Unit 5 Energetics Flashcards

1
Q

What is an enthalpy

A

term used for the heat released or absorbed by a system at constant pressure

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2
Q

What symbols represents the change in enthaply

A

Delta H

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3
Q

What do you call the enthaply change measured under standard conditions

A

standard enthalpy change of reaction (ΔHƟ).

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4
Q

Exothermic and endothermic reactions are

A

Exo: release heat
Endo: absorb heat

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5
Q

Examples of exothermic reactions

A

Combustion and Neutralisation reactions

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6
Q

Explain energy profile of exothermic reaction

A

the enthalpy change (∆H) is negative because heat has been transferred from the system to the surroundings, so the system has lost heat.
The products of the reaction have lower enthalpy than the reactants, so they are more energetically stable.

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7
Q

Examples of endothermic reaction

A

thermal decomposition of calcium carbonate.

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8
Q

Explain energy Diagram of endothermic reaction

A

The enthalpy change is positive because the reactants gain heat from their surroundings. For this reason, the products have higher enthalpy than the reactants and are therefore less energetically stable.

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9
Q

How are enthalpy changes calculated:

A

Q= mc deltaT, where…
q is heat absorbed or released in J
C is specific heat capacity of solution
M is mass of solution in gr
DeltaT is change in Temperature in C or K

Add or remove minus to results depending on exothermic or endothermic reaction.

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10
Q

Calculate enthalpy change (deltaH) for a reaction

Ex:

50.0 cm3 of a 0.500 mol dm−3 solution of aqueous copper(II) sulfate was reacted with 3.00 g of zinc powder, according to the equation:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

The results are shown in Table 1 ​below. Use this data to determine the enthalpy change (ΔH) for the reaction.

Volume of 0.500 mol dm-3 CuSO4 (aq) used (cm3) =50.00
Mass of Zn powder added (g) = 3.00
Initial temperature of solution (C) = 21.0
Maximum temperature of solution (C) = 43.5

A
  1. Determine the change in heat (q) for the reaction using the equation q=mc deltaT with values from table
    - delta T: 22.5
    - mass is 50.00 cm3 which is 50.00 gr
    -assume heat capacity is 4.18
  2. Determine limiting reactant in reaction
    - calculate mol of each reactant
    - divide each amount in mol by the coefficient in balanced equation
    - lower value is limiting reactant
  3. Calculate the enthalpy change per mole of limiting reagent.
    - DeltaH divided by moles of limiting reagent
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11
Q

What is the standard enthalpy of combustion or the molar enthalpy of combustion

A

ΔHƟc), for a substance is defined as the enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions.
Always exothermic

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12
Q

How to calculate enthalpy change of combustion of hexane
Ex. In an experiment to measure the enthalpy change of combustion of hexane (C6H14), a student heated a copper calorimeter containing 400.0 cm3 of water with a spirit lamp and collected the following data:

Initial temperature of water: 20.0 °C
Final temperature of water: 77.9 °C
Mass of hexane burned: 2.49 g

Use this data to calculate the enthalpy change of combustion, ΔHc, of hexane.

A
  1. Calculate mol of hexane burned
  2. Calculate q using mc deltaT
  3. Divide heat released by amount in mol of hexane burned.
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13
Q

How do you calcite the experimental error

A

Percentage error=
(experimental value- theoretical value) divided by theoretical values times 100.

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14
Q

What is Hess’s law

A

states that the total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place, as long as the initial and final conditions are the same.

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15
Q

What is Hess’s law used for?

A

be used to calculate enthalpy changes for reactions that cannot be determined experimentally
Ex, enthalpy change of formation of methane bc hydrogen doesn’t react with carbon under standard conditions

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16
Q

Define entalphy chnage of formation (ΔHf), or standard enthalpy change of formation (ΔH⦵f)

A

the enthalpy change when one mole of a compound is formed from the elements in their standard states under standard conditions.

17
Q

Outline the enthalpy cycle when using the enthalpy change of formation and how it can be summarized

A

Arrows going to products and reactants from intermediate elements

ΔH⦵ = ΣΔH⦵f (products) − ΣΔH⦵f (reactants)

A useful application of Hess’s Law is that the enthalpy change of any reaction can be calculated provided that we know the enthalpy of formation values of the reactants and products.

18
Q

What does the standard enthalpy of formation indicate

A

they indicate the stability of compounds in relation to their elements - the more negative the value, the greater the stability of the compound.

19
Q

How can the enthalpy change of a reaction be calculated

A

using standard enthalpy change of combustion (ΔH⦵c ) values, using the equation below:

ΔH⦵ = ΣΔH⦵c (reactants) − ΣΔH⦵c (products)

20
Q

Outline MexoBendo

A

Making bonds is exothermic
Breaking bonds is endothermic

21
Q

In terms of bonds how do you determine wether a reaction is exothermic or endothermic

A

if the energy absorbed while breaking bonds is less than the energy released when forming new bonds, the reaction is exothermic

If the energy absorbed while breaking bonds is greater than the energy released when forming new bonds, the reaction is endothermic

22
Q

Define bond enthalpy

A

the energy required to break one mole of chemical bonds in the gaseous state.

23
Q

Define average bond enthalpy

A

the enthalpy change when one mole of bonds are broken in the gaseous state averaged for the same bond in similar compounds.

24
Q

How to calculate enthalpy changes using bond enthalpies

A

ΔH = ΣE(bonds broken) − ΣE(bonds formed)

or

ΔH = Σ(bond enthalpies of reactants) – Σ(bond enthalpies of products)

Weird E means sum

25
Q

What is the equation for the Haber process

A

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

26
Q

Outline relative strengths of the bonding in oxygen and ozone

A

Oxygen, O2
Bonding: 0=0 (two electron pairs on each O)
Bond energy: 498
Dissociation: UV-C. Higher energy radiation of shorter wavelength

Ozone, O3
Bonding: triangle with O, one bond between two Os, has a dotted bond.
Bond energy: 364
~Dissociation: UV-B. Lower energy longer wavelength radiation.

27
Q

Outline the formation of free radicles in stratosphere

A

The strong double covalent bond in di-oxygen (O2) can be broken in the stratosphere by ultraviolet radiation with a wavelength shorter than 242 nm, producing two free oxygen atoms (or radicals).

O2 (g) → O• (g) + O• (g) UV light, λ < 242 nm

The oxygen atoms produced each have an unpaired electron; they are highly reactive free- radicals and will react with oxygen molecules to form ozone (tri-oxygen, O3).

O2 (g) + O• (g) → O3 (g)