chemical energetics Flashcards

(25 cards)

1
Q

Define activation energy

A

The activation energy of a reaction (Ea) is the minimum energy which the reacting particles must
possess in order to overcome the energy barrier before becoming products.

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2
Q

Define standard state, and thus standard enthalpy change.

A

The standard state of a substance at a specified temperature is its pure form at 1 bar. The standard enthalpy change of a reaction is the difference in enthalpy between the products in their standard states and the reactants in their standard
states, all at the same specified temperature. A common reference temperature used is 298 K, for pressure 1 bar/10000Pa, for concentration 1mol/dm3.
OR The enthalpy change when molar quantities of reactants as specified by the chemical equation
react to form products at 1 bar and 298 K (standard conditions).

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3
Q

Define standard enthalpy change of formation.

A

The enthalpy change when 1 mole of a substance is formed from its constituent elements in their
standard states at 298 K and 1 bar.
eg ½H2(g) + ½I2(s) → HI(g), Hf = +26.5 kJ mol–1

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4
Q

Explain the different signs for standard enthalpy change of formation.

A

From the definition, the enthalpy change of formation of elements in their standard state (at 298 K and 1 bar) is 0 kJ mol–1.
Positive: compound is energetically less stable than its constituent
elements, greater likelihood for compound to decompose into its constituent elements
Negative: compound is energetically more stable than its constituent
elements, less likelihood for compound to decompose into its constituent elements

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5
Q

Standard enthalpy change of combustion.

A

The heat evolved when 1 mole of a substance is completely burnt in excess oxygen at 298 K and
1 bar. It is always exothermic.

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6
Q

Standard enthalpy change of neutralisation

A

The heat evolved when 1 mole of water is formed in the neutralisation reaction between an acid and a base, at 298 K and 1 bar.
Neutralisation is an exothermic reaction since it involves the attraction of H+ and OH– ions to
form an O–H bond.

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7
Q

Why is standard enthalpy change of neutralisation ALMOST always the same?

A

The enthalpy change of neutralisation of a strong acid with a strong base is almost the same
for all strong acids and bases (–57.0 kJ mol–1). Strong acids and strong bases ionise completely in dilute aqueous solution. Reaction between them is effectively the reaction between aqueous H+ ions and OH– ions.
The enthalpy change of neutralisation involving a weak acid or weak base is slightly less
exothermic than –57.0 kJ mol–1.
Weak acids (or weak bases) do not ionise completely in dilute aqueous solution. During neutralisation, energy is absorbed to ionise the un-ionised weak acid (or weak base). Thus, less energy is released and the resulting HꝊneut is less exothermic.

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8
Q

What are precautions taken to improve the accuracy of a calorimeter?

A
  • Polystyrene is an excellent insulator so heat loss from the sides is minimised. The lid also helps
    to minimise heat exchange with the surrounding air.
  • The polystyrene cup is often supported in a 250 cm3 beaker or using a retort stand, to prevent
    toppling under the weight of the thermometer. The insulation can be improved by nesting two
    cups.
  • The temperature changes are often small, so for accurate work the thermometer should have
    a precision of ±0.1oC.
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9
Q

What are assumptions made in calculations in calorimeter?

A
  • No heat loss to (or heat gain from) surrounding air
  • Heat capacity of the calorimeter (e.g. polystyrene cup) is omitted
    In addition, it is more convenient to measure the volume of the solution. Hence to calculate the mass of the solution, we assume that the density of the solution is the same as that of water (1.00 g cm–3) since the solution is very dilute.
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10
Q

State Hess’ Law.

A

The enthalpy change (H) of a reaction is determined only by the initial and final states and is
independent of the reaction pathway taken.

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11
Q

Standard enthalpy change of atomisation

A

Hatom for an element is the energy required when 1 mole of gaseous atoms is formed from the
element at 298 K and 1 bar.
Hatom for a compound is the energy required to convert 1 mole of the compound into gaseous
atoms at 298 K and 1 bar.
* HꝊ atom values are always positive since atomisation involves breaking bonds.
* For gaseous diatomic elements, enthalpy change of atomisation and bond energy are related:
HꝊ atom (X2) = ½ x B.E. (X–X)

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12
Q

Bond energy

A

Bond energy is the energy required to break 1 mole of a covalent bond in the gaseous state. Bond energies are always positive because energy is needed to pull the two atoms apart (endothermic process) to break the covalent bond. As some bond energy data in the Data Booklet are average values, using them in calculations can result in discrepancies when compared to values obtained by other means e.g. from experiments or calculated from other enthalpy changes.

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13
Q

Lattice energy

A

The heat evolved when 1 mole of solid ionic compound is formed from its constituent gaseous ions.
* Lattice energy is always negative as heat is evolved in forming electrostatic forces of attraction
between oppositely charged ions (i.e. ionic bonding).
* Lattice energy is a measure of the strength of ionic bonding and the stability of the ionic compounds. The more exothermic the lattice energy, the stronger the ionic bonding and the more stable the ionic compound.

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14
Q

Factors affecting lattice energy

A

Charges on the ions, size of ions or inter-ionic distance
eg MgCl2 and NaCl have the same anion. Mg2+ has a higher charge and smaller ionic radius than
Na+. (quote formula for L.E), L.E. of MgCl2 is more exothermic than that of NaCl.

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15
Q

Ionisation energy

A

The first ionisation energy is the energy required to remove 1 mole of electrons from 1 mole of
gaseous atoms to form 1 mole of singly charged gaseous cations. I.E. is always positive since energy is required to remove an electron (endothermic process).

