Hydrogen iodide is not produced by the same method is for hydrogen chloride why with reaction
1 answer:
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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.
We have the following balanced equation:
They also give us the heat of reaction equal to -764 kJ, i.e. it is an exothermic reaction.
By observing the reaction, we can deduce that for this heat to be generated, one mole of methanol is needed. Now let's see how many grams that mole of methanol equals. We will use the molecular weight equal to 32.04 g/mol
Now we know the grams of methanol that generate 764 kJ, because the heat of reaction is directly proportional to the mass of the reactants, we can apply a rule of three to know the grams needed to produce a heat of reaction equal to 701 kJ:
So, 29.4 g of methanol must be found to produce 701 kJ of heat
Answer:
trans-4-tert-butyl-1-phenylcyclohexanol
Explanation:
The bulky tert-butyl group will occupy the equatorial position in the cyclohexanone ring.
It will hinder approach of the Grignard reagent from the bottom side of the molecule.
The reagent will attack from the top, so the product alcohol will be
<em>trans</em>-4-<em>tert</em>-butyl-1-phenylcyclohexanol.
