thermochemistry review Flashcards
Endo and Exothermic
endothermic: heat is absorbed and enter so the test tube becomes cold –> the reactants to products gets an energy input and a positive change in heat/heat enters
exothermic: heat is released and energy is gived off so the test tube feels warm –> reactants to products has a decrease in the energy/energy is released so negative change in heat
Energy
Energy:
- Energy is the ability to do work (both physical and chemical)
- There are several forms of energy: kinetic, potential, chemical, thermal, light, sound and electricity.
- Each form of energy can be converted into another.
- Kinetic energy: Energy of motion.
- Potential energy: Stored energy (for example, energy possessed by a body because of its position)
- The unit of energy used to measure the amount of energy involved in chemical reactions: Joules (J)
1J = kg m2
s2
Recall: atoms can move, which means that they possess kinetic energy.
Molecules can also move in 3 types of motion:
- Translational motion: molecules move in linear pathways from one place to another.
- Rotational motion: molecules rotate about an axis through its center of mass.
- Vibrational motion: the bonds within the molecules are always stretching and compressing.
Thermodynamics and definitions
The First Law of Thermodynamics: The law of conservation of energy:
- Energy is neither created or destroyed but is simply converted from one form to another.
Temperature (T): is a measure of the average kinetic energy of molecules, ions or atoms. Temperature is an intensive property because it is independent of mass.
Heat (Q): energy that flows from one substance to another because of a temperature difference. Heat is an extensive property because it is dependent of mass.
When there is a difference in temperature, heat transfer will take place. The key is that heat is lost or gained. Things NEVER become cold! Note: Heat is always transferred from a high temperature to a lower temperature. The bigger the difference in temperature between two objects, the higher the amounts of heat transfer.
Thermochemistry deals with the transfers of energy (heat) involved in chemical reaction. Almost all reactions involve changes in energy (endothermic or exothermic)
Examples:
Photosynthesis: 6co2 + 6H2O + energy –> glucose + 6CO2 (endo since energy in reactants)
Rocket fuel: ? –? ? + enegry
Combustion reactions: CxHy + O2 –> H2o + CO2 + energy
Cold Packs: energy + NH4CL + Ba(OH)2 –> BaCl2 + NH4OH
System: particular part of the universe that we wish to study (chemical components involved in the chemical reactions). (chemical reaction)
Surroundings: whatever is entirely outside defined system (i.e. everything except system itself). (air surrounding)
Boundary: Separates system from its surrounding. (test tube)
It is possible to prevent heat transfer by INSULATING the system from its surrounding.
Open system: can exchange mass and energy usually in the form of heat with its surroundings (ex. test tube)
close system: allows the transfer of energy (heat) but not mass (ex. with stopper)
isolated system: doesn’t allow the transfer of either mass or energy (like a coffee cup)
Specific heat capactiy
Specific heat capacity (symbol: c and unit: J/g°C) – amount of energy (J) required to change 1g of a substance by 1°C. The specific heat capacity depends on the physical state of the substance (solid, liquid, gas) and type of substance.
Three factors affect heat transfers:
- Temperature in° C.
- Mass (m) of substance (in grams).
- Substance type – Specific heat capacity (c) in J/g°C. Lists for specific heat capacities of various substances can be found in the data booklet.
Q (heat energy) = m c ΔT
Where : ΔT = Tf -Ti
ΔT results in a negative number than heat has been gained from the system.
ΔT results in a positive number than heat has been lost by the system.
Specific heat capacity = Q (heat energy) / m (mass of sample) * T (temp in Celsius) –> so units are J/gc
Other ways in which specific heat capacity can be expressed:
- Heat Capacity = specific heat * mass Units: J/°C
- Molar Heat Capacity = specific heat * molar mass Units: J/mol°C
Recall: Energy is always conserved! Remember the Law of conservation of energy.
- Therefore energy lost by the system is absorbed by surroundings.
- Therefore absorbed by system comes from surroundings.
Calorimetry
So far, we have covered the heat evolved or absorbed by a system can be calculated using mass, specific heat of the substance and temperature change. However, we need a system that will allow us to find the heat of a specific reaction.
A Calorimeter is used in experiments to collect heat data. The simplest calorimeter is made of a Styrofoam cup, a lid, stirrer and a thermometer (basically Tim Horton provides most of the material!).
In a calorimeter, a chamber is isolated from all other components. The chamber is where the reaction takes place. The chamber is immersed in water. The change in temperature of water dictates how much energy is lost (endo) or gained (exo) by the reaction.
heat lost by the reaction = heat gained by water (exo) so -Qreaction = +Qsink
heat gained by the reaction = heat lost by water (endo) so +Qreaction = -Qsink
ANY CHANGE IN THE SYSTEM IS ACCOMPANIED BY AN EQUAL AND OPPOSITE CHANGE IN THE SURROUNDINGS.
