Answer :
To calculate the change in entropy (ΔS) for the reaction:
[tex]\[ Fe_2O_3(s) + 2Al(s) \rightarrow Al_2O_3(s) + 2Fe(s) \][/tex]
we use the formula for the change in entropy of a reaction:
[tex]\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \][/tex]
Step 1: Determine the total entropy of the products.
- The entropy of one mole of [tex]\( Al_2O_3(s) \)[/tex] is 51 J/K·mol.
- The entropy of one mole of [tex]\( Fe(s) \)[/tex] is 27.2 J/K·mol, and since we have 2 moles, this becomes [tex]\( 2 \times 27.2 = 54.4 \)[/tex] J/K·mol.
Adding these values together gives us the total entropy for the products:
[tex]\[ S_{\text{products}} = 51 + 54.4 = 105.4 \, \text{J/K·mol} \][/tex]
Step 2: Determine the total entropy of the reactants.
- The entropy of one mole of [tex]\( Fe_2O_3(s) \)[/tex] is 90 J/K·mol.
- The entropy of one mole of [tex]\( Al(s) \)[/tex] is 28.3 J/K·mol, and since we have 2 moles, this becomes [tex]\( 2 \times 28.3 = 56.6 \)[/tex] J/K·mol.
Adding these values together gives us the total entropy for the reactants:
[tex]\[ S_{\text{reactants}} = 90 + 56.6 = 146.6 \, \text{J/K·mol} \][/tex]
Step 3: Calculate the change in entropy (ΔS).
Subtract the total entropy of the reactants from the total entropy of the products:
[tex]\[ \Delta S = 105.4 - 146.6 = -41.2 \, \text{J/K·mol} \][/tex]
Thus, the change in entropy for the reaction is [tex]\(-41.2 \, \text{J/K·mol}\)[/tex].
The correct answer is C. [tex]\(-41.2 \, \text{J/K·mol}\)[/tex].
[tex]\[ Fe_2O_3(s) + 2Al(s) \rightarrow Al_2O_3(s) + 2Fe(s) \][/tex]
we use the formula for the change in entropy of a reaction:
[tex]\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \][/tex]
Step 1: Determine the total entropy of the products.
- The entropy of one mole of [tex]\( Al_2O_3(s) \)[/tex] is 51 J/K·mol.
- The entropy of one mole of [tex]\( Fe(s) \)[/tex] is 27.2 J/K·mol, and since we have 2 moles, this becomes [tex]\( 2 \times 27.2 = 54.4 \)[/tex] J/K·mol.
Adding these values together gives us the total entropy for the products:
[tex]\[ S_{\text{products}} = 51 + 54.4 = 105.4 \, \text{J/K·mol} \][/tex]
Step 2: Determine the total entropy of the reactants.
- The entropy of one mole of [tex]\( Fe_2O_3(s) \)[/tex] is 90 J/K·mol.
- The entropy of one mole of [tex]\( Al(s) \)[/tex] is 28.3 J/K·mol, and since we have 2 moles, this becomes [tex]\( 2 \times 28.3 = 56.6 \)[/tex] J/K·mol.
Adding these values together gives us the total entropy for the reactants:
[tex]\[ S_{\text{reactants}} = 90 + 56.6 = 146.6 \, \text{J/K·mol} \][/tex]
Step 3: Calculate the change in entropy (ΔS).
Subtract the total entropy of the reactants from the total entropy of the products:
[tex]\[ \Delta S = 105.4 - 146.6 = -41.2 \, \text{J/K·mol} \][/tex]
Thus, the change in entropy for the reaction is [tex]\(-41.2 \, \text{J/K·mol}\)[/tex].
The correct answer is C. [tex]\(-41.2 \, \text{J/K·mol}\)[/tex].
