Thermochemistry Unit 10 Lesson 2

Thermochemistry  Study of energy changes that occur during chemical reactions and changes of state

Endothermic Process that absorbs/gains heat Heat flows into the system (+q) Heat removed from the surroundings (-q) Surroundings feel cold

Exothermic Process that releases/produces heat Heat flows out of the system (-q) Heat added to the surroundings (+q) Surroundings may feel warm/hot

assuming pressure is constant, ΔH = q Enthalpy (H)  the heat content of a system units for enthalpy are kJ assuming pressure is constant, ΔH = q ΔH  change in enthalpy Exothermic = - ΔH Endothermic = + ΔH

ΔH & Changes of State Change in state always involves a change in heat energy

Melting (fusion) Vaporization Sublimation Endothermic changes of state absorb heat energy, increasing in temperature Melting (fusion) Vaporization Sublimation

As more heat is added the temperature increases, meaning the particles are moving faster. To make a substance melt or vaporize you continuously add heat to: *increase the speed of the molecules *break the (IMFs) attractions between molecules causing them to separate

q = m·Hf Melting (fusion) The Heat of Fusion is the amount of heat needed to melt one gram of a substance at standard pressure. Heat of Fusion (Hf) of water is 334 J/g. LOOK in your ref. packet! q = m·Hf

How much heat is added when 15.7 moles of water melts? q = m·Hf 15.7 mol 1 mol 18.016 g = 282.85 g q = (282.85 g)(334 J/g) q = 94,471 J q = 94.5 kJ

In the diagram, how much heat is needed for the ice to melt? q = m · Hf Hf (water) = 334 J/g in ref. packet! q = 2000 g · 334 J/g q = 668000 J = 670 kJ

q = m·Hv Vaporization/Boiling The Heat of Vaporization is the amount of heat needed for one gram of the substance to change from liquid to gas at standard pressure. Heat of Vaporization (Hv) of water is 2260 J/g. LOOK in your ref. packet! q = m·Hv

You NEED to recognize that 100°C is the boiling point for water How much heat is absorbed when 24.8 g H2O at 100.0 oC is converted to steam at 100.0 oC? You NEED to recognize that 100°C is the boiling point for water q = m·Hv q = (24.8 g)(2260 J/g) q = 56,048 J q = 56.0 kJ

Freezing Condensation Deposition Exothermic changes of state release heat energy, decreasing in temperature Freezing Condensation Deposition

As heat is removed the temperature decreases, meaning the particles are moving slower. To make a substance freeze or condense you continuously remove heat to: *decrease the speed of the molecules *increase the (IMFs) attractions between molecules causing them to clump

Heating/Cooling Curves

Heat can be transferred even if there is no change in state What happens when you add heat to a solid? HEATING curve for water 100 Time Temp Heat can be transferred even if there is no change in state Ice at –30.0°C absorbs heat but stays SOLID while temperature rises to 0°C.

What happens as a solid melts? HEATING curve for water During a phase change, the temperature remains CONSTANT. This is because the heat being added is used to break IMFs holding the molecules together not to raise the temperature. 100 Time Temp At 0°C ice begins to melt

What happens when you add heat to a liquid? HEATING curve for water 100 Time Temp Water remains a LIQUID as it absorbs heat and goes from 0°C to 100°C.

What happens a liquid begins to change to a gas? HEATING curve for water 100 Time Temp At 100°C water begins to boil. The temperature will be constant until phase change is complete.

What happens you add heat to gas? HEATING curve for water 100 Time Temp After 100°C, the temperature of steam will continue to increase as heat is added.

Heating curves are endothermic as they absorb/gain heat. 100 Time Temp boiling gas liquid/gas Heating curves are endothermic as they absorb/gain heat. liquid melting solid/liquid solid

During a phase, KE is increasing (temp is increasing). What happens to Kinetic energy (KE) & potential energy (PE) at each stage of the curve? HEATING curve During a phase, KE is increasing (temp is increasing). 100 Time Temp KE constant PE KE KE During a phase change, KE is constant and PE is increasing. PE KE constant KE

Calculating heat for each segment of the heating curve for water. 100 Time Temp The temperature of the ice is increasing. The specific heat for ice is 2.05 J/g°C. LOOK in your ref. packet! q1 = mcpΔT

LOOK in your ref. packet for Cp for iron Calculate the amount of energy it takes to heat iron from 0°C to 557°C. (remains a solid the entire time) LOOK in your ref. packet for Cp for iron q = mcpΔT q = (2000 g) (0.449 J/g°C) (557°C) q = 500,000 J = 5.0 x 10 5J

Calculating heat for each segment of the heating curve for water. 100 Time Temp A phase change occurs at a constant temperature. Use the heat of fusion since ice is melting. Hf (water) = 334 J/g q2 = m·Hf LOOK in your ref. packet! q1 = mcpΔT

