Given the standard molar entropies measured at 25 C and 1 ATM pressure calculate delta S for the reaction 2al + 3 mg0 ---> 3mg + al2o3
You should learn to use the caps key. Co, CO, co all mean something different.
dSrxn = (n*dS products) - (n*dS reactants)
To calculate the change in entropy (ΔS) for a reaction, you need to use the molar entropies of the reactants and products. The change in entropy can be determined using the equation:
ΔS = Σn∙S(products) - Σm∙S(reactants)
Where:
ΔS is the change in entropy
n and m are the stoichiometric coefficients of the products and reactants, respectively
S(products) is the molar entropy of the products
S(reactants) is the molar entropy of the reactants
In this case, the reaction is: 2Al + 3MgO → 3Mg + Al2O3
First, you need to find the molar entropies (S) of each species involved in the reaction at 25°C and 1 atm pressure. These values are typically given in J/(mol·K). Let's assume the following molar entropies:
S(Al) = 28.3 J/(mol·K)
S(MgO) = 26.9 J/(mol·K)
S(Mg) = 32.7 J/(mol·K)
S(Al2O3) = 50.9 J/(mol·K)
Now, substitute these values into the equation to calculate ΔS:
ΔS = [3∙S(Mg) + S(Al2O3)] - [2∙S(Al) + 3∙S(MgO)]
ΔS = [3∙32.7 J/(mol·K) + 50.9 J/(mol·K)] - [2∙28.3 J/(mol·K) + 3∙26.9 J/(mol·K)]
ΔS = [98.1 J/(mol·K) + 50.9 J/(mol·K)] - [56.6 J/(mol·K) + 80.7 J/(mol·K)]
ΔS = 148.5 J/(mol·K) - 137.3 J/(mol·K)
ΔS = 11.2 J/(mol·K)
Therefore, the change in entropy (ΔS) for the given reaction is positive and equals to 11.2 J/(mol·K).