91Ó°ÊÓ

Magnesium oxide(MgO), often known as magnesia, is a hygroscopic white solid mineral that occurs naturally as periclase and is a magnesium source. It has a high melting point of ${\text{2,852}}^{\text{°}}\text{C}$${\text{(5,166}}^{\text{°}}\text{F;3,125)K}$.

MgCO₃decomposes to magnesium oxide and carbon dioxide at high temperatures. In the manufacture of magnesium oxide, this step is critical. Calcination is the term for this procedure.

Let us write and balance the equation for the decomposition ofMgCO₃

_{${\text{MgCO}}_{\text{3}}\text{(s)}→{\text{MgO(s)+CO}}_{\text{2}}\text{(g)}$}

This is the balanced equation for magnesite decomposition.

(b)First, solve for ${\text{Δ±á}}^{\text{°}}$using the enthalpy constants found in Appendix B. $\begin{array}{c}{\text{Δ±á}}^{\text{°}}{\text{=n}}_{\text{p}}{\text{Δ±á}}_{\text{f}}^{\text{°}}{\text{(product)-n}}_{\text{r}}{\text{Δ±á}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ±á}}_{\text{f}}^{\text{°}}{\text{(MgO(s))+nΔ±á}}_{\text{f}}^{\text{°}}\left({\text{CO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ±á}}_{\text{f}}^{\text{°}}\left({\text{MgCO}}_{\text{3}}\text{(s)}\right)\right]\\ \text{=([1(-601.2)+1(-393.5)]-[1(-1112)])kJ}\\ {\text{Δ±á}}^{\text{°}}\text{=117.3kJ}\end{array}$

Next, solve for ${\text{Δ³§}}^{\text{°}}$using the entropy constants found in Appendix B.

$\begin{array}{c}{\text{Δ³§}}^{\text{°}}{\text{=n}}_{\text{p}}{\text{S}}^{\text{°}}{\text{(product)-n}}_{\text{r}}{\text{S}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nS}}^{\text{°}}{\text{(MgO(s))+nS}}^{\text{°}}\left({\text{CO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nS}}^{\text{°}}\left({\text{MgCO}}_{\text{3}}\text{(s)}\right)\right]\\ \text{=([1(26.9)+1(213.7)]-[1(65.86)])J/K}\\ {\text{Δ³§}}^{\text{°}}\text{=174.74J/K}\end{array}$

Then solve for ${\text{Δ³Ò}}^{\text{°}}\text{.}$

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ \text{=117.3°ì´³Ã—}\frac{\text{1000J}}{\text{1kJ}}\text{-(298K)(174.4J/K)}\\ {\text{Δ³Ò}}^{\text{°}}\text{=65328.8J}\\ \text{=65.3kJ}\end{array}$

Therefore,${\text{Δ³Ò}}^{\text{°}}\text{at298K}$ is 65.3kJ.

c) To find the minimum temperature at which the reaction is spontaneous, we need to first solve for at equilibrium ${\text{Δ³Ò}}^{\text{°}}\text{=0.}$

$\begin{array}{l}{\text{Δ³Ò}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ {\text{0=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ {\text{TΔ³§}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}\\ \text{T=}\frac{{\text{Δ±á}}^{\text{°}}}{{\text{Δ³§}}^{\text{°}}}\\ \text{=}\frac{\text{117.3°ì´³Ã—}\frac{\text{1000J}}{\text{1kJ}}}{\text{174.4J/K}}\\ \text{T=674.1K}\end{array}$

Therefore, the temperature at where the reaction is spontaneous is at$\text{T>674.1K.}$

d) Let us calculate the equilibrium.

We know the ${\text{Δ³Ò}}^{\text{°}}\text{at298K.}$

Solving for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{65.3°ì´³Ã—}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=-26.36}\\ {\text{K=e}}^{\text{-26.36}}\\ {\text{°­=3.58×10}}^{\text{-12}}\end{array}$

Write the expression for K.

${\text{K=P}}_{{\text{CO}}_{\text{2}}}$

Therefore, ${\text{PCO}}_{\text{2}}{\text{=3.58×10}}^{\text{-12}}.$

e) Let us calculate the equilibrium.

First, solve for ${\text{Δ³Ò}}^{\text{°}}\text{at1200K.}$

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ \text{=117.3°ì´³Ã—}\frac{\text{1000J}}{\text{1kJ}}\text{-(1200K)(174.4J/K)}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-91980J-92.0kJ}\end{array}$

Then, solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-92.0°ì´³Ã—}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—1200°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=9.22}\\ {\text{K=e}}^{\text{9.22}}\\ {\text{°­=1.01×10}}^{\text{4}}\end{array}$

Write the expression for K.

${\text{K=P}}_{{\text{CO}}_{\text{2}}}$

Therefore,${\text{P}}_{{\text{CO}}_{\text{2}}}{\text{=1.01×10}}^{\text{4}}$