Q is dependent on the mass, temp and heat capacity (the more mass, higher temp and higher heat capacity the more energy contained)
look at calculations in note
Enthalpy
Every substance involved in a chemical reaction or physical process has particular heat content. Heat content is called enthalpy (H). Enthalpy is measured in terms of a heat differences between the initial and final states of any process at constant pressure.
(q/n)
A –> B ΔH = Hp - Hr
Recall:
ΔH= negative value indicates the reaction is exothermic
ΔH = positive value indicates the reaction is endothermic
Thermochemical equations: ΔH value is always based on molar ratios given in the equation. Change in coefficients will also change the heat values.
in units KJ or KJ/mol
if given the enthalpy of one reactant and asked for the enthaply of other reatants and products then do the enthalpy of one/moles of the needed but if asked for enthalpy of same type but different amount then use:
ΔH1 = ΔH2
n1 n2
(q is heat while h is enthalpy (heat content))
Calorimetry and Enthalpy Summary
Specific heat capacity (soup cup calaroimeter): small c –> c=q/mΔT units are in J/gdegreecelcius
Heat capacity (bomb calorimeter): big C so C = q/ΔT units are J/degree celcius
Molar heat capacity: Cm = q/nΔT
Because of the law of conservation of energy, the total thermal energy of the system and its surroundings remains constant therefore:
qsystem+ qsurroundings= 0
qsystem= -qsurroundings
qreaction = -qsink/H2O
Bomb calorimeter – a bomb containing the reactant is sealed and placed within a calorimeter containing a know quantity of water. The bomb has access to oxygen for combustion to take place. From this the heat capacity of the bomb calorimeter can be calculated.
Heat capacity of a bomb calorimeter (Ccal) : is determined by taking a specific mass of a substance that releases a known quantity of heat and measuring the change in temperature.
Ccal: Joules per degrees Celsius or Kelvin –> J/oC
Once, this value has been determined then one can measure the change in temperatures of other reactions in the bomb calorimeter and calculate the heat evolved.
qrxn = -Ccal x ΔT (the c is in units J/degree celcius so it becomes J as final unit for reaction)
equarion that is also used is ΔH = -mcΔT/n or just cΔT when moles are involved
Enthalpy change involving chemical reactions:
∆H=-(mc∆T)sink/nLR –> use if J/gdegree celcius
OR ∆H=-(C∆T)sink/nLR –> use if J/degree celcius and if bomb calorimeter
nLR means moles of limiting reagent
Important note: Most reactions take place as dilute solutions (dissolved in lots of water). Thus it is safe to assume that the solution has the same density and specific heat of water.
rmbr to convert any units –> 1KJ = 1000J etc.
j or jk/mol is ∆H
use mcat for normal problems and cat for bomb calorimeters
thermochemical equations are the same equation except you write + enthalpy or on side write ∆H = ?
calculate limiting moles by finding both substances moles and using ratios to see if you need more or less than what you have
distinguish between the sink and reactants, the reactants are responsible for change in temp, consider if bomb calorimeter or not (specific heat capacity if not and heat capacity if it is), find LR, consider type of reaction to find moles (either m/MM or c*v)
can find -mcat for one substance and get the ∆H of something then use it to find any other information. equation can also be ∆H = -(qcalorimeter + qreaction/h2o)/n if given such information. can also do -qrxn = qsink. Q is heat while ∆H is enthalpy. can do ratios of ∆H1/n1 = ∆H2/n2 to find out information on same substance or divide q or ∆H/ n if different substances. if asking for mass rearrange question to either get mass or moles (then to mass).
usually convert from kj to j by *1000j/kj
** heat(q) is the transfer of energy due to a temperature difference
** enthalpy (H) is a change in the amount of heat in a system at constant pressure
Enthalpy and Hess’s law
Chemical energy - form of potential energy that arises from the forces of attraction (chemical bonds) that bind atoms together in compounds. i.e. enthalpy of reaction.
Enthalpy (H) - heat content of a substance involved in a chemical reaction or a physical process. Measures Hinitial and Hfinal of a process at constant pressure.
Therefore, we are not worried about what happens in between or the path taken (state function).
Therefore: ∆H = Hf -Hi; where f = products and i = reactants.
