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The mass of HCl that is contained in the solution is 147 g HCl
Why?
To find the mass of HCl we have to apply what is called a conversion factor. In a conversion factor we put the units we don't want at the bottom, and the ones we want at the top.
For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:
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Answer:
(a) ΔG° = -474 kJ/mol; E° = 1.23 V
(b) ΔH° negative; ΔS° negative
(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.
Explanation:
Let's consider the following reaction.
2 H₂(g) + O₂(g) → 2 H₂O(l)
with an equilibrium constant K = 1.34 × 10⁸³
<em>(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.</em>
We can calculate the standard Gibbs free energy (ΔG°) using the following expression:
ΔG° = - R × T × lnK
ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol
To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.
Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻
Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻
The moles of electrons (n) involved are 4.
We can calculate E° using the following expression:
<em>(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative</em>
The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:
ΔG° = ΔH° - T.ΔS°
Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).
The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.
<em>(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?</em>
The maximum amount of work obtained depends on the standard Gibbs free energy.
wmax = ΔG° = ΔH° - T.ΔS°
Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.