Energetics.

Energetics

Exothermic and Endothermic reactions
Energy = force x distance (Joules) In chemical reactions, we need energy usually in the form of heat. Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products. Exothermic reactions Endothermic reactions

Heat and Temperature Heat is the energy transferred between objects that are at different temperatures. The amount of heat transferred depends on the amount of the substance. Energy is measured in units called joules (J).

It does not depend on the amount of the substance.
Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance. It does not depend on the amount of the substance. Do both beakers contain the same amount of heat?

Energy Changes in Chemical Reactions
All chemical reactions are accompanied by some form of energy change Exothermic Energy is given out Endothermic Energy is absorbed Activity : observing exothermic and endothermic reactions

Enthalpy (H)and Enthapy change(ΔH)
Enthalpy (H) is the heat content that is stored in a chemical system. We measure the change in enthalpy ∆H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in kilojoules per mole (kJmol-1). ∆H = H(products) – H(reactants)

Enthalpy Level Diagram -Exothermic Change
For exothermic reactions, the reactants have more energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants) ∆H is negative since H(products) < H(reactants) There is an enthalpy decrease and heat is released to the surroundings Enthalpy

Examples of Exothermic Reactions
Self-heating cans CaO (s) + H₂O (l)  Ca(OH)₂ (aq) Combustion reactions CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l) neutralization (acid + base) NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l) Respiration C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

Enthalpy Level Diagram -Endothermic Change
For endothermic reactions, the reactants have less energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants) ∆H is positive since H(products) < H(reactants) There is an enthalpy increase and heat is absorbed from the surroundings Enthalpy

Examples of Endothermic Reactions
Self-cooling beer can H ₂O (l)  H₂O (g) Thermal decomposition CaCO₃ (s)  CaO (s) + CO ₂ (g) Photosynthesis 6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)

Specific heat capacity
Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin. Uint : Jg-1 0C-1 The specific heat capacity of alminium is 0.90 Jg-1 0C-1 . If 0.90J of energy is put into 1g of aluminium, the temperature will be raised by 10C. Calculating heat absorbed and released q = c × m × ΔT q = heat absorbed or released c = specific heat capacity of substance m = mass of substance in grams ΔT = change in temperature in Celsius

Calorimetry Heat given off by a process is measured through the temperture change in another substance (usually water). Due to the law of conservation of energy, any energy given off in a process must be absorbed by something else, we assume that the energy given out will be absorbed by the water and cause a temperature change. calculate the heat through the equation Q = mcΔT

Example How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg-1 0C-1) from 500C to 700C?

Enthalpy change of combustion reactions
The standard enthalpy change of combustion for a substance is the heat released when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions. Example, CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHƟc=-698 kJmol-1 The heat given out is used to heat another substance,e.g. water with a known specific heat capacity. The experiment set-up can be used to determine the enthalpy change when 1 mole of a liquid is burnt. Example : refer to page 185

Problems with calorimetry
Loss of heat to the surroundings (exothermic reaction); absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter. Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for. Others include – e.g incomplete combustion. Some of the ethanol could be used to produce CO & soot & water (less heat is given out) Use bomb calorimeter – heavily insulated & substance is ignited electronically with good supply of oxygen

Example If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change when 1 mole of methanol is burnt. Note: Specific heat capacity of water is 4.18 Jg-1 0C-1 Cal no of moles of methanol = 1/32 = Q = 100 x 4.18 x 42 Enthalpy change, delta H = (100 x 4.18 x 42)/ -563kJmol-1 Practice questions page 187 #1-4

Enthalpy change in solutions
Enthalpy change of neutralisation (ΔHn) The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions. Example, NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1 The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water. Reaction between strong acid and strong base involves H+(aq) + OH-(aq)  H2O(l) ΔHƟ=-57 kJmol-1 For sulfuric acid, the enthalpy of neutralisation equation is ½ H2SO4(aq) + KOH(aq)  ½K2SO4(aq) + H2O(l) ΔHƟ=-57 kJmol-1 Example : refer to page 188

Enthalpy change in solutions
Enthalpy change of neutralisation (ΔHn) The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions. Example, NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1 The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water. Enthalpy change of solution (ΔHsol) The enthalpy change when 1 mol of solute is dissolved in excess solvent to form a solution of ‘infnite dilution’ under standard conditions. NH4 NO3(s) in excess water  NH4 + (aq)+ NO 3 -(aq) Example : refer to page 188

For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol-1 (less exothermic) CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) ΔHƟ=-55.2 kJmol-1 Some of the energy released is used to ionise the acid.

