thermo equations
thermo equations
internal energy:
\(\Delta U = U_{\rm f} - U_{\rm i}\)
(Note that U, is also shown as E in many books and often on Quest)
First Law of Thermodynamics
\(\Delta U = q + w\)
(this is a mathematical version of the first law)
Substances not changing phase:
\(q = m\cdot C_{\rm s}\cdot \Delta T\) (per gram)
\(q = n\cdot C_{\rm m}\cdot \Delta T\) (per mol)
Substances that are changing phase:
\(q = m\cdot\Delta H_{\rm trans}\) (per gram)
\(q = n\cdot\Delta H_{\rm trans}\) (per mol)
"trans" is for phase transition and will be one of the following: fusion, freezing, evaporation, condensation, sublimation, or deposition
\(w = - P_{\rm ext}\Delta V\) (constant external pressure)
or
\(w = -\Delta n_{\rm gas}RT\) (also constant external pressure)
Δngas = (#mol gas prod) - (#mol gas react)
Isothermal expansion requires a matched set of q and w values. This is because \(\Delta U = 0\) throughout the process due to the fact that the temperature never changes. Heat and work are perfectly matched in magnitude (opposite in signs). Pressure and volume are changing throughout but are linked via the ideal gas law and specifically: \(P = nRT/V\).
\(dw = - PdV\) (infintesimal amount of work)
\(\int_1^2 dw = -nRT \int_1^2 {dV\over V} \) (integrate state1 to state2)
This leads to both the answer for work and for heat for isothermal expansion:
\[w = -nRT \ln \left({V_2\over V_1}\right)\]
\[q = +nRT \ln \left({V_2\over V_1}\right)\]
\(H = U + PV\)
\(\Delta H = \Delta U + P\Delta V\) (constant pressure)
\(\Delta U = \Delta H - P\Delta V\)
\(\Delta U = \Delta H -\Delta nRT\)
conditional heat flow
\(\Delta H = q_{\rm P}\) (constant pressure)
\(\Delta U = q_{\rm V}\) (constant volume)
\(q_{\rm cal} = -q_{\rm sys}\)
\(q_{\rm cal} = C_{\rm cal}\Delta T\)
coffee-cup Calorimetry (constant pressure, qP)
\(q_{\rm cal} = m_{\rm water}\cdot C_{\rm s,water}\cdot \Delta T\)
bomb Calorimetry (constant volume, qV)
\(q_{\rm cal} = \)
\(m_{\rm water}\cdot C_{\rm s,water}\cdot \Delta T + C_{\rm hardware}\Delta T\)
\(\Delta H_{\rm rxn} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots\)
flip and scale various reactions to match the target reaction
\(\Delta H^\circ_{\rm rxn} = \sum n\Delta H^\circ_{\rm f}{\rm (products)} \)
\(- \sum n\Delta H^\circ_{\rm f}{\rm (reactants)} \)
look up heat of formation data from a table
\(\Delta H^\circ_{\rm rxn} = \sum n\Delta H_{\rm bond}{\rm (breaking)} \)
\(- \sum n\Delta H_{\rm bond}{\rm (making)} \)
look up bond energies from a table
entropy is a measure of energy dispersal and the Second Law states that universal entropy must increase for any spontaneous process:
\(\Delta S_{\rm univ} = \Delta S_{\rm sys} + \Delta S_{\rm surr}\)
entropy (S)
\(S = k \ln \Omega\) Boltzmann formula
k is the Boltzmann Constant with a value of 1.38 × 10−23 J/K which is the same as the universal gas constant, R, divided by Avogadro's number, NA.
entropy change via reversible heat flow
\(\displaystyle\Delta S = {q_{\rm rev}\over T}\) (constant T)
entropy change with changing T
\(\displaystyle\Delta S = n\;C_{\rm m} \ln\left(T_{\rm f}\over T_{\rm i}\right)\) (changing T, constant P)
entropy change for phase change
\(\displaystyle\Delta S = {\Delta H_{\rm trans}\over T_{\rm trans}}\)
entropy change for isothermal expansion
\(\displaystyle\Delta S = nR \ln \left({V_2\over V_1}\right)\)
third law: the entropy of a perfectly crystalline substance is zero
\(\Delta S^\circ_{\rm rxn} = \sum n S^\circ{\rm (products)} \) \(- \sum n S^\circ{\rm (reactants)} \)
Note: S° is standard absolute entropy and is value you find in a thermodynamic table. It is NOT a "formation" reaction - notice the fact that there is no Δ there or a subscript "f". All values for substances that are solids, liquids, or gases are positive - aka: "absolute"
\(G = H - TS\) (definition)
\(\Delta G = \Delta H - T\Delta S\) (constant pressure)
using formation data from a table...
\(\Delta G^\circ_{\rm rxn} = \sum n\Delta G^\circ_{\rm f}{\rm (products)} \) \(- \sum n\Delta G^\circ_{\rm f}{\rm (reactants)} \)
equilibrium is a state in which \(\Delta S_{\rm univ}=0\). We generally "recast" that idea into the free energy of the system and say that equilibrium is achieved when the free energy of the system is at a minimum and \(\Delta G = 0\).
So IF you have equilibrium in play...
\(\Delta H = T_{\rm eq}\Delta S\)
\(T_{\rm eq}\) is the temperature at which equilibrium occurs. It is also the transition temperature where a process/reaction goes from spontaneous to non-spontaneous and vice-versa.
A chemical system will tend to change from one state to another via whatever means are available such that universal entropy is increased. Maximum universal entropy change will coincide with the final state that tends to have the lowest possible total free energy of all the components. If that state is reached, then at that exact composition, all the reactants have the exact total free energy as all the products. That also means that ∆G = 0 at this point. This is the point at which equilibrium has been reached. There is no longer any tendency for the reaction to go in a net forward direction or net reverse direction. Either way would actually increase the free energy of the system. The equilibrium condition is basically what all reactions are "trying" to achieve. The "drive" to get to that position/condition is the change in free energy for the reaction, ∆G. If that quantity is positive, then the reaction will be driven in reverse from the way it is written on the page - non-spontaneous forward direction means spontaneous reverse direction. If the quantity is negative (-∆G) then the reaction will be driven forward.
One last thing... as reactions proceed forward, the amounts of reactants and products are constantly changing. The reactants decrease and the products increase in amount for a reaction going forward. Realize that free energy is tied to the amounts of substance as well as what the substance is. When a compound is decreasing, its free energy is decreasing as well because there is less and less of it. If it all reacts, then the free energy of that substance is now zero because it is gone. Of course, on the other side of the arrow are the products. They were at zero free energy when the reaction started (because nothing was there), but then the products begin to appear and the free energy climbs. The reaction will effectively stop when the total of all the free energies of the reactants equals the total free energy of the products. If the free energies match, then ∆G must be equal to zero, and the system has reached equilibrium.
So what does that mean? Well, no matter what your ∆G is for a reaction - whether it is positive or negative, the absolute value of ∆G will decrease and continue to decrease as the reaction proceeds forward (or backwards). Once the reaction hits that spot where ∆G = 0, then equilibrium is now "in play". Lots and lots of wonderful relationships and equations become important once you have equilibrium. We save all that "wonderfulness" for you in CH302.
Read the the previous paragraphs over and over and TRY to understand what they are saying. Understanding this is the KEY to having a good understanding of just what drives everything around us. Thermodynamics rules the driving force of all things.
