Bellwork Wednesday Determine if the following are endothermic or exothermic. H = -226.55 kJ/mol H = -6.89 kJ/mol H = +588.457 kJ/mol H = +0.2275 kJ/mol

Unit 10 Ch 17 Part 4

Heat of vaporization and heat of condensation. They are equal amounts! Heat of fusion and heat of solidification. They are equal amounts!

Heating/Cooling Curve shows energy changes Heating/Cooling Curve shows energy changes. All matter follows curve when energy is added or lost. Csteam= 2.1 J/g oC or .5 cal/goC ∆Hvap =∆Hcond =2260J/g Cwater= 4.18 J/g oC or 1.00 cal/goC ∆Hfus = ∆Hsolid= 334J/g CIce= 2.0 J/g oC or .5 cal/goC

At these points, temperature is CONSTANT!!!

17. 3 Phase Changes -change of physical state Ex 17.3 Phase Changes -change of physical state Ex. Melting, freezing, evaporation, condensation, sublimation, deposition -The temperature of a substance does not change during a phase change.

Question Why is the slope between ice and vapor so long on the heating curve?

Sublimation- Change of a solid to a gas without going through the liquid phase Deposition is the opposite Ex. Iodine, ice cube “shrinkage”, freeze drying, freezer burn

Exothermic - energy is a product -∆H Endothermic - energy is a reactant +∆H  H is also called the heat of reaction or Enthalpy

Heat Horizontal portions of the curve indicate a physical state change. Notice that the temperature does not change in these regions! However, there a a change in particle position resulting in a change in potential energy. Heating/Cooling Curve -Shows energy changes of matter. All matter follows a curve when energy is added or lost. Slope portions show temperature change which indicates a change in kinetic energy as well.

Ex. How much heat, in calories, is needed to melt 150. g of ice at 0oC? No temp change (just phase change) Molar heat of fusion = 334 J/g for H2O q=(150.g)(334 J/g) = 50100J 1 cal =11985.645J 4.18 J = 1.20 x 104 calories

Ex. How much heat, in calories, is needed to heat the liquid water in the above problem to 20.oC? No phase change (just temp. change) q = mC T q = (150.g)(1.00 cal/goC)(20.C) q = 3.00 x 103 calories

Ex. A 50. g sample of ice is held at -10. oC Ex. A 50. g sample of ice is held at -10.oC. Will 270 calories of heat be sufficient to raise the temperature of the ice to 0oC? just temp change q = mC T q = (50.g) (0.5 cal/goC)(10.oC) q = 250 calories, Yes

Ex. How many calories are released when 36 g of steam at 100 Ex. How many calories are released when 36 g of steam at 100.oC condenses to water at 100.oC? just phase change q=mHv q= (36g) (2260 J/g)= 81360J 1 cal =19464.11 4.18J = 1.9 x 104 calories

Ex. How many calories are needed to convert 5 Ex. How many calories are needed to convert 5.0 g of ice at -15oC to steam at 130oC? Phase changes and temp changes q = (5.0g)(0.5cal/goC)(15oC) = 38 calories q= (5.0g)(334 J/g)= 1670 J 1 cal = 399.5 4.18J = 4.0 x 102 calories (continued)

q =( 5. 0g) (1. 0 cal/goC)(100oC) = 500 calories q= (5 q =( 5.0g) (1.0 cal/goC)(100oC) = 500 calories q= (5.0g )(2260 J/g)= 11300J 1cal = 2703.349 4.18J = 2.7 x 103 calories q = (5.0g) (1.0cal/goC)(30oC) = 150 calores q = 38 + 400 + 500 + 2700 + 75 = 3713 calories

Standard Heat of Formation of a compound ( Hfo) * Hfo of a free element in its standard state is zero. *This is another way to calculate  H for a reaction.  H =  Hfo products -  Hforeactants *(a table of standard heats of formation will be provided)

∆H = 178.4 kJ (endothermic-absorbs energy) Ex. 1. Calculate  H for the following reaction: CaCO3(s)  CaO(s) + CO2(g) ∆Hfo values: CaCO3 = -1207.0 kJ/mol CaO = -635.1 kJ/mol CO2 = -393.5 kJ/mol ∆H = [1∙ -635.1 + 1 ∙ -393.5)] – [1 ∙ -1207.0] ∆H = 178.4 kJ (endothermic-absorbs energy)

∆H = [2 ∙ (-241.8)] – [2∙ 0 + 1 ∙0] ∆H = -483.6 kJ Calculate the heat of reaction for the following reaction: 2H2(g) + O2(g)  2H2O(g) ∆Hfo values: H2O(g) = -241.8 kJ/mol ∆H = [2 ∙ (-241.8)] – [2∙ 0 + 1 ∙0] ∆H = -483.6 kJ (exothermic- releases Energy) Remember to multiply heat values by coefficients!!!

Bellwork Summary Part 4 first!!! Friday Calculate the enthalpy of formation for the reaction below. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) H HCl = -199.0 kJ/mol H ZnCl2 = -766.5 kJ/mol

Review day Another multi step problem How much energy in joules is required to convert 50.5 g of ice at -5.5 °C to water at 85.0 °C?