Example 200.0cm3 of M HCl is mixed with 100.0cm3 of M NaOH. The temperature rose by 1.360C. If both solutions were originally at the same temp, calculate the enthalpy change of neutralisation. Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is 4.18J Jg-1 0C-1. Total vol = 300 Density of water = 1g/cm3 Enthalpy = 300 x 4.18 x 1.36 = J No. of moles of HCl = 0.2 x 0.15M = 0.03 No. of moles of NaOH = .100 x 0.35M = 0.035 Therefore, no. of moles of water formed= 0.03 Heat released = /0.03 = kJmol-1 Enthalpy change of neutralisation = kJmol-1 (exothermic ) -56.8kJmol-1

Possible errors The experimental change of neutralisation is kJmol-1 The accepted literature value is kJmol-1 (1) Heat loss to the environment. Assumptions that the denisty of NaOH and HCl solutions are the same as water. the specific heat capacity of the mixture are the same as that of water

Example When 3 g of sodium carbonate are added to 50 cm3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is 4.18J Jg-1 0C-1. Vol of water = 50 cm3 Q = 50 x 4.18 x (28.5 – 22) No. of moles of HCl = 0.05 No. of moles of Na2CO3 = 3/106 = No. of moles of water = Heat of change = /0.0283= 48kJmol-1 Example : refer to page 189 dissolving ammonium chloride

Possible errors (page 189)
The experimental change of solution is kJmol-1 The accepted literature value is 15.2 kJmol-1 (1) Absorption of heat from the environment. Assumptions that the specific heat capacity of the solution is the same as that of water The mass of ammonium chloride is not taken into consideration when working out the heat energy released.

Example 100.0 cm3 of mol dm-3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction. First step Make sure you understand the graph. Extrapolate to determine the change in temperature. Actual Highest temp reached could be 26 deg Extrapolate to get the temp reached if no heat is lost to surroundings Delta T = 28 – 17 The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used?

100.0 cm3 of mol dm-3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction. Determine the limiting reactant Calculate Q Calculate the enthalpy for the reaction.

Enthalpy changes of combustion of fuels
The following measurements are taken: Mass of cold water (g) Temperature rise of the water (0C) The loss of mass of the fuel (g) We know that it takes 4.18J of energy to raise the temperature of 1g of water by 10C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg-1K-1. Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules) If one mole of the fuel has a mass of M grams, then: Enthalpy transfer = m x 4.18 x T x M/y where y is mass loss of fuel.

Example Given that: Vol of water = 100 cm3 Temp rise = 34.50C Mass of methanol burned = 0.75g Specific heat capacity of water = 4.18 Jg-10C-1 Calculate the molar enthalpy change of the combustion of methanol. What is the big assumption made with this type of experiment?

Hess’s Law States that If a reaction consists of a number of steps, the overall enthalpy change is equal to the sum of enthalpy of individual steps. the overall enthalpy change in a reaction is constant, not dependent on the pathway take.

Standard enthalpy changes, ΔHƟ
measured under standard conditions: pressure of 1 atmosphere (1.013 x 105 Pa), temperature of 250C (298K) and concentration of 1 moldm-1. e.g. N2(g) + 3H2(g)  2NH3(g) ΔHƟ = -92 kJmol-1 The enthalpy change of reaction is -92 kJmol-1 92 kJ of heat energy are given out when 1 mol of nitrogen reacts with 3 mols of hydrogen to form 2 mols of ammonia.

