Calculate the entropy change when 2.42 g of
H2 reacts with O2 according to the reaction
2 H2(g) + O2(g) → 2 H2O(ℓ)
at 298 K and 1 atm pressure. The standard molar enthalpy of formation of H2O(ℓ)
at 298 K is −285.8 kJ/mol. The corresponding free energy of formation is −237.2 kJ/mol.
Answer in units of J/K
see the post above.
To calculate the entropy change (ΔS) for the given reaction, you need to use the equation:
ΔS = ΣnS(products) - ΣnS(reactants)
Where:
Σn represents the stoichiometric coefficient of each substance in the reaction
S represents the molar entropy
(products) represents the products in the reaction
(reactants) represents the reactants in the reaction
To find the molar entropy of each substance, you can use the formula:
ΔS = ΔH/T
Where:
ΔS is the molar entropy change
ΔH is the molar enthalpy change
T is the temperature in Kelvin
Step 1: Calculate ΔS for the reaction
The given reaction is:
2 H2(g) + O2(g) → 2 H2O(ℓ)
Since there are two moles of H2O(ℓ) formed, we multiply the molar entropy of H2O(ℓ) by 2.
ΔS(H2O(ℓ)) = 2 × ΔS(H2O(ℓ))
Step 2: Calculate the molar entropy change for H2O(ℓ)
Given that the standard molar enthalpy of formation of H2O(ℓ) at 298 K is -285.8 kJ/mol, we convert it to J/mol:
ΔH(H2O(ℓ)) = -285.8 kJ/mol = -285,800 J/mol
Since the reaction occurs at 298 K, we can directly use these values in the equation. Remember to convert the values to Kelvin if necessary.
ΔS(H2O(ℓ)) = ΔH(H2O(ℓ)) / T = (-285,800 J/mol) / (298 K)
Step 3: Calculate the total ΔS for the reaction
Now that we have ΔS(H2O(ℓ)), we can substitute the values into the overall equation:
ΔS = (2 × ΔS(H2O(ℓ))) - (2 × ΔS(H2))
We need to know the molar entropy of H2 to complete the calculation. However, that information has not been provided in the question. The entropy data for H2 can be found in reference books or online databases.
Once you have the molar entropies for H2O and H2, you can substitute the values in and calculate the ΔS for the reaction. The final answer will be in units of J/K.