Question

The thermite reaction was once used to weld rails: 2Al (s) + Fe2O3 (s)  Al2O3 (s) + 2Fe (s) a) Using heat of formation data (see your textbook), calculate ΔH for this reaction b) Take the specific heats of Al2O3 and Fe to be 0.77 and 0.45 J/g∙°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) What assumption(s) do you have to make in answering part b)? d) Will the reaction produce molten iron (melting point Fe = 1535°C, ΔHfus = 270 J/g) ?

Answer 1

a) ∆H(reaction) = -849.7 kJ/mol

b) T2 = 6623.6 °C

c) The products will be change from phase, and will be present as gas

d) Since this final temperature >>> melting point iron, we can say that the reaction will melt iron

Step-by-step explanation:

Step 1: Data given

ΔHf Al2O3 = −1675.7 kJ/mol

ΔHf Fe2O3 = −826 kJ/mol

the specific heats of Al2O3 and Fe to be 0.77 and 0.45 J/g∙°C

melting point Fe = 1535°C

ΔHfus = 270 J/g

Step 2: The balanced equation

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

Step 3: Calculate ΔH for this reaction

(ΔHf of elements = 0)

∆H(reaction) = ∆Hf(products) - ∆Hf(reactants)

∆H(reaction) = −1675.7 kJ/mol) - (−826 kJ/mol) = -849.7 kJ/mol

Step 4: 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.

molar specific heat Al2O3 = 0.77 J/g°C * 101.96 g/mol

molar specific heat Al2O3 = 78.51 J/mol*C°

molar specific heat Fe = 0.45 J/g°C * 55.85 g/mol

Molar specific heat Fe = 25.13 J/mol °C

2Al(s) + Fe2O3(s) --> Al2O3(s) + 2Fe(s)

For 2 mol Al we need 1 mol Fe2O3 to produce 1 mol Al2O3 and 2 moles Fe

Q(total) = Q(Al2O3) + Q(Fe)

Q = n*c*∆T = n(Al2O3)*c(Al2O3)*∆T(Al2O3) + n(Fe)*c(Fe)*∆T(Fe)

Q= ∆T*((n(Al2O3))*(c(Al2O3)) + (n(Fe))*(c(Fe)))

Q= ∆H(reaction) (without sign)= 849.7 kJ/mol

∆T = T2 - T1 = T2 - 25 °C

∆T = (Q(total))/(((n(Al2O3))*(c(Al2O3)) + (n(Fe))*(c(Fe)))

T2 - 25 °C = (849.7 kJ/mol)*(78.51 J/mol °C) + (2 mol)*(25.13 J/mol °C) )

T2- 25 °C = (849.7 kJ/mol)/(78.51 J/°C + 50.26 J/C°)

T2 - 25 °C = 6598.6 °C

T2 = 6623.6 °C

What assumption(s) do you have to make in answering part b

The products will be change from phase, and will be present as gas

d) Will the reaction produce molten iron (mp Fe=1535 degrees C, deltaH fusion = 270 J/g) ?

The final temperature >> melting point of iron

ΔHfus = 270 J/g

ΔHfus = 270 J/g * 55.85 g/mol

ΔHfus = 15079.5 J/mol = 15.0795 J/mol

Energy needed to melt iron for the reaction

15.0795 J/mol * 2 mol = 30.159 kJ

Heat = ∆H(reaction) - Q(fusion) = -849.7 kJ + 30.159 kJ = -819.541 kJ

Calculate the temperature

T2 - 25 °C = (819541 J/mol)/(128.77 J/°C)

T2 - 25 °C = 6364.4 °C

T2= 6364.4 °C + 25 °C = 6389.4 °C

Since this final temperature >>> melting point iron, we can say that the reaction will melt iron