Reaction in aqueous soln
Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(s) + H2O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔHƟ1=-43kJmol-1 NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ2=-57kJmol-1 NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l). Indirect path + HCl(aq) + H2O(l) ΔH2 ΔH1 Direct path -100kJmol-1

Combustion reaction (using cycles)
Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide. C(s) + O2(g)  CO2(g) ΔHƟ =-394 kJmol-1 2C(s) + O2(g)  2CO(g) ΔHƟ = -222kJmol-1 2CO(s) + O2(g)  2CO2(g) 2CO(g) + O2(g) 2CO2(g) ΔHƟ -111kJmol-1 The enthalpy for the reaction C(s) + 1/2O2(g) -> CO(g) cannot be found directly by experiment because CO2 is always formed when carbon reacts with only a limited amount of oxygen – unavoidable. However, the enthalpy changes of combustion of C and CO can be found experimentally. The reactions and their enthalpy changes can be linked using Hess law ΔHƟ = -(-222)+2(-394) = -566kJmol-1

Combustion reaction (manipulating equations)
Example : refer to page 196 evaporation of water & 197 formation of ethanol from ethene

*Example : Decomposition reaction
Calculate the enthalpy change for the thermal decomposition of calcium carbonate. CaCO3(s)  CaO(s) + CO2(g) CaCO3(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-17 kJmol-1 CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-195kJmol-1 CaCO3(s) CaO(s) +CO2(g) CaCl2(aq) + H2O(l) +CO2(g) ΔH Direct path + 2HCl(aq) + 2HCl(aq) +178kJmol-1 The reaction is slow and a temp is required to bring it to completion. Direct measurement is not practical. 2 reactions that take place at rm temp are carried out nd their enthalpy change used to find the enthalpy of decomposition of calcium carbonate. -17 kJmol-1 -195kJmol-1 Indirect path

*Example : Enthalpy of hydration of an anhydrous salt.
Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change. CuSO4(s) +5H2O(l)  CuSO4.5H2O (s) CuSO4(s) +5H2O(l) CuSO4.5H2O (s) Cu2+(aq) + SO42- (aq) ΔH Direct pathway ΔH1 ΔH2 It cannot be found directly because when 5 moles of water is added, hydrated copper (II) sulphate is not produced in a controlled way. Can only be produced by crystallisation from a solution. The enthalpy change can be determined indirectly by finding the enthalpy of solution of both the anhydrous and the hydrated copper(II) sulfates. Indirect pathway

*Example From the following data at 250C and 1 atmosphere pressure: Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1 Eqn 2: 3CO(g) + O3(g) 3CO2(g) ΔHƟ=-992 kJmol-1 Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O2(g) O3 (g) +143kJmol-1

*Example Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O2(g) CO2 (g) ΔHƟ=-393 kJmol-1 C (s, diamond) + O2(g) CO2(g) ΔHƟ=-395 kJmol-1 +2kJmol-1

Practice questions page 199 #7-9

Bond enthalpies (Bond energies)
Enthalpy changes can also be calculated directly from bond enthalpies. The bond enthalpy is the amount of energy required to break one mole of a specified covalent bond in the gaseous state. For diatomic molecule the bond enthalpy is defined as the enthalpy change for the process X-Y(g) X(g) + Y(g) [gaseous state]

Bond Enthalpies Bond enthalpy can only be calculated for substances in the gaseous state. Br2(l)  2Br(g) ΔHƟ= 224 kJmol-1 atomisation 2 x ΔH Ɵat Br2(l) Br(g) Br2(g) ΔH Ɵvap enthalpy change of vaporisation Br-Br bond enthalpy Energy must be supplied to break the van der Waals’ forces between the Bromine molecules and to break the Br-Br bonds. Endothermic process

Average bond enthalpies
Ave bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH4 , alkanes and other hydrocarbons.

