LDA makes an anion (base step)

The epoxysilane rearranges (ring opens + Si moves to O)

That rearranged anion attacks MeI → methylation

I would love some help with these, thank you so much!

Hello, I would appreciate some help with this!

Hello, I would appreciate some help with this!

Hello, I would appreciate some help with this!

Hello, I would appreciate some help with this!

Quote from: Siddharth_Pendyala on April 29, 2025, 02:29:54 PMHi, I was confused on two questions I got wrong from the quiz 6.1. Could you please explain how to solve these?

The right answer is the reaction between a strong acid and a strong base (HCl + NaOH). Since both fully ionize in water, the reaction is essentially H⁺ + OH⁻ → H₂O and goes to completion, producing the maximum heat release (largest heat transfer from the system to the surroundings).

Quote from: Siddharth_Pendyala on April 29, 2025, 02:29:54 PMHi, I was confused on two questions I got wrong from the quiz 6.1. Could you please explain how to solve these?

Correct answer: a. All of the lost heat is gained by the surroundings.
As the pressure increases, a gas in a non-rigid container can change its volume to match the external pressure. If the gas expands, it does PV work on the surroundings.Work is also the form of energy so all this energy is gained by the surroundings.
So, when the system "loses energy," it can lose it as heat and/or work, but any heat that leaves the system goes to the surroundings

This question is based on bond enthalpies. Remember:

Bond breaking is endothermic (energy is absorbed).

Bond formation is exothermic (energy is released).

So, the enthalpy change of the reaction is found by taking the total energy required to break all bonds in the reactants and subtracting the total energy released when new bonds form in the products.
Part 1: Bond dissociation enthalpy (BDE) method — step by step (no answers)
Step 1: Write the balanced reaction (already given)
Step 2: Count bonds broken (reactants)

In 2 CH₄: each CH₄ has 4 C–H bonds → total C–H bonds broken
= 2 ×4
In 1 O₂: one O=O bond
→ total O=O bonds broken = 1
Step 3: Count bonds formed (products)
Each CH₃OH has: 3 C–H  1 C–O  1 O–H
So in 2 CH₃OH: C–H formed = 2×3
C–O formed = 2 × 1
O–H formed = 2× 1
Use the BDE equation
ΔHrxn�≈∑(BDE of bonds broken)−∑(BDE of bonds formed)
Plug in your counts with the provided BDE table:
 
Broken: (#"C-H" )(413)+(#"O=O" )(498)
Formed: (#"C-H" )(413)+(#"C-O" )(358)+(#"O-H" )(463)
Step 5: Keep track of units and sign
Your result will be in kJ per reaction as written (for producing 2 mol CH₃OH).
Negative value means exothermic.

Part 2: Enthalpies of formation method — step by step (no answers)
Step 1: Use the standard formula


Step 2: Multiply each ΔH_f^∘by its coefficient
Step 3: Substitute and simplify

Step 4: Units and meaning
   Result is in kJ per reaction as written.
   Negative → exothermic, positive → endothermic.

 
Step 1: Identify the path (state changes + temperature changes)

You are going from:H2O(𝑠)at 0∘C  →H2O(g) at 100 ∘C

That requires three stages:

1) Melt ice at 0°C → liquid at 0°C (fusion) B to C in heating curve)
    q1 = m ΔH⁰ fusion ( m is the mass of water)

2) Heat liquid water 0°C → 100°C (temperature rise from C to D in heating curve)
q2 = mcΔt (m is mass of water)

3) Vaporize liquid at 100°C → steam at 100°C (vaporization)
q3 = m Δ 𝐻 vap �

Total amount of heat needed = q1 + q2 + q3