Almost every substance involved in a chemical reaction or a physical process has a particular heat content (∆H: units = kJ). (mol is optional except for formation)
Enthalpy in physical processes:
∆Hvap, ∆Hcond, ∆Hmelt aka ∆Hfus, ∆Hfre, ∆Hat (amount of energy needed to rupture all the chemical bonds in one mole of gaseous molecules to give gaseous atoms as products –> energy needed to break bonds breaking into initial element (atomization energy)
Enthalpy in chemical processes:
∆Hsol, ∆Hcomb, ∆Hneut,
- ∆Hdegreef = enthalpy of formation (kJ/mol) –> degree means at satp –> its the energy required to create one mole of a substance from its elements in their most energetically stable state (look at chart values of formation) –> you start by writing the product of the equation, and then figure out the reactants by using any coefs to balance
Any chemical equation written with ∆H value is called: Thermochemical Equation. if negative ∆H value or is written on right side of equation then exothermic. if on left or positive value then endothermic.
THERMOCHEMICAL EQUATIONS: Guidelines in writing and interpreting:
- Stoichiometric coefficients refer to the number of moles of each substance.
- When you reverse an equation - you are really changing the roles of the reactants and products. Therefore the magnitude of ∆H remains the same but its sign changes.
- If you multiply both sides of the equation by a factor n, then the enthalpy value must be multiplied by the same amount. ( n x ∆H = n∆H)
- When writing thermochemical equations, we must specify the physical states of all reactants and products (help to determine enthalpy changes).
Standard heats of reaction: Expressing the heat/enthalpy of a reaction at a specific standard of reference: 25 degrees Celsius and 101.3 kPa.
HESS’S LAW (Law of heat summation)
Hess’s law: for any reaction that can be written in steps, the standard heat of reaction is the same as the sum of the standard heats of reaction for the steps.
- use the given equations to manipulate them into giving the rxn/product/needed item enthalpy. cancel things out that are on opp sides of arrow, then add (make sure to separate the enthalpy from the equation it self)
Standard Heats of Formation
Standard heat of formation: is the amount of heat absorbed or evolved when ONE MOLE of the compound is formed from its ELEMENT in their standard (NATURAL) state. ∆Hf expressed in kJ/mol
Recall: The diatomic gases: I Bring Clay For Our New House; these gases are in their elemental form. Important Note: ∆Hf for any element in its standard state = ZERO so for diatomic elements ∆Hf is zero
Direct method: HESS’S LAW WITH STANDARD HEATS OF FORMATION
Recall: ∆Hrxn = Hproducts - Hreactants
Using ∆Hdegreef can calculate the ∆Hr of a reaction.
aA + bB 🡪 nN + mM
Note: a, b, n, m = coefficients. A,B,N,M = substances
Since the values for the enthalpy of formation is in kJ/mol and the enthalpy of reaction is expressed in kJ, it is important to take the coefficients (moles) into account during the calculations.
∆H = [ n∆Hf (N) + m∆Hf (M) ] - [ a∆Hf (A) + b∆Hf (B) ]
Can be given such chemical equation and asked to find the ∆H of the reaction.
NH3(g) + 7/4O2(g) –> NO2(g) + 3/2H2O(g) ∆H =?
Given ∆H: NH3(g) = -46.15 KJ, NO2(g) = 33.18 kJ, H2O(g) = -241.6kJ
Using the Indirect Method…or the Long way. You would write formation reactions for NH3, NO2, and H2O to find target equation. So create the formation reactions for the necessary compounds with the ∆H written on the side then figure out the ways to manipulate them to match the moles and sides with the balanced equation given. Then cancel and add them up including the ∆H to get the ∆H of the target equation.
Using the Direct Method: you would calculate the ∆Hreaction by using the equation ∆Hreaction = ∆H = [ n∆Hf (N) + m∆Hf (M) ] - [ a∆Hf (A) + b∆Hf (B) ] making sure that ∆Hf is written in kj/mol even if given as kj before. the answer should match with the answer from the indirect method. the ∆H values for the elements may be given or have to find on sheet.
always cancel out units, rmbr sig figs.
Enthalpy of Combustion
In enthalpy diagram of combustion the products are shown at lower Ep value and ∆H is negative since combustion is exothermic.
if given enthalpy of reaction and asked for enthalpy change for a mole of each gas you do ∆H/n and n is dependent on the coef in the equation –> if asked for the enthalpy change of same substance then do ∆H1/n1 = ∆H2/n2
use stoich to convert from different units like moles and grams
∆H combustion can be used in enthalpy of formation questions –> write out the equations and manipulate however to match the target equation, then cancel and add to get the target equation and ∆H value –> do not use direct method for combustion values
It is often difficult to experimentally determine the formation of values of certain combination of elements. For example, combining carbon and hydrogen will not simply create a hydrocarbon. However by burning the compound in question, energy data can be of use in finding enthalpies of formation.
Note: * both enthalpy of formation and enthalpy of combustion can be used to describe the formation of carbon dioxide from its elements and for water from its elements.