Some average bond enthalpies
Ave bond enthalpy, ΔHƟ (Kjmol-1) Bond length (nm) H-H 436 0.07 C-C 348 0.15 C-H 412 0.11 O-H 463 0.10 N-H 388 N-N 163 C=C 612 0.13 O=O 496 0.12 C Ξ C 837 NΞN 944 Refer to page 201

Bond breaking and Forming
When a hydrocarbon e.g. methane (CH4) burns, CH4 + O2  CO2 + H2O What happens?

Energy Level Diagram CH4 + 2O2  CO2 + 2H2O Enthalpy Level (KJ)
Bond Breaking O O C H ENERGY O H ENERGY O H C H + O H Bond Forming C H O 4 C-H 2 O=O 4 H-O 2 C=O CH4 + 2O2  CO2 + 2H2O Progress of Reaction Energy Level Diagram

Bond breaking and Forming
CH4 + 2O2  CO2 + 2H2O C H + O C H O Why is this an exothermic reaction (produces heat)?

CH4 + 2O2  CO2 + 2H2O Break  Form C-H 412 H-O 463 O=O 496 C=O 743
Bond Ave Bond Enthalpy (kJ/mol) C-H 412 H-O 463 O=O 496 C=O 743 C H + O C H O Energy absorbed when bonds are broken = (4 x C-H + 2 x O=O) Energy given out when bonds are formed = ( 2 x C=O + 4 x H-O) = 4 x x 496 = kJ/mol = 2 x x 464 = 3338 kJ/mol

Energy absorbed when bonds are broken (a) = 2640 kJ/mol
Energy released when bonds are formed (b) = kJ/mol Enthalpy change, ΔH = ∑(bonds broken) - ∑(bonds made) = a + (-b) = 2640 – 3338 = -698 kJ/mol The forming of the bonds (2 x C=O and 4 x H-O) between the atoms in water gives out more heat than is required to break the bonds (4 x C-H and 2 x O=O) . The bonds in the products (2 x C=O and 4 x H-O) are stronger than those in the reactants (4 x C-H and 2 x O=O) since they require more energy to break them. Why is this an exothermic reaction (produces heat)?

Example What can be said about the hydrogenation reaction of ethene?
H H C=C (g) + H-H (g)  H-C-C-H (g) H H H H H H

Example What can be said about the combustion of hydrazine in oxygen?
H H N-N (g) + O=O (g)  NΞN (g) O (g) H H H H

The bond enthalpy from an enthalpy change of reaction
Example Calculate the mean Cl-F bond enthalpy given that Cl2(g) + 3F2(g)  2ClF3(g) ΔHƟ= -164 kJmol-1 Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1

Using bond enthalpies & enthalpies of atomisation
Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions. Example C(s)  C(g) Calculate the enthalpy change for the process 3 C(s) + 4H2(g)  C3H8(g) ΔHƟ= -164 kJmol-1 Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1 (ΔH Ɵat ) ΔH Ɵat = 715 kJmol-1 -103kJmol-1 Practice questions page 206 #10,12,13

Hess’s Law - example The combustion of both C and CO to form CO2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle. ΔHx = -393 – (-283) = kJmol-1 ΔHx C(s)+ ½O2(g) CO(g) CO2(g) ½O2(g) -393kJmol-1 ½O2(g) -283kJmol-1

Hess’s Law - example Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbon, hydrogen and methane are -393, -286 and -890 kJmol-1 respectively. -75kJmol-1

Entropy : degree of disorder
Dissolving sugar . Sugar molecules are dispersed throughout the solution and are moving around. More disordered or random. Other examples: melting ice H2O(s)  H2O(l) evaporating water H2O(l)  H2O(g)

The sign of entropy, S Increasing entropy
Entropy (S) : amount of disorder Unit : JK-1mol-1 SƟ : standard entropy Δ SƟ : entropy change If Δ SƟ > 0 => increase in entropy => increase in disorder E.g. H2O(l )  H2O(g) Δ SƟ =+119JK-1mol-1 If Δ SƟ < 0 => decrease in entropy => decrease in disorder E.g. NH3(g ) + HCl(g)  NH4Cl(s) Δ SƟ = - 285JK-1mol-1

state of matter temperature number of molecules
What are the factors that affect ENTROPY? state of matter temperature number of molecules

