To start, energy is the ability to do work or transfer heat. Its SI unit is the joule (J). Heat is energy in transit. It is represented by the variable q. Work is equal to force multiplied by distance and is represented by the variable w. Additionally, work is equivalent to the negative pressure multiplied by change in volume (this will come into play later). Heat energy can be used to do work. Likewise, work can be transformed into heat. When referring to the transfer of energy, one must consider the system and the surroundings. The system is the actual chemical reaction that is taking place. The surroundings are everything outside of the reaction. The energy inside a system is referred to as a system's internal energy. Changes in a system's internal energy can be modeled by the following equation:
∆E = q + w
In other words, the change in internal energy is equivalent to heat plus work.
Enthalpy is related to internal energy. If a reaction takes place at constant pressure and the only work done is pressure-volume work, the heat flow of the process can be accounted for by measuring the enthalpy of the system. Simply put, enthalpy is the internal energy plus the product of pressure and volume. This can be modeled by the following equation:
H = E + PV
When the system changes at constant pressure, the change in enthalpy (∆H) is equivalent to:
∆H = ∆E + P∆V
Since it is known that ∆E = q + w, "q + w" (heat plus work) can be substituted into the equation for the change in enthalpy. It is also known that w = -P∆V, so this can also be substituted into the equation for change in enthalpy. Thus:
∆H = (w + q) - w
∆H = w - w + q
∆H = w - w + q
∆H = q
At constant pressure, the change in enthalpy is equivalent to the heat gained or lost. When a reaction is endothermic (takes in heat) ∆H is positive. When ∆H is negative, the reaction is exothermic (heat leaves the system). There are four methods for determining the change in enthalpy, the first being using average bond energies. To do this, one must calculate the energy needed to break bonds in the reactants and then form bonds in the products. This process is modeled by the equation:
∆H = ∑BE(bonds broken) + ∑BE(bonds formed)
Where the energy of the bonds broken is always negative
∑BE(bonds broken) is always positive as breaking bonds is an endothermic process, while ∑BE(bonds formed) is always negative as the formation of bonds in exothermic.
∆H can also be found using a calorimetry. Calorimetry is the measurement of heat transfer. Heat flow is measured using a calorimeter, which measures heat flow by measuring the before and after a reaction. Calculating change in enthalpy using this method requires values of heat capacity, the amount of energy required to raise the temperature of a substance by 1 K, as well as specific heat, the heat capacity of a substance per unit mass (side note: this website helped me to differentiate between specific heat and heat capacity). The heat capacity of a substance as well as the mass of the sample are two influential factors in temperature change. Thus, the change in enthalpy can be determined using this equation:
∆H = m x Cp x ∆T
The change in enthalpy is equivalent to the mass of the sample multiplied by the specific heat capacity of the substance multiplied by the change in temperature. However, calorimetry uses the energy gained or lost in the surroundings to calculate the energy gained or lost in the reaction. Due to the first law of thermodynamics (also known as the law of conservation of energy) the energy gained or lost by the system is is equal and opposite to the energy gained or lost by the surroundings.
∆Hreaction = -∆Hsurroundings
The third method of obtaining the change in enthalpy value is through Hess' Law which states that if a reaction is carried out in a series of steps, the overall change in enthalpy will be equal to the sum of enthalpy for each of the individual steps. Hess' Law is closely related to the concept of state functions. A state function is a property that depends on the present state of the system, not the path taken to arrive at that state.
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| An example of a state function. |
The fourth and final method of calculating ∆H is through enthalpies of formation. Enthalpies of formation are hypothetical values that indicate how much heat would be lost or gained during the formation of one mole of a compound from its elements in their standard states (standard states meaning 25˚C and 1 atm). Heat of formation reactions are always written so that all reactants exist as they would under standard conditions and if there was one mole of product. Using the method of enthalpy of formation, ∆H is calculate using the equation:
The heat gained or lost in the reaction is equal to the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the products minus the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the reactants. Methods of finding ∆H were covered in several thermodynamics lectures as well as in the Thermodynamics I and II worksheets.
Entropy is a statistical measure of the number of most probable distinguishable microstates or degrees of freedom available in a system. When there are a greater number of distinguishable microstates (i.e. greater entropy) the reaction is referred to as favorable or spontaneous. Thus, when the change in entropy, ∆S, is positive, the value is favorable. The change in entropy is equal to the sum of the change in entropy of the products minus the sum of the change in entropy of the reactants.
∆H = ∑n∆H˚f(products) - ∑n∆H˚f(reactants)
The heat gained or lost in the reaction is equal to the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the products minus the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the reactants. Methods of finding ∆H were covered in several thermodynamics lectures as well as in the Thermodynamics I and II worksheets.
Entropy is a statistical measure of the number of most probable distinguishable microstates or degrees of freedom available in a system. When there are a greater number of distinguishable microstates (i.e. greater entropy) the reaction is referred to as favorable or spontaneous. Thus, when the change in entropy, ∆S, is positive, the value is favorable. The change in entropy is equal to the sum of the change in entropy of the products minus the sum of the change in entropy of the reactants.
∆S = ∑S(products) - ∑S(reactants)
The second law of thermodynamics states that the entropy of the universe is constantly increasing. Reactions may have a reduction in entropy as long as the entropy of the surroundings increase. Entropy is a state function. ∆S is greater than 0 when heat is added (such as in processes like melting and vaporization). ∆S is less than zero in certain cases, such as when substances are dissolved and when solutions of gases and liquids are made. The concept of Entropy was covered in a lecture as well as in the Thermodynamics III worksheet.
Entropy plays a part in determining thermodynamic favorability. A thermodynamically favored process (or a spontaneous process) is a reaction that proceeds without any assistance from the surroundings. In reversible reactions, the thermodynamic favorability determines the direction of the reaction. There are two methods of determining thermodynamic favorability. The first, ∆Suniverse states that ∆Suniverse = ∆Ssystem + ∆Ssurroundings. If ∆Suniverse is positive, then the reaction is favorable. If a reaction increases the entropy of the system (∆Ssystem > 0) and is exothermic, then the process must be thermodynamically favored. The second method of determining favorability is through Gibbs free energy. Gibbs free energy equals the change in enthalpy minus the system temperature multiplied by the entropy change.
∆G = ∆H - T∆S
If ∆G is negative, then the reaction is thermodynamically favored. Gibbs free energy is a state function. ∆G˚f is the energy change that occurs when one mole of a compound is made from its elements in their standard states. ∆G˚f values can be used to determine the ∆G value for an equation using the equation above.
Combing over these concepts and writing this blog post, none of these concepts seem incredibly difficult. However, when it comes to applying these concepts and choosing which equation or formula to use, I find myself at a complete loss. I struggled to complete the review task chains. In the Thermodynamics II worksheet as well as in the task chains, I found myself struggling the most on problems that involved stoichiometry. I feel I have adequate understanding of Hess' law and finding ∆H based on enthalpies of formation. I have had trouble determining whether or not reactions are endothermic or exothermic. The concept of specific heat and calorimetry also baffled me a bit. As I do not feel prepared in the slightest for the upcoming test this week, I will be going over the task chains multiple times and revisiting worksheets that were completed in class. I'm sure the test will be a sobering experience that will test the self inflated image of my self worth and knowledge.
