Answer:
An amide may be produced by reacting an acid chloride with ammonia.
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
First, we have to see how K2O behaves when it is dissolved in water:
K2O + H20 = 2 KOH
According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:
c(KOH) : c(K2O) = 1 : 2
So,
c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)
Now we can calculate pH:
pOH= -log c(OH⁻) = -log 0.01 = 2
pH= 14-2 = 12
Hello my friend thank you