Factors affecting entropy
(1) State of matter Gas particle motion is more random in a gas Liquid particle motion is less random than in a liquid than a gas but more than a solid Solid particle motion is restricted. Examples (changing state) H2O(l)  H2O(g) (changing state) H2O(s)  H2O(l) ΔS(gas) > ΔS(liquid) > ΔS(solid)

(2) Temperature Comparing two gasses, one at 20 C and one at 80 C
Molecules in the 80 C gas have more kinetic energy, they are moving more and colliding more

(4) More complex molecules have higher entropy values
(3) The number of molecules More molecules means more possible positions relative to the other molecules (more moles and change of state) Li2CO3(s)  Li2O(s) + CO2(g) (more moles) MgSO48H2O  Mg2+(aq) + SO42-(aq) + 8H2O(l) (4) More complex molecules have higher entropy values

Predict the sign of ΔSƟ Is there an increase or decrease in disorder of the system? Is there an increase or decrease in the no. of moles of gas? Reaction Entropy (increase/ decrease) ΔSƟ ( + / - ) Explanation N2(g) + 3H2(g)  2NH3(g) 4 moles of gas to 2 moles of gas CaCO3(s)  CaO(s) + CO2(g) 1 mole of solid to 1 mole of solid + 1 mole of gas CH4(g) + 2O2(g)  CO2(g) +2H2O(l) 3 moles of gas to 1 mole of gas C2H4(g) + H2(g)  C2H6(g) 2 moles of gas to 1 mole of gas decrease - increase + decrease - decrease - For reaction where the no. of moles of gas is the same on both sides, the ΔS = 0. E.g F2(g) + Cl2(g)  2ClF(g) Practice Qn 24from pg 230 (textbook

Example Which of the following reactions has the largest ΔS value?
CO2(g) + 3H2(g)  CH3OH(g) + H2O(g) 2Al(s) + 3S(s)  Al2S3(s) CH4(g) + H2O(l)  3H2(g) + CO(g) 2S(s) + 3O2(g)  2SO3(g) 3rd

Calculate ΔSƟ Entropy change = total entropy of products – total entropy of reactants ΔSƟ =∑ ΔSƟproducts - ∑ ΔSƟreactants E.g. Calculate the standard entropy change for the reaction CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l) Use standard entropy from pg 230 (textbook)

Spontaneity-Driving forces
Some reactions are spontaneous because they give off energy in the form of heat (H < 0). Others are spontaneous because they lead to an increase in the disorder of the system (S > 0). Calculations of H and S can be used to probe the driving force behind a particular reaction.

Spontaneity Spontaneous reaction – one that occurs without any outside influence (no input of energy) A spontaneous reaction does not have to happen quickly. E.g. 4Na(s) + O2(g)  2Na2O(s) can happen by itself.

Example Calculate H and S for the following reaction and decide in which direction each of these factors will drive the reaction. N2(g) + 3 H2(g) 2 NH3(g) Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information: Compound Hfo(kJ/mol) S°(J/mol-K) N2(g) H2(g) NH3(g) The reaction is exothermic ( H° < 0), which means that the enthalpy of reaction favors the products of the reaction: kJ The entropy of reaction is unfavorable, however, because there is a significant increase in the order of the system, when N2 and H2 combine to form NH J/K

Gibbs free energy, G What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the system, which reflects the balance between these forces. ΔG= ΔH – TΔS ΔG : free energy change ΔGƟ : standard free energy change

Temperature, T should be in Kelvin, K
0°C = 273K (273.15K) Check the units of ΔS, entropy is often given in JK–1mol–1 but must be converted to kJK–1mol–1 ΔGө = ΔHө – TΔSө ө = standard conditions, 25°C (298K) and 1 atm (101.3 kPa)

Gibbs free energy, G measures the balance between the two driving forces that determine whether a reaction is spontaneous. the enthalpy and entropy terms have different sign conventions. When heat is released in a chemical reaction, the surrounding is hotter and particles move around more => entropy increases. Favourable Unfavourable ΔHƟ < 0 ΔHƟ > 0 ΔSƟ > 0 ΔSƟ < 0

