Understanding Thermodynamics: Changes of State, Heat Capacity, and Enthalpy Reactions

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°univwill 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

Understanding Thermodynamics: Changes of State, Heat Capacity, and Enthalpy Reactions