Thermochemistry Study Guide (page 3)
Thermodynamics is the study of energy and its processes. Energy can exist aspotential energy (an object at rest) or kinetic energy (an object in motion). Chemical energy, a form of kinetic energy, can be calculated with calorimetry, the science of measuring heat, and enthalpy, the change of thermal energy of a system. These two components of thermodynamics are essential to understanding heat and thermal energy.
First Law of Thermodynamics
The first law of thermodynamics states that energy is neither created nor destroyed. The energy in the universe is constant and can only be converted from one form to another.
Second Law of Thermodynamics
The second law states that there is an increase in entropy (randomness) in a spontaneous process.
Heat Energy and Calorimetry
The heat of vaporization is the heat required to evaporate 1 g of a liquid. Water's large heat of vaporization (540 cal/g) requires large amounts of heat in order to vaporize it into gas. During perspiration, water evaporates from the skin and large amounts of heat are lost. Heat of fusion is the heat required to fuse or melt a substance. Water's heat of fusion is 6.02 kJ/mol.
Heat capacity is the amount of energy required to raise the temperature of a substance by 1° C. Water has a high heat capacity, absorbing and releasing large amounts of heat before changing its own temperature. It thus allows the body to maintain a steady temperature even when internal and/or external conditions would increase body temperature. The heat energy of a substance can be calculated using the following equation:
q = C *ΔT
(heat = heat capacity * change in temperature)
However, because the amount of the substance determines how much heat is released or absorbed, it is more useful to use the specific heat capacity or molar heat capacity. The specific heat capacity is the heat capacity per mole and a known value for substances. The molar heat capacity is the heat capacity per mole. The equation can be modified to
- Energy = s * m * ΔT
(energy or heat = specific heat * mass of substance * change in temperature)
Calculate the amount of heat released when 150 g of liquid water cools from 100° C (boiling) to 20° C (room temperature).
Energy = s * m * ΔT= (4.18 J/g° C) (150 g) (20° C –100° C)
Notice that the change in temperature is final minus initial.
Energy = s * m * ΔT= (4.18 J/g° C) (150 g) (–80° C) = – 50,160 J = – 50.160 kJ =
The final answer has a negative value that indicates a release of energy.
Standard Enthalpy of Formation
Chemical reactions can release or absorb energy. Exothermic reactions are energy-releasing reactions like most catabolic and oxidative reactions. Endothermic reactions are reactions that consume energy in order to take place like anabolic reactions. This energy can be calculated using standard enthalpies of formations. The standard enthalpy of formation () is the energy required to form one mole of a substance from its elements in their standard states. Water can be theoretically formed from its elements, hydrogen and oxygen. Hydrogen is H2(g) and oxygen is O2(g) in their standard states. The enthalpy of this exothermic (negative ΔH) reaction is –286 kJ, hence making the standard enthalpy of formation () for liquid water–286 kJ/mol.
The standard enthalpy of reaction () can be calculated by subtracting the standard enthalpies of formation for each reactant from the standard enthalpies of formation for each product.
Calculate the standard enthalpy of reaction for the combustion of methane:
CH4(g) + O2(g) → H2O(g) + CO2(g)
Balance the chemical equation:
CH4(g) + 2O2(g) → 2H2O(g) + CO2(g)
Set up the mathematical equation:
= ( products) – ( reactants)
so, = (2 [H2O(g)] + [CO2(g)]) – ( [CH4(g)] + 2 [O2(g)])
= (2(–242) + (–393.5)) – ((–75) + 2(0))
[Note: O2 is oxygen's standard state.)
= –877.5 – (–75) = –802.5 = –803 kJ
As expected, the combustion (burning) of methane is exothermic.
Many reactions are difficult to calculate with the standard enthalpies of formation or calorimetry. Enthalpy is a state function that means that the calculation is independent of the pathway or calculation. Hess's Law states that an enthalpy of a reaction can be calculated from the sum of two or more reactions.
Calculate the enthalpy for ice, H2O(s), yielding steam, H2O(g), given the following equations:
|H2O(s) → H2O(I)||6.02|
|H2O(I) → H2O(g)||44.0|
Knowing we need to sum the previous equations to produce H2O(s) → H2O(g), we get
|H2O(s) → H2O(I)||6.02|
|H2O(I) → H2O(g)||44.0|
|H2O(s) → H2O(g)||50.0|
The yield arrow (→) acts like an equals sign. Therefore, identical substances on both the right and left sides, regardless of which equations, can be canceled. The sum of the reaction and sum of the enthalpies are added to create the wanted equation and final answer.
Calculate the enthalpy change for C(s) + 2H2(g) → CH4(g) given the following equations:
|C(s) + O2(g) →CO2(g)||–393.5|
|CH4O2(g) →CO2(g) + 2H2O(l)||–890.3|
This problem requires more effort. The two rules of Hess's Law must be applied:
- Equations can be reversed if the sign on ΔH is changed.
- Equations can be multiplied by any number to be able to cancel out substances in other equations. The ΔH must also be multiplied by the same value.
We want C(s) + 2H2(g) → CH4(g). So,
|C(s) + O2(g) → CO2(g)||– 393.5|
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) + 890.3
(equation is reversed to get CH4 on the right)
|C(s) + 2H2(g) → CH4(g)||– 74.8 kJ|
(–393.5 + 890.3 + 2(–285.8)= –74.8)
Free energy (G) of a system relates the enthalpy (H), temperature in Kelvin (T), and entropy (S) of a system. A negative ΔG means the process is spontaneous, and a positive ΔG means the process is not spontaneous.
ΔG = ΔH –TΔS
When is a process spontaneous (i.e., – ΔG)?
|– ΔH + ΔS||spontaneous at all temperatures|
|+ ΔH + ΔS||spontaneous at high temperatures|
|– ΔH – ΔS||spontaneous at low temperatures|
|– ΔH + ΔS||not spontaneous at any temperature|
Practice problems for these concepts can be found at - Thermochemistry Practice Questions
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