THERMODYNAMICS!!!! Nick Fox Dan Voicu

Changes of State Order of phase changes based on temperature (lower-higher) Solid Melting/solidification Liquid Boiling/Condensation Gas For more info on the changes of state look at Review of Section 2

Specific Heat Heat capacity: amount of heat required to raise its temperature by 1K Molar heat capacity Cm: heat capacity of one mole of a substance Specific heat: the heat capacity of one gram of a substance Formula= (quantity of heat transferred)/(grams of substance) X (temperature change) CS= q/m*∆T Water Specific heat= 4.18 J/g*K

Heating Curves This is the heating curve of water (H2O) 1st plateau: 0 °C = 273.15 K 2nd plateau: 100 °C = 373.15 K http://bhs.smuhsd.org/science-dept/marcan/apchemistry/cool_phase_changes_explain.html

Calorimetry Calorimetry: experiments measuring the amount of heat transferred between the system and the surroundings Calorimeter: measures temperature change accompanying a process http://www.livingston.org/153220824141414520/blank/browse.asp?a=383&BMDRN=2000&BCOB=0&c=62270&153220824141414520Nav=%7C&NodeID=4484

Covalent Bond Energies Covalent Bond: when two atoms share a pair of valence electrons The energy that is released when covalent bonds are formed=∆H (kJ) Forming bonds releases energy which makes ∆H negative Breaking bonds takes energy in which makes ∆H positive

Enthalpy of Formation/Reaction ∆Hf It is the enthalpy change for the reaction where the substance is formed from the reactants To solve the Enthalpy of a reaction you would use this forumla: ∆Horxn= ∑n∆Hof (products) – ∑m∆Hof (reactants)

Entropy and Spontaneity Three Types of Motion Translational motion is the movement of an entire molecule in one direction Vibrational motion is the periodic movement of atoms in a molecule Rotational motion is the spinning movement of a molecule about an axis

Entropy and Spontaneity (Cont.) Three Main Factors Affecting Entropy Temperature Increasing the temperature increases the kinetic energy of the molecules, therefore the there is more movement of the molecules, increasing the randomness/disorder; decreasing the temperature has the opposite effect Volume When the molecules occupy a greater volume, there is more disorder to the system because they are more spread out and have more motional energy; less volume means more order because molecules are less spread out and have less motional energy Number or independently moving particles The more molecules there are present in the system, the greater the entropy because there is more motional energy present in a greater number of molecules The Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero. S(0 K) = 0

Entropy and Spontaneity (Cont.) One way for a reaction to be spontaneous is for the change in entropy to be positive (and the change in enthalpy to be negative) ∆S° = ∑nS°(products) - ∑mS°(reactants) where n and m are coefficients in the chemical equation We also can look at the change in entropy of the surroundings ∆Ssurr = -q /T = - ∆H/T ∆S°univ = ∆S°sys + ∆S°surr ∆S°univ will be positive for any spontaneous reaction

Gibb’s Free Energy and Spontaneity ∆G = ∆H – T∆S, under standard conditions ∆G° = ∆H° – T∆S° (standard temperature is 298K) If Gibb’s free energy is negative, the reaction is spontaneous in the forward direction If Gibb’s free energy is zero, the reaction is in equilibrium If Gibb’s free energy is positive, the forward reaction is nonspontaneous, and work must be supplied by surroundings to make it occur, however the reverse reaction is spontaneous Standard free energy of formation can be tabulated, similar to how it is done for standard enthalpy of formation ∆G° = ∑n∆G°f(products) - ∑m∆G°f(reactants) The above equation gives the standard free energy change of a reaction

Gibb’s Free Energy and Equilibrium ∆G° = -RT lnQ can be used to find the reaction quotient To find equilibrium constant: ∆G = ∆G° + RT lnK Gibb’s free energy is zero when reaction is in equilibrium 0 = ∆G° + RT lnK RT lnK = -∆G° lnK = -∆G°/RT Raise both sides as powers of e K = e-∆G°/RT A positive value for lnK means K>1, therefore the more negative the standard Gibb’s free energy, the larger the equilibrium constant If Gibb’s free energy is positive, then lnK is negative (note lnK is negative, not K) which means that K<1 (K cannot equal zero or have a negative value)

Hess’s Law Hess’s Law states that if a reaction is carried out in a series of steps, the enthalpy for the overall reaction will equal the sum of the enthalpy changes for the individual steps C(s) + O2(g)  CO2(g) ∆H1 = -393.5 kJ CO(g) + 1/2 O2(g)  CO2(g) ∆H2 = -283.0 kJ Calculate the enthalpy for the combustion of C to CO. C(s) + 1/2 O2(g)  CO(g) ∆H3 = ? In order to cancel some of the molecules, one reaction must be reversed. The second reaction should be reversed because the enthalpy will remain negative if this is done C(s) + O2(g)  CO2(g) ∆H1 = -393.5 kJ CO2(g)  CO(g) + 1/2 O2(g) ∆H2 = 283.0 kJ C(s) + 1/2 O2(g)  CO(g) ∆H3 = -110.5 kJ