89CP - Chemistry: Molecular Nature Of Matter And Change

Question

Magnesia (MgO) is used for fire brick, crucibles, and furnace linings because of its high melting point. It is produced by decomposing magnesite (MgCO₃) at around ${\mathbf{\text{1200}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$

a)  Write a balanced equation for magnesite decomposition. 

b) Use ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}{\mathbf{\text{ and S}}}^{\mathbf{\text{°}}}$ values to find ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}\mathbf{\text{ at 298 K.}}$ 

c)  Assuming ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}{\mathbf{\text{ and S}}}^{\mathbf{\text{°}}}$ do not change with temperature, find the minimum temperature at which the reaction is spontaneous. 

d) Calculate the equilibrium ${\mathbf{\text{P}}}_{{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}}$above ${\mathbf{\text{ MgCO}}}_{\mathbf{\text{3}}}\mathbf{\text{ at 298 K.}}$

e) Calculate the equilibrium ${\mathbf{\text{P}}}_{\mathbf{\text{CO}}}$  above ${\mathbf{\text{MgCO}}}_{\mathbf{\text{3}}}\mathbf{\text{ at 1200 K.}}$

Step-by-Step Solution

Answer

a) A balanced equation for magnesite decomposition is, ${\text{MgCO}}_{\text{3}}\text{(s) }\to {\text{ MgO(s)+CO}}_{\text{2}}\text{(g)}$ 

b) ${\text{ΔG}}^{\text{°}}\text{ at 298 K}$  is 65.3kJ.

c) The minimum temperature at which the reaction is spontaneous is $\text{T > 674.1 K. }$

d) The equilibrium is ${\text{P}}_{{\text{CO}}_{\text{2}}}{\text{ = 3.58×10}}^{\text{-12}}\text{ atm.}$ 

e) The equilibrium is ${\text{P}}_{{\text{CO}}_{\text{2}}}{\text{ = 1.01×10}}^{\text{4}}\text{ atm.}$

Step 1: Concept Introduction

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}$.

Step 2: Writing a balanced equation for magnesite decomposition

MgCO₃ decomposes 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 of MgCO₃

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

This is the balanced equation for magnesite decomposition.

Step 3: Finding ΔG °  at 298 K

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

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

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{ = ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ \text{= 117.3kJ × }\frac{\text{1000J}}{\text{1kJ}}\text{ - (298K)(174.4J/K)}\\ {\text{ΔG}}^{\text{°}}\text{ = 65328.8J}\\ \text{= 65.3 kJ}\end{array}$

Therefore, ${\text{ΔG}}^{\text{°}}\text{ at 298 K}$ is 65.3kJ.

Step 4: Finding the minimum temperature at which the reaction is spontaneous

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

$\begin{array}{l}{\text{ΔG}}^{\text{°}}{\text{ = ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{0 = ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{TΔS}}^{\text{°}}{\text{ = ΔH}}^{\text{°}}\\ \text{T = }\frac{{\text{ΔH}}^{\text{°}}}{{\text{ΔS}}^{\text{°}}}\\ \text{=}\frac{\text{117.3 kJ×}\frac{\text{1000 J}}{\text{1 kJ}}}{\text{174.4 J/K}}\\ \text{T = 674.1 K }\end{array}$

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

Step 5: Calculating the equilibrium P CO 2 above MgCO 3 at 298K

d) Let us calculate the equilibrium.

We know the ${\text{ΔG}}^{\text{°}}\text{ at 298K.}$

Solving for K.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{ = -RTlnK}\\ {\text{K = e}}^{\frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{ = }\frac{\text{65.3 kJ×}\frac{\text{1000 J}}{\text{1 kJ}}}{\text{-8.314 J/mol×K×298 K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{ = -26.36}\\ {\text{K = e}}^{\text{-26.36}}\\ {\text{K = 3.58×10}}^{\text{-12}}\end{array}$

Write the expression for K.

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

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

Step 6: Calculating the equilibrium P C O 2 above MgCO 3 at 1200K.

e) Let us calculate the equilibrium.

First, solve for ${\text{ΔG}}^{\text{°}}\text{ at 1200 K.}$

$\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{ = ΔH}}^{\text{°}}{\text{ - TΔS}}^{\text{°}}\\ \text{= 117.3kJ × }\frac{\text{1000J}}{\text{1kJ}}\text{ - (1200K)(174.4J/K) }\\ {\text{ΔG}}^{\text{°}}\text{ = -91980J - 92.0kJ}\end{array}$

Then, solve for K.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{ = -RTlnK }\\ {\text{K = e}}^{\frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{ = }\frac{\text{-92.0 kJ×}\frac{\text{1000 J}}{\text{1 kJ}}}{\text{-8.314 J/mol×K×1200 K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{ = 9.22}\\ {\text{K = e}}^{\text{9.22}}\\ {\text{K = 1.01×10}}^{\text{4}}\end{array}$

Write the expression for K.

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

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