Calculate the entropy change when 4.44g of H2 reacts with O2 according to the reaction

2H2(g) + O2(g)→2H2O(ℓ)
at 298 K and 1 atm pressure. The stan-
dard molar enthalpy of formation of H
2O(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

I would rewrite the equation as

H2 + 1/2 O2 ==> H2O
dGof = dHof - TdSof
You know dGo and dHo and T, solve for dS.
That gives you dS for 1 mol H2O from 1 mol H2. You have 4.44g/2 = 2.22 mols
dS for 1 mol x 2.22 mol = ?

To calculate the entropy change (ΔS) for this reaction, you need to use the formula:

ΔS = ∑S(products) - ∑S(reactants)

First, let's find the entropy values for the reactants and products.

For H2(g), we need to find its molar entropy (S°) value at 298 K. We can use a reference table or use statistical thermodynamics to calculate it. Let's assume the value is x J/K.

For O2(g), we do the same and assume its molar entropy at 298 K is y J/K.

For H2O(ℓ), we are given the standard free energy of formation (ΔG°f) which is -237.2 kJ/mol. The relationship between free energy and entropy is:

ΔG° = ΔH° - TΔS°

By rearranging the equation and solving for ΔS°, we get:

ΔS° = (ΔH° - ΔG°)/T

Substituting the given values, we have:

ΔS° = (-285.8 kJ/mol - (-237.2 kJ/mol))/(298 K)

Now, convert the given enthalpy and free energy values to J/mol:

ΔS° = (-285.8 kJ/mol - (-237.2 kJ/mol))/(298 K) * (1000 J/1 kJ)

Next, determine the stoichiometric coefficients from the balanced equation:

2H2(g) + O2(g) → 2H2O(ℓ)

This means that for every 2 moles of H2O(ℓ) formed, we have 2 moles of H2(g) reacting.

Now that we have the molar entropy values (x and y) and the stoichiometric coefficients, we can calculate the entropy change (ΔS):

ΔS = (2 * x + y) - (2 * 0 + 0)

Since there are no reactants present (0 moles of H2(g) and O2(g)), their contribution to the entropy change is zero. Thus, we are left with:

ΔS = 2 * x + y

Finally, calculate the entropy change by substituting the value of x, y, and the derived ΔS° value that we calculated earlier:

ΔS = 2x + y

This will give you the entropy change (ΔS) for the given reaction at 298 K and 1 atm pressure in units of J/K.