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16
Q

Electron affinity

A

The first electron affinity is the enthalpy change when 1 mole of electrons is added to 1 mole of
gaseous atoms to form 1 mole of singly charged gaseous anions.
* 1st E.A. is usually negative as the effective nuclear charge of the atom leads to an attraction of the incoming electron. When the attraction is stronger, the energy given off is greater and hence
E.A. is more negative.
* 2nd and subsequent E.A. are always positive because energy is required to overcome the electrostatic repulsion between the incoming electron and the anion.

17
Q

Describe and explain the difference between experimental and theoretical lattice energies.

A

Experimental lattice energy refers to the value found from experimental results using the
Born-Haber cycle. Theoretical lattice energy refers to the value calculated based on a model which assumes that the compound is completely ionic. There is always a difference between the two values which suggests that no compound is completely ionic. A small difference shows that the ionic model is a good one for the compound. A large difference shows that there is covalent character in the ionic compound. This is most apparent when a cation with a high charge density distorts an anion with a large electron cloud.

18
Q

Standard enthalpy change of hydration

A

The heat evolved when 1 mole of free gaseous ions is dissolved in an infinite volume of water at
298 K and 1 bar. Always negative as heat is evolved in forming ion-dipole interactions between the ions and the polar water molecules.
Magnitude depends on charge density. The higher the charge density of the ion, the stronger the ion-dipole interaction and Hhyd will be more exothermic.

19
Q

Standard enthalpy change of solution

A

The enthalpy change when 1 mole of solute is completely dissolved in an infinite volume of solvent
at 298 K and 1 bar. Can be either positive or negative.
Positive: likely insoluble
Negative: likely soluble
BUT there are exceptions ie salts with positive Hsol and are soluble due to positive entropy change.

20
Q

Relationship between lattice energy, enthalpy changes of hydration and enthalpy change of solution

A

The dissolution of an ionic solid MX(s) can be divided into 2 processes:
1) Breaking up the solid ionic lattice to form isolated gaseous ions.
Process is endothermic (overcoming ionic bonding) and the enthalpy change is ‘–LE’.
MX(s) → M+(g) + X–(g) H = –LE
2) Hydration of the gaseous ions.
Process is exothermic (forming ion-dipole interactions between ion and water).
M+(g) + X–(g) + aq → M+(aq) + X–(aq) H = Hhyd(M+) + Hhyd(X–)
Overall: HꝊsol = – L.E. + sum of HꝊhyd

21
Q

Define a spontaneous process.

A

A spontaneous process is one that takes place naturally in the direction stated. In other words, the
change occurs without a need for continuous input of energy from outside the system.

22
Q

Entropy (S)

A

Entropy ‘S’ is a measure of the randomness or disorder in a system, reflected in the number of ways that the energy of a system can be distributed through the motion of its particles. For the same amount of a substance, entropy of solid < liquid &laquo_space;gas.
* If a reaction or process results in more ways to disperse or distribute the energy, entropy
increases (S > 0).
* If a reaction or process results in less ways to disperse or distribute the energy, entropy
decreases (S < 0).

23
Q

How to predict the sign of S for processes

A
  1. As temperature increases, the average kinetic energy of the particles and the range of energies
    increase. There are more ways to disperse the energy among
    the particles. Hence, entropy increases (S increases, S is positive).
  2. Changes in state: The particles in the solid state vibrate about their fixed positions. The energy is thus the least dispersed and the solid has the lowest entropy. When temperature increases, entropy increases gradually as the kinetic energy of the particles increases. When the solid melts, the particles move more freely in the liquid state and become more disordered. Hence there is an abrupt increase in entropy as there are more ways to
    distribute the particles and their energy in the liquid state. Likewise, as the liquid is heated, its entropy increases as the particles gain kinetic energy. During vaporisation, the liquid converts to a gas where the particles are able to move even more freely. Hence there is a large increase in entropy as there are more ways to distribute the particles and their energy in the gaseous state.
  3. Change in number of particles: When a chemical reaction results in an increase in the number of gas particles, there is a large
    increase in entropy. This is because the particles in gas are the most disordered so the number
    of ways that the particles and the energy can be distributed increase greatly. If there is no change in the number of gas particles, entropy may increase or decrease but S
    will be relatively small numerically.
  4. Dissolution of an ionic solid: Two entropy terms operate when an ionic solid is dissolved in water:
    * Entropy increases because the ions in the solid are free to move in solution.
    * Entropy decreases because water molecules that were originally free to move become restricted in motion as they arrange themselves around the ions.
    If the first factor is more significant, the overall entropy change is positive.
24
Q

Gibbs free energy change

A

G = H – TS. The standard Gibbs free energy change of reaction, GꝊ, is the Gibbs free energy change needed to convert reactants into products at 1 bar and at constant temperature (usually at 298 K). If G<0, The reaction takes place spontaneously, ie exergonic. If G=0, The system is at equilibrium.
There is no net reaction in
the forward or backward
direction. E.g. G = 0 during melting
and boiling at the melting
point and boiling point
respectively. If G>0, the reaction cannot take place spontaneously.
It is spontaneous in the reverse
direction, ie endergonic.

25
What are the limitations in the use of G to predict spontaneity of a reaction?
1. Non-standard condition Go can only be used to predict spontaneity of a reaction under standard conditions. Under non- standard conditions, G must be calculated. 2. Kinetics consideration Just because a reaction is spontaneous does not mean that it will occur at an observable rate. While the Gibbs free energy change can be used to determine the spontaneity of a reaction, it does not take into account the kinetics of the reaction, ie the rate at which the reaction takes place. The reacting species may have to overcome a large energy barrier (Ea) before reaction can occur. Some reactions are thermodynamically (or energetically) favourable (G is negative) but kinetically not favourable (occur very slowly). Such reactions are spontaneous but slow, for example rusting.