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

At thermal equilibrium, the tea and the ice/water mixture are at the same final temperature.

Since the final temperature is 0.0°C, the tea cools from 20.0°C → 0.0°C, so the tea loses heat:

q_tea = m_tea · c · (ΔT)
where ΔT = T_final − T_initial = 0.0 − 20.0 = −20.0°C, so q_tea is negative (heat released).

That same amount of heat is gained by the ice (law of conservation of energy):

q_tea + q_ice = 0
So, q_ice = −q_tea (this will be positive).

Because the ice starts at 0°C and ends at 0°C, the ice does not warm up. The heat it gains is used only for melting (phase change).

Important point: Final temperature being 0°C does NOT mean all the ice melts.
A mixture of ice + liquid water can remain at 0°C at equilibrium. If the tea does not supply enough heat to melt all the ice, some ice will remain.

Decision check (this is the key step)

Calculate heat released by tea
Use density to convert volume to mass:
m_tea = (volume in mL) × (1.00 g/mL)

Then compute:
q_tea = m_tea · c · (−20.0°C)

Compute heat required to melt all the ice
q_{melt all} = m_ice,total · ΔH_fus

Compare

If q_ice < q_{melt all}, then not all ice melts → final remains 0°C, and some solid ice is left.

If q_ice > q_{melt all}, then all ice melts and the leftover heat warms the water above 0°C (you would then use the remaining heat to find T_final).

If some ice remains (q_ice < q_{melt all})

Use:
q_ice = m_melted · ΔH_fus

So:
m_melted = q_ice / ΔH_fus

Then:
m_{ice remaining} = m_ice,total − m_melted

This question is based on formation enthalpies.
First right the balanced equation
C3H8(g)+5O2(g) →  3CO2(g)+4H2O(g)
Use the formation-enthalpy  equation

here apply the coefficients of the balanced equation
Plug in the values from the given table and you will get ΔH⁰comb in kJ / mol
Answer should be in kJ/g so convert 1mol into grams.
All questions based on formation enthalpies are done by using the equation given above.

I am confused about how to do this kind of question. How do I approach this?