Gibbs free energy, G Because of the way the free energy of the system is defined, Go is negative for any reaction for which Ho is negative and So is positive. For a Favorable, or spontaneous reactions: Go < 0

When a reaction is favored by both enthalpy (Ho < 0) and entropy (So > 0), there is no need to calculate the value of Go to decide whether the reaction should proceed. Similarly, for reactions favored by neither enthalpy (Ho > 0) nor entropy (So < 0). Free energy calculations become important for reactions favored by only one of these factors.

Calculate ΔGƟ using ΔGƟ = ΔHƟ – TΔSƟ
Given that the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then state whether the reaction is spontaneous at 25 C C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)

Example Calculate H and S for the following reaction:
NH4NO3(s) + H2O(l)  NH4+ (aq) + NO3- (aq) Use the results of this calculation to determine the value of Go for this reaction at 25o C, and explain why NH4NO3 spontaneously dissolves is water at room temperature. Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information: Compound Hfo(kJ/mol) S°(J/mol-K) NH4NO3(s) NH4+ (aq) 113.4 NO3- (aq) 146.4 This reaction is endothermic, and the enthalpy of reaction is therefore unfavorable: 28.05kJ The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction: J/K To decide whether NH4NO3 should dissolve in water at 25o C we have to compare the Ho and TSo to see which is larger. Before we can do this, we have to convert the temperature from oC to kelvin: TK = 25o C = K We also have to recognize that the units of Ho for this reaction are kilojoules and the units of So are joules per kelvin. At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis. Perhaps the easiest way of doing this is to convert Ho to joules. We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term: Go = Ho - T So = 28,050 J - ( K x J/K) = 28,050 J - 32,410 J = J At 25o C, the standard-state free energy for this reaction is negative because the entropy term at this temperature is larger that the enthalpy term: Go = -4.4 kJ The reaction is therefore spontaneous at room temperature.

The Effect of Temperature on the Free Energy of a Reaction
The balance between the contributions from the enthalpy and entropy terms to the free energy of a reaction depends on the temperature at which the reaction is run. Predict whether the following reaction is spontaneous at 250C: N2(g) + 3 H2(g) NH3(g) Go = kJ According to this calculation, the reaction should be spontaneous at 25°C.

Predict whether the following reaction is still spontaneous at 500C:
The equation suggests that the entropy term will become more important as the temperature increases. Go = Ho - TSo Since the entropy term is unfavorable, the reaction should become less favorable as the temperature increases. Predict whether the following reaction is still spontaneous at 500C: N2(g) + 3 H2(g) NH3(g) Assume that the values of Ho and S used in the previous example are still valid at this temperature. Go = 61.4 kJ Because the entropy term becomes larger as the temperature increases, the reaction changes from one which is favorable at low temperatures to one that is unfavorable at high temperatures

Standard free energy Go
What does the value of Go tell us about the following reaction? N2(g) + 3 H2(g) NH3(g) Go = kJ The value of Go for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard- state conditions. describes this reaction only when all three components are present at 1 atm pressure. The fact that Go is negative for this reaction at 25oC means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium.

Calculate ΔGƟ using standard free energy of formation
ΔGƟ =∑ ΔGfƟproducts - ∑ ΔGfƟreactants standard free energy of formation : free energy change for the formation of 1 mole of substance from its elements in their standard states & under standard conditions. Calculate ΔGƟ for the reaction CaCO3(s)  CaO(s) + CO2(g) given that Substance ΔGfƟ /kJmol -1 CaCO3 - 1129 CaO - 604 CO2 - 395 Practice Qn from pg 235 (textbook

Spontaneity of reactions – Gibbs Free Energy, ΔG
ΔH : Enthalpy Change ΔS : Entropy Change For a spontaneous reaction ΔG is negative (–) For a non–spontaneous reaction ΔG is positive (+)

Energetics.

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Energetics.

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