Q22.52P - Chemistry: Molecular Nature Of Matter And Change

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 changing  T.) 

(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?

Step-by-Step Solution

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 1: 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.

Step 2: 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,

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

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 $∆{\mathrm{G}}_{\mathrm{rxn}}^{°}=-141.60 \mathrm{kJ}$.

Step 3: 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.

Step 4: 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 $773 \mathrm{K}\left(500°\mathrm{C}\right)$ :

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

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

Step 5: 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

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

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

Step 6: 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  $1055.993 \mathrm{K},$.