Bond Dissociation Energy
Calorimeter: ∆H = -mc∆t/nlr or -c∆T/nlr
∆H –> indirect method is sub equations –> target
∆H –> direct method is sum of n *∆Hdegreef products - sum of n * ∆Hdegreef reactants
∆Hdegreef = via ∆H combustion
Now ∆H bia bond dissociation energies (given values)
Energy must be supplied when chemical bonds are broken.
Energy must be released when chemical bonds are formed. (bond energy)
Principle of Additivity of Bond Energies:
Bond Dissociation Energy (B.D.E.): is the energy required to break all bonds in one mole of gaseous molecules that are in the lowest/most stable energy states. The bond energy of a molecule is the sum of the bond energies between the individual atoms making up the molecule.
Bond energy is an average of bond energies for a given bond found in a variety of different molecules. Because of this, enthalpy of reaction found using the bond energy data can be somewhat different from the enthalpy of reaction found using heat of formation data.
Note: The bond energies of ELEMENTS is NOT equal to ZERO
ΔH = Σ Bond energies of bonds broken (input or reactants) – Σ Bond Energies of bonds formed (output or products)
breaking a bond is input of energy (+) and forming a bond is output of energy (-)
Bond Energy: the amount of energy needed to break one mole of a chemical bond to give electrically neutral fragments (atoms not diatomics), assuming reactants are in the gaseous state.
To figure out the enthalpy of an reaction follow steps:
- Draw Lewis structures for all compounds.
- Count the number of bonds broken and the number of bonds formed.
- Obtain total energy input and total energy released.
- substitute into equation –> ΔH = Σ Bond energies of bonds broken (input or reactants) – Σ Bond Energies of bonds formed (output or products)
(draw out lewis structure –> multiply the number of moles, number of bonds and the given bond enthalpy to start totalling)
- can either do bond enthalpies or can even use heats of formation (products minutes reactants)
Enthalpy of Solvation
Enthalpy of solvation – is a physical change
- the change in enthalpy when 1 mol of a substance is dissolved in a large excess of a pure solvent
- ∆Hsoln
Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration.
ion-dipole forces are created which creates a stable state
Energy change associated with creating solutions by adding water to solid compounds.
The steps involved:
- breaking intermolecular attractions between the solvent (break hydrogen bonds between water molecules. –> Energy input, endothermic
- Breaking attractive forces (ionic bonds) within the solute to separate ions (lattice energy). –> Energy input, endothermic –> Lattice enthalpy –> Separating an ionic solute in the solid state into its separated anions and cations in the gaseous state
- Creating the solution or hydrated ions or interactions between the ions and solvent –> Energy output, exothermic –> Enthalpy of hydration: the enthalpy change associated with 1 mol of the gaseous ion when added to water to form a dilute solution –> This is when water molecules orient themselves to create ion-dipole attractions to create a hydration shell (solvation shell).
Note:
Magnitude of enthalpy of hydration is influenced by the charge and size of the ion.
- Size: as you move down a group in the Periodic table, the larger the ionic radius the lower the enthalpy of hydration.
- Charge: an increase in charge for CATIONS combined with a decrease in size results in a significantly larger enthalpy of hydration –> Al3+> Mg2+> Na+
- larger anion/size, lower values of hydration but smaller radius, stronger attraction and higher values
The overall enthalpy change that accompanies the formation of a solution, ΔHsoln, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (ΔHsoln < 0) processes favor solution formation.
…or in other words, overall enthalpy of solvation depends on how endothermic the energy input is to how exothermic the energy of solvation
(overall process enthalpy is the sum of the solvent and solute enthalpies and mixture enthalpy)
Summary of equations
products having more potential enegry is pos enthalpy and endo while them having less compared to reactants is negative and exo
q= mc∆T
qsink = msinkcsink∆tsink
-qrxn = qsink
qrxn = -msinkcsink∆tsink
∆H - -msink/h2oc∆T/nLR
∆H for bomb calorimter is ∆H = -Ccal∆T/nLR
moles is of the actual thing causing a reaction while sink is water or surroundings –> mass of something other than h2o is moles
if given two concentration and volumes, add the volumes together to get mass (because assuming it dilute so D = 1g/1mL). C is of water since dilute. n is found by c*v and figure out which is limiting (mole ratios).
Moles can be found by n = m/M, n = c*v, n = PV/RT, n = # of partcles/ avagadros number
∆H –> calorimeter –> ∆H = -mc∆T/n
—> bomb –> ∆H = -C∆T/n
—> nuetralization –> ∆H = -mc∆T/n
∆Hrxn = [ n∆Hf products] - [ n∆Hf reactants ] –> formation values given and is the creation of one mole a substance form its elements
∆Hrxn = input BDE - output BDE
∆H*f from combustion rxns (given combustion equations or have to create some formation ones for water and co2, then manipulate for target equation, then add and cancel)
∆H hess’s law –> target equation (formation equations to target)
