Thermodynamics: Study of heat and its relationship with other forms of energy

Transcription

1 Unit 6 The 6 th planet in our solar system is Saturn Ch. 5: Thermodynamics: Study of heat and its relationship with other forms of energy Two types of energy: Kinetic: movement, active energy Potential: stored energy, gravitational and electrostatic are two examples KE=1/2mv 2 ` unit: Joules (J) =Newtons (N) times meters (m) W=Force (N) Distance (m) unit: Joules (J) Energy is the capacity to do work or to transfer heat A calorie is the energy required to raise 1g of water 1 C 1cal = J 1Cal = 1000 cal Energy is always conserved. It is never truly lost, only transferred. When energy moves into a measured system, it is considered positive change in energy When energy moves into the surroundings from the measured system, it is negative change. ΔH=final-initial +ΔH=endothermic reaction -ΔH=exothermic reaction State function: value of a state function depends on its present condition, not on the history of the sample

2 Enthalpy: thermodynamic function accounting for the heat flow in chemical changes occurring at constant pressure when no forms of work are performed other than pressure or volume work H=enthalpy ΔH=ΔE+PΔV Calorimetry: measurement of flow q= mcδt C=specific heat of the substance The specific heat of toluene (C 7 H 8 ) is 1.13J/gK. How many Joules of heat are needed to raise the temperature of 62.0g of toluene from 16.3 C to 38.8 C? This problem is simply a matter of plugging in numbers. Q=(62.0g)(1.13J/gK)( ) Q= J I found this problem in the textbook, ch. 5 number 48 Heat capacity: temperature change of an object when it absorbs a certain amount of energy Dependent on amount of material present and the nature of the material Specific heat is the heat capacity of 1g of the substance C H2O = 4.184J/gK Latent heat: energy released or absorbed during a process, like a phase change; temperature is constant Enthalpy is an extensive property- it varies depending on the amount of reactant The enthalpy of reverse reactions is the opposite of the enthalpy of the reaction; just flip the sign Enthalpy also changes depending on the state of the reactants

3 Hess s Law If a reaction is expressed as the algebraic sum of two or more reactions, then the heat of the overall reaction is the sum of the heats of the individual reactions Use Hess s Law to determine the ΔH for the reaction: CH 3 OH(g) CO(g) + 2H 2 (g) Using these reactions and their enthalpies: 2CH 3 OH(l) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(l) 2CO 2 (g) O 2 (g) + 2CO(g) 2H 2 O(g) 2H 2 (g) + O 2 (g) CH 3 OH(g) CH 3 OH(l) H 2 O(l) H 2 O(g) ΔH= kj ΔH= kj ΔH= kj ΔH= kj ΔH= kj First begin with the combustion reaction, since you need methanol on the left side, but you only want one, so divide the equation by 2, including the enthalpy. CH 3 OH(l) + 3/2 O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH= kj Unfortunately, the methanol is in liquid form, and we need it to be a gas. So we can use the 4 th equation to find the enthalpy of vaporization. CH 3 OH(g) CH 3 OH(l) ΔH= kj Next, you need to convert CO 2 to CO, so you would use the next equation, but you would divide this one by 2 as well. Now, the CO 2 cancels, and you are left with 1 O 2 on the left and 1 CO on the right CO 2 (g) ½ O 2 (g) + CO(g) ΔH= 283 kj The H 2 O can be converted to H 2 and O 2 next. 2H 2 O(g) 2H 2 (g) + O 2 (g) ΔH= kj However, the H 2 O doesn t match up, as the one above is in gaseous form and in the combustion reaction it is in liquid form. However, we need to multiply by 2. 2H 2 O(l) 2H 2 O(g) ΔH= kj

4 Next, we need to add up all the enthalpies kj kj kj kj kj = ΔH The Assyrians gave Saturn the nickname Lubadsagush, meaning oldest of the old ΔH= 90.7kJ I found the resultant reaction in the textbook in ch. 5 number 34, and calculated each of the enthalpies of the individual reactions. Enthalpy of formation: tabulated change in the enthalpy that is the result of the formation of one mole of a compound from its constituent elements at standard conditions ΔH=ΣnΔH f (products) ΣnΔH i (reactants) Standard conditions: 1atm and 298K Above, when using Hess s Law, I calculated each of the enthalpies of the individual reactions using that equation. Find the enthalpy of the combustion of methanol, as I had done previously. 2CH 3 OH(l) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(l) ΔH=ΣnΔH f (products) ΣnΔH i (reactants) ΔH=[4( ) + 2(-393.5)] [2(-238.6) + 3(0)] ΔH=( ) ( ) ΔH= Find the enthalpies of formation for the substance, and multiply by the coefficient in the BCE. Then add together the Ch. 19: Spontaneous: a process that will proceed without any outside interventions What makes a reaction spontaneous: Change in enthalpy (H)

5 Change in entropy (S) Enthalpy: reactions prefer to be exothermic; -ΔH Entropy is the degree of randomness. To calculate entropy: ΔS = ΣnS products ΣnS reactants Systems prefer chaos (+ΔS) Gases have greater entropy than liquids, and liquids have greater entropy than solids When S=0, this is a perfect crystalline shape at 0K, molecules can t move at all Entropy of the universe is increasing; it increases with any spontaneous reaction Unlike energy, enthalpy is not conserved. ΔS 0 ΔS 0 *at equilibrium ΔH 0 ΔH 0 ΔS 0 Always spontaneous ΔH 0 ΔS 0 Spontaneous at low temps. ΔH 0 ΔH 0 ΔS 0 Spontaneous at high temps. ΔH 0 ΔS 0 Never spontaneous* Given the ΔS and the ΔH, determine whether the reaction is sometimes, always, or never spontaneous, and describe the effect of temperature on the reaction s spontaneity. ΔS 0, ΔH 0 Sometimes spontaneous; as temperature increases, the likelihood of spontaneity increases ΔS 0, ΔH 0 Never spontaneous; temperature has no effect Determine if the reaction is spontaneous at standard conditions using the tabulated values of the substances and the equation ΔG=ΣnΔG f (products) ΣnΔG i (reactants). 2HNO 3 (g) + Cl 2 (g) N 2 (g) + 3O 2 (g) + 2HCl(g) Simply plug in the numbers and solve. ΔG o = [2(-95.27)] [2(-73.94)]

6 ΔG o = Spontaneous I created this question. Gibb s Free Energy Equation: relates entropy temperature and enthalpy to predict spontaneity at standard conditions; the o indicates at standard conditions ΔG o =ΔH o -TΔS o If ΔG o 0, the reaction is spontaneous If ΔG o 0, the reaction is not spontaneous* (*the forward reaction isn t spontaneous, the reverse reaction is spontaneous) If ΔG o =0, the reaction is at equilibrium Determine whether the reaction is spontaneous at 247 C using Gibb s free energy equation, given that ΔH= kJ/mol and ΔS= and doesn t vary with changes in temperature. N 2 (g) + 3H 2 (g) 2NH 3 (g) = 520 K ΔG o =ΔH o -TΔS o ΔG= (-92.38) (520)( ) ΔG= kj Not spontaneous I created this question and calculated the change in enthalpy and entropy used. To find ΔG not at standard conditions: ΔG=ΔG o + RTlnQ R=8.314 J/molK Q=reaction quotient

7 This can also relate to equilibrium: 0=ΔG o + RTlnK eq K eq = e -ΔG/RT Jupiter was named after the titan depicted here on the left. He was the titan of time, and the oldest and most powerful. Legend has it that he ate his children.