Calculating heat for each segment of the heating curve for water. 100 Time Temp q3 = mcpΔT The temperature of the water is increasing. The specific heat for water is 4.18 J/g°C. q2 = m·Hf LOOK in your ref. packet! q1 = mcpΔT

Calculating heat for each segment of the heating curve for water. 100 Time Temp q4 = m·Hv q3 = mcpΔT A phase change occurs at a constant temperature. Use the heat of vaporization. Hv (water) = 2260 J/g q2 = m·Hf LOOK in your ref. packet! q1 = mcpΔT

Calculating heat for each segment of the heating curve for water. q5 = mcpΔT 100 Time Temp q4 = m·Hv q3 = mcpΔT The temperature of the steam is increasing. The specific heat for steam is 2.02 J/g°C. q2 = m·Hf LOOK in your ref. packet! q1 = mcpΔT

Calculating heat for each segment of the heating curve for water. q5 = mcpΔT 100 Time Temp q4 = m·Hv q3 = mcpΔT Use q = mcpΔT when the temperature is changing. Use q = m·Hf or q = m·Hv when there is a phase change. q2 = m·Hf q1 = mcpΔT

The total amount of heat absorbed is the sum of all five equations: Calculating heat for each segment of the heating curve for water. q5 = mcpΔT 100 Time Temp q2 = m·Hf q3 = mcpΔT The total amount of heat absorbed is the sum of all five equations: qtot= q1+q2+q3+q4+q5 q2 = m·Hf q1 = mcpΔT

q4 = m·Hv Hf (water) = 334 J/g Hv (water) = 2260 J/g q2 = m·Hf NOTICE! It takes more time (longer line) & energy for vaporization of water to occur than melting. WHY? Look at the pictures to the right.  100 Time Temp q4 = m·Hv Hf (water) = 334 J/g Hv (water) = 2260 J/g q2 = m·Hf

Cooling curves are exothermic as they release/lose heat. gas 100 Time Temp condensation liquid/gas liquid Hv may be used to determine amount of heat released during condensation freezing solid/liquid solid Hf may be used to determine amount of heat released during freezing

How much heat is released when 275 grams of water freezes? q = m·Hf q = (275 g)(334 J/g) q = 91,850 J q = 91.9 kJ

During a phase, KE is decreasing (temp is decreasing). What happens to Kinetic energy (KE) & potential energy (PE) at each stage of the curve? COOLING curve KE During a phase, KE is decreasing (temp is decreasing). 100 Time Temp PE KE constant KE During a phase change, KE is constant and PE is decreasing. PE KE constant KE

ΔH & Chemical Reactions Chemical reactions always involves a change in heat energy

Endothermic Reactions C(s) + H2O(g) + 113 kJ  CO2(g) + H2(g) Heat energy is written in the equation as a reactant since it is coming in/being used. Surrounding area feels cool DHrxn = +113 kJ meaning 113kJ are absorbed

Potential energy diagram Endothermic because…. 1. Products have more energy than the reactants Activation Energy DH 2. ΔH is positive

Exothermic Reactions -ΔH C3H8 + 5O2  3CO2 + 4H2O + 2043 kJ Heat energy is written in the equation as a product since it is released/produced. Surrounding area feels warm ΔHrxn = -2043 kJ . . . meaning 2043 kJ of heat is released

Potential energy diagram Exothermic because…. 1. Products have less energy than the reactants Activation Energy 2. ΔH is negative DH

Thermochemistry & Stoichiometry If you know the ΔH for a balanced equation, you may determine the amount of energy used or released by a reaction.

ΔH (enthalpy) is proportional to the coefficients for a balanced equation, therefore they may be used to write conversion factors. 2 H2O2(l)  2 H2O(l) + O2(g) Hrxn = -190 kJ 2 mole H2O2 = -190 kJ 1 mole O2 = -190 kJ Or Or

Example How much heat will be released if 1.0 g of hydrogen peroxide (H2O2) decomposes in a bombardier beetle to produce a steam spray? 2 H2O2(l)  2 H2O(l) + O2(g) Hrxn = -190 kJ 1.0 g H2O2 34.016 g 1 mol 2 mol H2O2 -190 kJ = - 2.8 kJ

C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) Example How much heat is transferred when 9.22 g of glucose (C6H12O6) in your body reacts with O2 according to the following equation? C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) Hrxn = -2803 kJ 9.22 g C6H12O6 180.156 g 1 mol 1 mol C6H12O6 -2803 kJ = - 143 kJ

Example How much energy will be required to extract 59.5 grams of tin? SnO2(s) + 4NO2(g) + 2H2O(l) + 192 kJ  Sn(s) + 4HNO3(aq) Hrxn = +192 kJ 118.71 g 1 mol 59.5 g Sn 1 mol Sn 192 kJ = + 96.2 kJ