Chapter 8, Problem 87QAP

The thermite reaction was once used to weld rails:

$\text{2Al}\left(s\right)+{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{2Fe}\left(s\right)$(a) Using heat of formation data, calculate ΔH for this reaction.

(b) Take the specific heats of Al₂O₃ and Fe to be 0.77 and 0.45 J/g $\cdot$

°C, respectively. Calculate the temperature to which the products of this reaction will be raised, starting at room temperature, by the heat given off in the reaction.

(c) Will the reaction produce molten iron $\left(\text{mpFe}=1535°\text{C},\Delta H_{\text{fus}}=270\text{J}/\text{g}\right)$?

Interpretation Introduction

(a)

Interpretation:

The enthalpy change (ΔH) for the termite reaction needs to be deduced.

Concept introduction:

• The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

i.e. $\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(1)}$

Where n_p and n_r are the number of moles of the products and reactants.

${\text{The values of ΔH}}_{\text{f}}^{\text{0}}\text{ are measured under standard conditions of 25}\text{C}\text{ and 1 atm pressure}$

Since ΔH° is independent of pressure and varies very slightly with temperature, we can interpret that ΔH° is equal to the reaction enthalpy, ΔH.

Answer to Problem 87QAP

ΔH = -851.5 kJ

Explanation of Solution

The thermite reaction can be represented as:

${\text{2Al(s) + Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\to {\text{ Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 2Fe(s) }\Delta \text{H = ? kJ}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1) we have:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe(s))] - [2ΔH}}_{\text{f}}^{\text{0}}{\text{(Al(s)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)) ]}\end{array}$

Substituting the values:

$\begin{array}{l}\Delta \text{H = [1 mole }\times (\text{-1675}\text{.7 kJ/mol) + 2 moles }\times \text{(0 kJ/mol)]-[2 moles }\times \text{(}0\text{ kJ/mol) + 1 mole }\times (\text{-824}\text{.2 kJ/mol) ]}\\ \text{ = -1675}\text{.7 kJ + 0 kJ + 0 kJ -824}\text{.2 kJ = -851}\text{.5 kJ }\end{array}$

Interpretation Introduction

(b)

Interpretation:

The final temperature of the products (Al₂ O₃ and Fe) needs to be calculated.

Concept introduction:

• Chemical reactions in general involve exchange of heat between the system (reactants and products) and the surroundings. The heat (q) absorbed or evolved is given as: $\begin{array}{l}\text{q = m ×c×ΔT ------(2)}\\ \text{where:}\\ \text{m and c are the mass and specific heat of the substance respectively}\text{.}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}\\ {\text{where T}}_{\text{2}}{\text{ and T}}_{\text{1}}\text{ are the final and initial temperatures of the solution}\text{.}\end{array}$

Answer to Problem 87QAP

$\text{Final temperature is 6629}\text{C}$

Explanation of Solution

Given:

Specific heat of Al₂ O₃ = $0.77\text{ J/g}\text{C}$

Specific heat of Fe = $0.45\text{ J/g}\text{C}$

Initial temperature T₁ = $25\text{C}$

Calculating the mass of Al₂ O₃ and Fe:

The thermite reaction can be represented as:

${\text{2Al(s) + Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\to {\text{ Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 2Fe(s) }\Delta \text{H = -851}\text{.5 kJ}$

Based on the reaction stoichiometry:

1 mole of Al₂ O₃ and 2 moles of Fe are formed.

Molar mass of Al₂ O₃ = 102 g/mol

Molar mass of Fe = 56 g/mol

Therefore:

Mass of Al₂ O₃ = 102 g

Mass of Fe = 2(56) = 112 g

Calculating the total heat absorbed by the products (Al₂ O₃ and Fe):

Heat absorbed by Al₂ O₃ is:

$\begin{array}{l}\text{q = m ×c×ΔT }\\ \text{where:}\\ {\text{m = mass of Al}}_{\text{2}}{\text{O}}_{\text{3}}\\ {\text{c = specific heat of Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 0}\text{.77 J/g}\text{C}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}={\text{T}}_{\text{2}}-25\\ {\text{q}}_1\text{ = 102 g}\times \text{0}\text{.77 J/g}\text{C}\times ({\text{T}}_{\text{2}}-25)\text{C}\text{ = 78}\text{.54}\times ({\text{T}}_{\text{2}}-25)\end{array}$

Heat absorbed by Fe is:

$\begin{array}{l}\text{q = m ×c×ΔT }\\ \text{where:}\\ \text{m = mass of Fe}\\ \text{c = specific heat of Fe = 0}\text{.45J/g}\text{C}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}={\text{T}}_{\text{2}}-25\\ {\text{q}}_2\text{ = 112 g}\times \text{0}\text{.45 J/g}\text{C}\times ({\text{T}}_{\text{2}}-25)\text{C}\text{ = 50}\text{.4}\times ({\text{T}}_{\text{2}}-25)\end{array}$

$\begin{array}{l}{\text{Total heat absorbed = q}}_{\text{1}}{\text{ + q}}_{\text{2}}\\ =\text{78}\text{.54}\times ({\text{T}}_{\text{2}}-25)+\text{50}\text{.4}\times ({\text{T}}_{\text{2}}-25)=128.94\times ({\text{T}}_{\text{2}}-25)\end{array}$

Calculating the final temperature of the products:

Heat released by the reaction = Heat absorbed by the products

$\begin{array}{l}851500\text{ = }128.94\times ({\text{T}}_{\text{2}}-25)\\ {\text{T}}_{\text{2}}\text{ = 6629}\text{C}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The possibility of the production of molten iron under the given conditions needs to be examined.

Concept introduction:

• The heat energy absorbed when a solid melts is termed as the enthalpy of fusion (ΔH_fus ) and that when a liquid vaporizes is termed as the enthalpy of vaporization (ΔH_vap ).
• The change in enthalpy (ΔH) for a given reaction is equal in magnitude but opposite in sign for the reverse reaction.

Answer to Problem 87QAP

Yes, the reaction will produce molten iron

Explanation of Solution

The thermite reaction can be represented as:

${\text{2Al(s) + Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) }\to {\text{ Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 2Fe(s) }\Delta \text{H = -851}\text{.5 kJ}$

Based on the given information, the melting point of Fe is 1535 °C. As per the calculations in part (b) the final temperature of the products is 6629 °C. Therefore, at this temperature, the iron formed will be in its molten state.