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
just forget about it dude. it's all pointless now
Oh, calculating entropy changes, huh? Let's dive into it (or should I say, let's dive into the Clownfish Tank of Thermodynamics)!
First, we need to determine the moles of H2 that reacted. To do that, we divide the given mass by the molar mass of H2 (2 g/mol):
2.42 g / 2 g/mol = 1.21 mol
According to the balanced equation, we see that 2 mol of H2 are involved in the reaction. So, the moles of H2O produced will also be 1.21 mol.
Now, we can calculate the entropy change (ΔS) using the formula:
ΔS = n * ΔS°
where:
n = number of moles of H2O produced
ΔS° = standard molar entropy change of the products minus the standard molar entropy change of the reactants
The standard molar entropy change of water can be found in thermodynamic tables or obtained from a really thirsty clownfish who knows all about entropies:
ΔS° = 69.95 J/(mol·K)
So, plugging in the values:
ΔS = 1.21 mol * 69.95 J/(mol·K)
ΔS = 84.7945 J/K
Rounding off to the appropriate significant figures, the entropy change is approximately 84.79 J/K.
Now, let's all do a little happy dance to celebrate the successful calculation of entropy! 🎉
To calculate the entropy change (ΔS) for the given reaction, we can use the equation:
ΔS = ΣnS(products) - ΣnS(reactants)
First, we need to find the entropy values of the reactants and products. The entropy values are typically given in tables or can be calculated using statistical mechanics, but here we will use the standard entropy values provided. The standard entropy of a substance (S) is normally given in units of J/mol·K.
The entropy values (S) for the reactants and products are as follows:
H2(g): Given value is not provided. We can assume its entropy to be zero, as it is the reference state.
O2(g): 205 J/mol·K (from tables)
H2O(ℓ): 70 J/mol·K (from tables)
Now we can calculate the entropy change:
ΔS = ΣnS(products) - ΣnS(reactants)
= (2 mol × 70 J/mol·K) - (2 mol × 205 J/mol·K) - 0
Note that we multiply the entropy values by the coefficients in the balanced equation to account for the molar amounts.
ΔS = (140 J/K) - (410 J/K)
= -270 J/K
The entropy change (ΔS) for the given reaction is -270 J/K.
I would calculate dH rxn as
dHorxn = (n*dHproducts) - (n*dHreactants)
Also dGorxn = (n*dGproducts) - (n*dGreactants)
Then dG = dH - TdS.
You know dG, dH and T, solve for dS rxn. That will be dS for 4g H. Adjust with a ratio to obtain dS for 2.42g.