Unit 6 - Thermodynamics

Question 1

System

Answer 1

The portion of the experiment we’re focusing on; consists of the reactants and products

Question 2

Surroundings

Answer 2

Everything other than the system; consists of the container, air, thermometer, human, etc

Question 3

First law of thermodynamics

Answer 3

Total energy of an isolated system stays constant

Energy cannot be created or destroyed; it can only be converted from one form to another

Question 4

Energy

Answer 4

The capacity to do work or transfer heat

△E = q + w
Internal energy of system = heat + work

Question 5

Difference between temperature and heat?

Answer 5

Temperature represents the average kinetic energy present in a substance.

Heat represents the transfer of thermal energy from one substance to another because of a difference in temperature. Heat spontaneously flows from the warmer object to the cooler object

Question 6

Enthalpy

Answer 6

Enthalpy (△H): a measure of the energy that is released or absorbed by the substance when bonds are formed or broken during a reaction for a constant-pressure system

q = magnitude of △H

Enthalpy is measured in [kJ/mol] but you must relate it back to the specific quantities given in the question or the molar coefficients in the reaction equation

Question 7

Thermal equilibrium

Answer 7

Collisions between particles in thermal contact can result in the transfer of energy. Heat is transferred from the warmer object to the cooler object. The particles of the warmer object slow down while the particles of the cooler object speed up.

Eventually, the system reaches thermal equilibrium in which each substance has the same average kinetic energy and thus the same temperature

The magnitude of the heat lost by the warmer object is equal to the magnitude of the heat gained by the cooler object. This is because of the first law of thermodynamics and the assumption that the container is perfectly insulated so no heat is lost to the surroundings

Question 8

Specific heat capacity

Answer 8

If a substance has a high specific heat capacity, this means it can absorb a lot of heat without experiencing a large raise in temperature.

There is an inverse relationship between specific heat capacity (c) and temperature change (△T)

Water’s high specific heat capacity arises from its strong hydrogen bonds which take a lot of heat to break before the molecules can start moving faster to increase the temperature

Question 9

How do you convert from specific heat capacity (in grams) to molar heat capacity (in moles)

Answer 9

Multiple the specific heat capacity by the molar mass of the compound. Cancel out the units accordingly

Question 10

Heating curve

Answer 10

Slanted lines = temperature is increasing

Flat lines = phase changes (temperature remains constant because all the added energy/heat is being used to break the intermolecular forces between particles instead of adding extra kinetic energy to increase the temperature).

Vapor already has the highest energy so there won’t be any more phase changes after that even if you keep increasing the temperature

Enthalpy of vaporization > enthalpy of fusion

The energy absorbed during a phase change is equal to the energy released during the corresponding phase change in the opposite direction. (EX: Enthalpy of condensation = negative enthalpy of vaporization).

Question 11

When a solid melts, its density decreases. Why?

Answer 11

In the liquid state, the distance between the particles is greater than in the solid state. Thus, the same number of particles occupies a greater volume, decreasing the density because of their inversely proportional relationship (d = m/V)

Question 12

Forming vs breaking bonds

Answer 12

Forming bonds = releases energy = exothermic process

Breaking bonds = absorbs energy = endothermic process

Question 13

Why is dissolution an exothermic process?

Answer 13

Dissolution involves ion-dipole forces between the solute ions and the polar H2O molecules (solvent).

This results in dissolution being an exothermic process because bonds are being formed

Question 14

Bond enthalpy

Answer 14

The energy required to break a bond or the energy released when a bond is formed

Question 15

Hess’ Law

Answer 15

If you add two or more reactions together, the value of △H for the overall reaction is equal to the sum of the △H values for the individual reactions

1. If a reaction is reversed, △H stays constant in magnitude but becomes reversed in sign.
2. If an equation is multiplied by N, multiply the value of △H by N
3. If two (or more) equations are added together, add up the △H values for each reaction to obtain the net enthalpy of the overall reaction.

Question 16

Enthalpy of formation (△Hf°)

Answer 16

Change in energy (△H) when one mole of a compound is formed from its component pure elements under standard conditions (1 atm and 298 K)

Question 17

Enthalpy of combustion (△Hc°)

Answer 17

Amount of energy released (△H) when one mole of a hydrocarbon combusts (combustion is always exothermic)

Question 18

Hydration energy

Answer 18

ΔH2 + ΔH3 = hydration energy
(always negative since ΔH3 > ΔH2)

Hydration energy is a Coulombic energy like lattice energy and thus increases as the ions increase in charge or decrease in size

If hydration energy (ΔH2 + ΔH3) > lattice energy (ΔH1), enthalpy of solution (ΔH) is negative and vice versa

Question 19

What is the effect of an error in a calorimeter in which some heat is lost and not absorbed by the cooler substance?

Answer 19

q will be artificially low (since there was a lot more heat available originally). This will lower the value of the calculated specific heat since they have a directly proportional relationship.

Question 20

What are the most favorable conditions for ΔH and ΔG?

Answer 20

1. Low energy (released), decreasing enthalpy, exothermic reaction, ΔH < 0
2. High disorder, increasing entropy, thermodynamically favorable, ΔG < 0

Question 21

How do make an object have a higher temperature at thermal equilibrium?

Answer 21

Use objects that have a smaller specific heat and smaller mass.

This will create a larger temperature change to keep q constant (q = mcΔT). ΔT = Tf - Ti, or Tf = ΔT + Ti. With a larger ΔT, Tf (final temperature) will be higher at thermal equilibrium

Question 22

How does increasing the mass in a calorimetry experiment affect final temperature?

Answer 22

Increasing the mass increases the value of q (more heat is absorbed/released). As a result, the final temperature (Tf) will be higher with the increased mass