You have been asked to calculate the enthalpy change in the following
reaction: 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
The reaction occurs in the following stages:
Chemical Change Assignment 2
1. 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) ΔH = -732.5kJ mol–1
2. Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ΔH = -27.6kJ mol–1
3. Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ΔH = -91kJ mol–1
Note stages 2 and 3 look the same, but the reaction is cooling from 1700oC to 25oC
and hence the enthalpy change there. Calculate the overall enthalpy change for the
reaction. You must show your workings within your submission, and explain your
reasoning
To calculate the overall enthalpy change for the reaction, we need to add the enthalpy changes of all the stages.
ΔH_total = ΔH1 + ΔH2 + ΔH3
ΔH1 = -732.5 kJ mol–1 (given)
ΔH2 = -27.6 kJ mol–1 (given)
ΔH3 = -91 kJ mol–1 (given)
Therefore,
ΔH_total = -732.5 kJ mol–1 + (-27.6 kJ mol–1) + (-91 kJ mol–1)
ΔH_total = -851.1 kJ mol–1
The overall enthalpy change for the reaction is -851.1 kJ mol–1. This means that the reaction is exothermic, releasing 851.1 kJ of energy per mole of the reaction. The negative sign indicates that the reaction releases energy, and the magnitude of the enthalpy change indicates the amount of energy released.
To calculate the overall enthalpy change for the reaction, we need to sum up the enthalpy changes for each stage. Let's calculate each stage separately:
Stage 1:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) ∆H = -732.5 kJ mol–1
Stage 2:
Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ∆H = -27.6 kJ mol–1
Stage 3:
Al2O3(s) + 2Fe(l) → Al2O3(s) + 2Fe(s) ∆H = -91 kJ mol–1
Now, we need to consider the enthalpy change for stage 2, where the reaction is cooling from 1700°C to 25°C. Since the reaction is exothermic, the enthalpy change will be negative.
To calculate the enthalpy change due to cooling, we can use the specific heat capacity formula:
q = mcΔT
Where:
q = Heat energy transferred
m = Mass of the substance
c = Specific heat capacity
ΔT = Change in temperature
The enthalpy change due to cooling can be calculated using the equation:
ΔH = -cqΔT
Given that the specific heat capacity of Fe(s) is 0.45 J g–1 °C–1 and the mass of Fe(s) is 2 mol (from the balanced equation), we can calculate the change in temperature (ΔT) as follows:
ΔT = 1700°C - 25°C = 1675°C
Converting ΔT to Kelvin, we have:
ΔT = 1675°C + 273.15 = 1948.15 K
Now, substituting the values into the equation:
ΔH = -0.45 J g–1 °C–1 * 1948.15 K * 2 mol
Calculating ΔH:
ΔH = -0.45 J g–1 °C–1 * 1948.15 * 2 mol = -1763.0175 J/mol
Therefore, the enthalpy change due to cooling (stage 2) is -1763.0175 J/mol.
Now, let's calculate the overall enthalpy change for the reaction by summing up the enthalpy changes for each stage:
Overall ∆H = ∆H_stage1 + ∆H_stage2 + ∆H_stage3
Overall ∆H = -732.5 kJ/mol + (-27.6 kJ/mol) + (-91 kJ/mol) + (-1.7630175 kJ/mol) = -852.8630175 kJ/mol
Therefore, the overall enthalpy change for the reaction is approximately -852.9 kJ/mol.