Ch. 16 Thermochemistry Notes Heat changes that occur during a chemical reaction

What is Thermochemistry? The study of heat changes in chemical reactions and physical states. Law of Conservation of Energy Calorimetry Temperature Heat Heat Capacity Specific Heat

Energy The capacity for doing work or supplying heat Energy is weightless, colorless, and odorless Energy that is stored in the bonds of chemical substances is called chemical potential energy. (Example: gasoline)

Forms of Energy Potential Kinetic Work Enthalpy stored energy Energy of motion Work energy used in a process Enthalpy Heat energy lost or gained in a process

Law of Conservation of Energy Energy is neither created or destroyed only transferred.

What is work and heat? Work = when a force is used to move an object Heat = energy that is transferred from one object to another, because of a temperature difference. (q) Heat flows from warm to cold until equilibrium is reached

Heat flow into a system is an endothermic process (+q) When describing the flow of heat: Heat flow into a system is an endothermic process (+q) Heat flow out of a system is an exothermic process (-q) (Calorimeter—an instrument that measures heat changes in a closed system)

What is a calorie? The quantity of heat required to raise 1 gram of water 1 °C. The SI unit of measurement for heat is the joule. 1 joule = 0.2390 calories 1 calorie = 4.184 joules

Specific Heat Capacity (Specific Heat) The amount of heat needed to raise 1 g of a substance 1 °C. Table 1 pg 533 Heat capacity depends upon the mass and the composition of the object.

Definition: Specific Heat The amount of heat it takes to raise 1 gram of a substance 1 °C.

Examples Water has the highest specific heat capacity. It absorbs a lot of heat before finally increasing temperature. Metals have the lowest specific heat capacity. A small amount of absorbed energy immediately changes temperature

There is a calculation that can be performed to determine the specific heat of an object. q = C x m x ∆T C = Specific Heat (J/g°C or cal/g°C) q = heat (J or cal) m = mass (g) ∆T = T final – T initial (°C)

Sample Problems

When 435 J of heat is added to 3 When 435 J of heat is added to 3.4 grams of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of olive oil? C = ? q = 435 J m = 3.4 g ∆T = Tfinal – Tinitial = 85°C – 21°C = 64 °C q = C x m x ∆T

C = ? q = 435 J m = 3.4 g ∆T = Tfinal – Tinitial = 85°C – 21°C = 64 °C q = C x m x ∆T Solve for C C = q / (m x ∆T) C = 435 J / (3.4 g x 64 °C) = 2.0 J/g°C

How much heat is required to raise the temperature of 250 How much heat is required to raise the temperature of 250.0 g of mercury 52 C°? (specific heat of mercury is 0.14 J/g°C) C = 0.14 J/g • °C q = ? m = 250.0 g ∆T = 52 °C q = C x m x ∆T q =(0.14 J/g°C)(250.0 g)(52 °C)= 1820 J Is this endothermic or exothermic?

Remember… Heat flow out of a system is an exothermic process (-q) Heat flow into a system is an endothermic process (+q) q =(0.14 J/g°C)(250.0 g)(52 °C)= 1800 J Positive = Endothermic

Endo vs. Exo? If the temperature of a reaction increases then the reaction is endothermic= heat in! This makes a (+q) If the temperature of a reaction decreases the then reaction is exothermic= heat out! This makes a (-q)

Now you solve the problems Practice --Heat problems 1. Define each variable in the equation for specific heat: q = C x m x ∆T 2. How many Joules of heat energy are required to raise the temp of 45 g of water from 10.2 C to 26.8 C? 3. How many calories of heat are released when 275 g of water cool from 85.2 C to 38.4 C? 4. What will the change in temperature be if 80 g of water absorb 7220 J of heat energy? 5. What is the specific heat of lead if a 30 g sample undergoes a 250 C temp change while absorbing 963.6 Joules of energy?

Now do the practice problem worksheet!! 20 minutes to do then we review and continue notes

C = Specific Heat (J/g°C or cal/g°C) q = heat (J or cal) m = mass (g) 1. Define each variable in the equation for specific heat: q = C x m x ∆T C = Specific Heat (J/g°C or cal/g°C) q = heat (J or cal) m = mass (g) ∆T = T final – T initial (°C)

2. How many Joules of heat energy are required to raise the temp of 45 g of water from 10.2 C to 26.8 C? C = 4.18 J/g • °C q = ? m = 45 g ∆T = 26.8 °C -10.2 °C =16.6 °C q = C x m x ∆T q =(4.18 J/g°C)(45 g)(16.6 °C)= 3123J

3. How many calories of heat are released when 275 g of water cool from 85.2 C to 38.4 C? C = 1.0 cal/g • °C q = ? m = 275 g ∆T = 38.4 °C - 85.2 °C = -46.8 °C q = C x m x ∆T q =(1.0 cal/g•°C)(275g)(-46.8°C)=-12870 cal

4. What will the change in temp 4. What will the change in temp. be if 80 g of water absorb 7220 J of heat energy? C = 4.18 J/g • °C q = 7220 J m = 80 g ∆T = ? ∆T= . q . C x m ∆T= ( 7220 J ) (4.18 J/g • °C )(80 g)= 21.6 °C

5. What is the specific heat of lead if a 30 g sample undergoes a 250 C temp change while absorbing 963.6 Joules of energy? C = ? q = 963.6 J m = 30 g ∆T = 250 °C C = q . m x ∆T C = ( 963.6 J ) (30 g)(250 °C)= 0.1284 J/g°C

Enthalpy q= ∆H = C x m x ∆T ∆H = enthalpy H=Heat of substance at a constant pressure q= ∆H = C x m x ∆T ∆H = enthalpy C = Specific Heat (J/g°C or cal/g°C) q = heat (J or cal) m = mass (g) ∆T = T final – T initial (°C)

Enthalpy + ∆H = endothermic rxn - ∆H = exothermic rxn

Enthalpy of Reaction The quantity of energy transferred as heat during a chemical reaction. ΔH = Hreactants- Hproducts

Thermochemical Reactions ∆ H = Hreactants- Hproducts Heat of Reaction – the heat required or released in a reaction (coefficients are mole quantities) Exothermic Reaction CaO (s) + H2O (l) → Ca(OH)2 (aq) + 65.2 kJ ∆H= -65.2 kJ (heat is released) See graph page 536 Endothermic Reaction NaHCO3(s) + 129 kJ → Na2CO3 (s) + H2O (g) + CO2(g) ∆H= + 129 kJ (heat absorbed) See graph page 537

Practice 1) Calculate the ∆ H of the reaction. 2NaHCO3 (s) + 129 kJ→ Na2CO3(s) + H2O(g) + CO2(g) 1) Calculate the ∆ H of the reaction. ∆ H = Hreactants- Hproducts ∆H= 129 kJ – 0 kJ= + 129 kJ 2) Calculate the kJ of heat required to decompose 2.24 mol NaHCO3(s). 2.24 mol NaHCO3 x (+ 129 kJ) = 144 kJ 2 mol NaHCO3

More Practice 2NaHCO3 (s) + 129 kJ→ Na2CO3(s) + H2O(g) + CO2(g) 3) How much energy is absorbed when 25g of NaHCO3 decomposes? 25g NaHCO3 x 1 mol NaHCO3 x 129 kJ = 19.2 kJ 84g NaHCO3 2 mol NaHCO3