International Baccalaureate Chemistry Topic 5 - Energetics

Thermochemistry * Thermochemistry is the study of energy changes associated with chemical reactions. Studies the amount of energy in a chemical reaction Energy is measured in Joules (J) Energy evolved or absorbed in a chemical reaction has nothing to do with the rate of the reaction (how fast or slow the reaction takes place. *

Potential Energy Energy an object possesses by virtue of its position or chemical composition.

Kinetic Energy Energy an object possesses by virtue of its motion. 1 2 KE =  mv2

Energy The ability to do work or transfer heat. * The ability to do work or transfer heat. Work: Energy used to cause an object that has mass to move. Heat: Energy used to cause the temperature of an object to rise.

Enthalpy Also called heat content Energy stored by the reactants and products Units kJ mol-1 Bonds are broken and made in chemical reactions but the energy absorbed in breaking bonds is almost never exactly the same as the energy that is released in making new bonds. Bonds broken = required energy Bonds formed = released energy *

Enthalpy Change * All reactions have a change in potential energy of the bonds. This change in potential energy is known as an enthalpy change. Enthalpy changes can only be measured for chemical reactions, not for state changes of substances. Given the symbol ΔH

Endothermic and Exothermic * A process is endothermic when H is positive.

Endothermicity and Exothermicity * A process is endothermic when H is positive. A process is exothermic when H is negative.

Endothermic Reactions Chemical reaction where the total enthalpy of the reactants is less than the total enthalpy of the products Positive H values Heat energy is absorbed from surroundings They get cooler or external heat must be added to make the reaction work. Will not occur as a spontaneous reaction. Bonds broken are stronger than bonds formed *

Exothermic Reactions Chemicals lose energy as products are formed. Negative H values Heat energy is lost and surroundings get warmer. Most spontaneous reactions are exothermic. Products are more stable than the reactants. Bonds made are stronger than bonds broken. *

Summary Type of Reaction Heat Energy Change Temperature Change Relative Enthalpies Sign of ΔH Exothermic Heat energy evolved Becomes Hotter Hp < Hr Negative Endothermic Heat energy absorbed Becomes colder Hp > Hr Positive

Enthalpies of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants *

Enthalpy Conditions * By definition, an enthalpy change must occur at constant pressure. Thermochemical standard conditions have been defined as a temperature of 25°C, a pressure of 101.3 kPa, and all solutions have a concentration of 1 mol dm-3. Thermochemical quantities that relate to standard conditions are indicated with a standard sign (θ). ΔH θ – Standard thermochemical conditions.

Types of Enthalpies ΔHrxn – Heat produced in a chemical reaction. ΔHcomb – Heat produced by a combustion reaction. ΔHneut – Heat produced in a neutralization reaction. ΔHsol – Heat produced when a substance dissolves. ΔHfus – Heat produced when a substance melts. ΔHvap – Heat produced when a substance vaporizes. ΔHsub – Heat produced when a substance sublimes.

Calculation of Enthalpy Changes Temperature: A measure of the average kinetic energy of the molecules. Heat: The amount of energy exchanged due to a temperature difference between two substances. An exchange of heat causes a change in temperature. *

Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. *

Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K. *

Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change c = q m  T q = mC  T *

Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g K), we can measure H for the reaction with this equation: q = mC  T *

Example * Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is 0.902 J g-1 oC-1. Solution q = mCDT = (400 g) (0.902 J g-1 oC-1)(200oC – 20oC) = 64,944 J

Example * What is the final temperature when 50 grams of water at 20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-1 Solution: Dq (Cold) = -Dq (hot) mCDT= -mCDT Let T = final temperature (50 g) (4.184 J g-1 oC-1)(T- 20oC) = -(80 g) (4.184 J g-1 oC-1)(T-60oC) (50 g)(T- 20oC) = -(80 g)(T-60oC) 50T -1000 = – 80T + 4800 130T = 5800 T = 44.6 oC

Bomb Calorimetry Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is very small.

First Law of Thermodynamics * Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Use Fig. 5.5

Hess’s Law * H is well known for many reactions, and it can be inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using H values that are published and the properties of enthalpy.

Hess’s Law * Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Hess’s Law * Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Standard Enthalpies of Formation Standard enthalpies of formation, Hθf, are measured under standard conditions (25°C and 1.00 atm pressure).

Hess’s Law Calculations * If a chemical reaction involves two or more steps, the overall change in enthalpy equals the sum of the enthalpy changes of the individual steps. If a reaction is reversed, then the sign of H must be reversed. If a reaction is multiplied by an integer, then H must also be multiplied by the integer. Attention must be paid to physical states.

Example 1: Calculate the heat of combustion (H comb) of C(s) to CO(g) * Calculate the heat of combustion (H comb) of C(s) to CO(g) Given: H (kJ) C(s) + O2(g) → CO2(g) -393.5 CO(g) + ½O2(g) → CO2(g -283 C(s) + ½O2(g) → CO(g) ???? (-110.5 kJ)

Example 2 Calculate ΔH for the reaction: 2C(s) + H2(g) → C2H2(g) * Calculate ΔH for the reaction: 2C(s) + H2(g) → C2H2(g) Given: ΔH (kJ) C2H2(g) +5/2O2(g) → 2CO2(g) + H2O(l) -1299.6 C(s) + O2(g) → CO2(g) -393.5 H2(g) + ½O2(g) → H2O(l) -285.9 (+226.7 kJ)

Example 3 Given: ΔH(kJ) NO(g) + O3(g) → NO2(g) + O2(g) -198.9 * Given: ΔH(kJ) NO(g) + O3(g) → NO2(g) + O2(g) -198.9 O3(g) → 3/2 O2(g) -142.3 O2(g) → 2O(g) +495.0 Find ΔH for the reaction: NO(g) + O(g) → NO2(g) (-304.1 kJ)

Try This! * Calculate ΔH for the synthesis of diborane from its elements according to the equation: 2B(s) + 3H2(g) → B2H6(g) Given: ΔH(kJ) 2B(s) + 3/2O2(g) → B2O3 -1273 B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) -2035 H2(g) + ½O2(g) → H2O(l) -286 H2O(l) → H2O(g) +44 (+36 kJ)

Bond Enthalpies note: Cl2 and H2 are not formed. Bond enthalpy is the amount of energy required to convert a gaseous molecule into gaseous atoms. HCl(g) → H(g) + Cl(g) note: Cl2 and H2 are not formed. Limitations: Can only be used if all the reactants and products are gaseous Values have been obtained from a number of similar compounds and therefore, will vary slightly in different compounds. Enthalpy changes (ΔH) can be calculated using bond enthalpies. Bond formation: negative (exothermic) Bond breaking: positive (endothermic) ΔH = H bonds broken – H bonds formed

Example 1 C2H4(g) + H2(g) → C2H6(g) ΔH = 2696 - 2820 = -125 kJ mol-1 Bonds Broken (kJ mol-1) Bonds Formed(kJ mol-1) C=C 612 C-C 348 4(C-H) 4(412) 6(C-H) 6(412) H-H 436 Total: 2696 kJ 2820 kJ ΔH = 2696 - 2820 = -125 kJ mol-1

C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g) Example 2 Estimate the ΔH for the following reaction from bond energies. C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g) Bonds Broken (kJ mol-1) Bonds Formed (kJ mol-1) 6 C-H 6(413) = 2478 4 C=O 4(745) = 2980 7/2 O=O 7/2(495) = 1732.5 6 O-H 6(467) = 2802 1 C-C 1(347) = 347 Total = 4557.5 Total = 5782 ΔH = 4557.5 – 5782 = -1224.5 kJ mol-1