $∆{\mathrm{G}}_{\mathrm{rxn}}^{°}=-141.60\mathrm{kJ}$, so

$\begin{array}{rcl}{\mathrm{lnK}}_{25}& =& \frac{-141.60\times {10}^3\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\times 298\mathrm{K}}\\ {\mathrm{lnK}}_{25}& =& 57.153\\ {\mathrm{K}}_{25}& =& 6.6257\times {10}^{24}\end{array}$

At $500°\mathrm{C}$from $∆{\mathrm{G}}_{\mathrm{rxn}}^{°}=-53.2306\mathrm{kJ}$so

role="math" localid="1663332726881" $\begin{array}{rcl}{\mathrm{lnK}}_{500}& =& \frac{-53.2306\times {10}^3\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\times 773\mathrm{K}}\\ {\mathrm{lnK}}_{500}& =& 8.2827\\ {\mathrm{K}}_{500}& =& 3.955\times {10}^3\end{array}$

Therefore, at a higher temperature, less product is formed as the K is lower.

Calculating the Highest T

e) The reaction is spontaneous when $∆\mathrm{G}<0$At equilibrium, Gibb's energy is equal to 0. Thus, the reaction is spontaneous at all temperatures, below $∆{\mathrm{G}}_{\mathrm{Txn}}=0$.

Thus,

$\begin{array}{rcl}∆{\mathrm{G}}_{\mathrm{rxn}}& =& ∆{\mathrm{H}}_{\mathrm{rxn}}^{°}-\mathrm{T}\times ∆{\mathrm{S}}_{\mathrm{rxn}}^{°}\\ & =& 0\\ ∆{\mathrm{H}}_{\mathrm{rxn}}^{°}& =& \mathrm{T}\times ∆{\mathrm{S}}_{\mathrm{rxn}}^{°}\\ \mathrm{T}& =& \frac{∆{\mathrm{H}}_{\mathrm{rxn}}^{°}}{∆{\mathrm{S}}_{\mathrm{rxn}}^{°}}\\ & =& \frac{-198.40\times {10}^3\mathrm{J}}{-187.88\frac{\mathrm{J}}{\mathrm{K}}}\\ & =& 1.0559\times {10}^3\mathrm{K}\end{array}$

At all temperatures below $1055.993\mathrm{K},$the reaction is spontaneous.

Therefore, the highest temperature at which the reaction is spontaneous is $1055.993\mathrm{K},$.